Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf ·...

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Probability Probability in Simple Sample Spaces Conditional Probability Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh Department of Political Science University California, San Diego November 11, 2010 Sebastian M. Saiegh Probability Theory

Transcript of Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf ·...

Page 1: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Probability TheoryPOLI 270 - Mathematical and Statistical Foundations

Sebastian M. Saiegh

Department of Political ScienceUniversity California, San Diego

November 11, 2010

Sebastian M. Saiegh Probability Theory

Page 2: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Introduction to Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

Page 3: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

Page 4: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck

The radical concept of replacing randomness with systematicprobability and its implicit suggestion that the future might bepredictable and even controllable to some degree, is whatseparates the modern times and the past.

Yet, probability theory was not developed until the XVIIcentury.

And, it was a result of the efforts to analyze games of chance!

Sebastian M. Saiegh Probability Theory

Page 5: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck

The radical concept of replacing randomness with systematicprobability and its implicit suggestion that the future might bepredictable and even controllable to some degree, is whatseparates the modern times and the past.

Yet, probability theory was not developed until the XVIIcentury.

And, it was a result of the efforts to analyze games of chance!

Sebastian M. Saiegh Probability Theory

Page 6: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck

The radical concept of replacing randomness with systematicprobability and its implicit suggestion that the future might bepredictable and even controllable to some degree, is whatseparates the modern times and the past.

Yet, probability theory was not developed until the XVIIcentury.

And, it was a result of the efforts to analyze games of chance!

Sebastian M. Saiegh Probability Theory

Page 7: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The game of craps provides a useful illustration of therelationship between gambling and the laws of probability:

Throwing a pair of six-sided dice will produce, not eleven (fromtwo to twelve), but thirty-six possible combinations, all theway from snake eyes (two ones) to box cars (double six).

There is only one way for each of double-one and double-six toappear. While seven, the key number in craps, is the easiest tothrow. There are six different ways to arrive at seven. Theprobability of a seven-throw is 6

36 or 16.66%.

Sebastian M. Saiegh Probability Theory

Page 8: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The game of craps provides a useful illustration of therelationship between gambling and the laws of probability:

Throwing a pair of six-sided dice will produce, not eleven (fromtwo to twelve), but thirty-six possible combinations, all theway from snake eyes (two ones) to box cars (double six).

There is only one way for each of double-one and double-six toappear. While seven, the key number in craps, is the easiest tothrow. There are six different ways to arrive at seven. Theprobability of a seven-throw is 6

36 or 16.66%.

Sebastian M. Saiegh Probability Theory

Page 9: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The game of craps provides a useful illustration of therelationship between gambling and the laws of probability:

Throwing a pair of six-sided dice will produce, not eleven (fromtwo to twelve), but thirty-six possible combinations, all theway from snake eyes (two ones) to box cars (double six).

There is only one way for each of double-one and double-six toappear. While seven, the key number in craps, is the easiest tothrow. There are six different ways to arrive at seven. Theprobability of a seven-throw is 6

36 or 16.66%.

Sebastian M. Saiegh Probability Theory

Page 10: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The probability of an outcome is the ratio of favorableoutcomes to the total opportunity set.

The odds on an outcome are the ratio of favorable outcomesto unfavorable outcomes.

Sebastian M. Saiegh Probability Theory

Page 11: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The probability of an outcome is the ratio of favorableoutcomes to the total opportunity set.

The odds on an outcome are the ratio of favorable outcomesto unfavorable outcomes.

Sebastian M. Saiegh Probability Theory

Page 12: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The odds obviously depend on the probability, but the oddsare what matter when you are placing a bet:

If the probability of throwing a 7 in craps is one out of sixthrows, the odds on throwing a number other than 7 are 5 to 1.This means that you should bet no more than $1 that 7 willcome up the next throw when the other guy bets $5 that itwill not.

Games of chance must be distinguished from games in whichskill makes a difference. The odds are all you know for bettingin a game of chance.

Sebastian M. Saiegh Probability Theory

Page 13: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The odds obviously depend on the probability, but the oddsare what matter when you are placing a bet:

If the probability of throwing a 7 in craps is one out of sixthrows, the odds on throwing a number other than 7 are 5 to 1.This means that you should bet no more than $1 that 7 willcome up the next throw when the other guy bets $5 that itwill not.

Games of chance must be distinguished from games in whichskill makes a difference. The odds are all you know for bettingin a game of chance.

Sebastian M. Saiegh Probability Theory

Page 14: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The odds obviously depend on the probability, but the oddsare what matter when you are placing a bet:

If the probability of throwing a 7 in craps is one out of sixthrows, the odds on throwing a number other than 7 are 5 to 1.This means that you should bet no more than $1 that 7 willcome up the next throw when the other guy bets $5 that itwill not.

Games of chance must be distinguished from games in whichskill makes a difference. The odds are all you know for bettingin a game of chance.

Sebastian M. Saiegh Probability Theory

Page 15: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Lady Luck (cont.)

The odds obviously depend on the probability, but the oddsare what matter when you are placing a bet:

If the probability of throwing a 7 in craps is one out of sixthrows, the odds on throwing a number other than 7 are 5 to 1.This means that you should bet no more than $1 that 7 willcome up the next throw when the other guy bets $5 that itwill not.

Games of chance must be distinguished from games in whichskill makes a difference. The odds are all you know for bettingin a game of chance.

Sebastian M. Saiegh Probability Theory

Page 16: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark

Gambling has been a popular pastime and often an addictionsince the beginning of recorded history. However, until theXVII century, people had wagered and played games of chancewithout using any system of odds to determine winnings andlosings.

The Greeks understood that more things might happen in thefuture than actually will happen.

In fact, the word, εικoς, meaning plausible or probable, hadthe same sense as the modern concept of probability: “to beexpected with some degree of certainty.”

However, they never made any advances in measuring thisdegree of certainty.

Sebastian M. Saiegh Probability Theory

Page 17: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark

Gambling has been a popular pastime and often an addictionsince the beginning of recorded history. However, until theXVII century, people had wagered and played games of chancewithout using any system of odds to determine winnings andlosings.

The Greeks understood that more things might happen in thefuture than actually will happen.

In fact, the word, εικoς, meaning plausible or probable, hadthe same sense as the modern concept of probability: “to beexpected with some degree of certainty.”

However, they never made any advances in measuring thisdegree of certainty.

Sebastian M. Saiegh Probability Theory

Page 18: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark

Gambling has been a popular pastime and often an addictionsince the beginning of recorded history. However, until theXVII century, people had wagered and played games of chancewithout using any system of odds to determine winnings andlosings.

The Greeks understood that more things might happen in thefuture than actually will happen.

In fact, the word, εικoς, meaning plausible or probable, hadthe same sense as the modern concept of probability: “to beexpected with some degree of certainty.”

However, they never made any advances in measuring thisdegree of certainty.

Sebastian M. Saiegh Probability Theory

Page 19: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark

Gambling has been a popular pastime and often an addictionsince the beginning of recorded history. However, until theXVII century, people had wagered and played games of chancewithout using any system of odds to determine winnings andlosings.

The Greeks understood that more things might happen in thefuture than actually will happen.

In fact, the word, εικoς, meaning plausible or probable, hadthe same sense as the modern concept of probability: “to beexpected with some degree of certainty.”

However, they never made any advances in measuring thisdegree of certainty.

Sebastian M. Saiegh Probability Theory

Page 20: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

When the Greeks wanted a prediction about what tomorrowmight bring, they turned to the oracles instead of consultingtheir wisest philosophers.

Up to the time of the Renaissance, people perceived the futureas little more than a matter of luck or the result of randomvariations, and most of their decisions were driven by instinct.

The inappropriateness of the numbering system was importantreason of why Europeans were not induced to explore themastery of risk.

Without numbers, there are no odds and no probabilities;without odds and probabilities, the only way to deal with riskis to appeal to the gods and the fates.

Sebastian M. Saiegh Probability Theory

Page 21: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

When the Greeks wanted a prediction about what tomorrowmight bring, they turned to the oracles instead of consultingtheir wisest philosophers.

Up to the time of the Renaissance, people perceived the futureas little more than a matter of luck or the result of randomvariations, and most of their decisions were driven by instinct.

The inappropriateness of the numbering system was importantreason of why Europeans were not induced to explore themastery of risk.

Without numbers, there are no odds and no probabilities;without odds and probabilities, the only way to deal with riskis to appeal to the gods and the fates.

Sebastian M. Saiegh Probability Theory

Page 22: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

When the Greeks wanted a prediction about what tomorrowmight bring, they turned to the oracles instead of consultingtheir wisest philosophers.

Up to the time of the Renaissance, people perceived the futureas little more than a matter of luck or the result of randomvariations, and most of their decisions were driven by instinct.

The inappropriateness of the numbering system was importantreason of why Europeans were not induced to explore themastery of risk.

Without numbers, there are no odds and no probabilities;without odds and probabilities, the only way to deal with riskis to appeal to the gods and the fates.

Sebastian M. Saiegh Probability Theory

Page 23: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

When the Greeks wanted a prediction about what tomorrowmight bring, they turned to the oracles instead of consultingtheir wisest philosophers.

Up to the time of the Renaissance, people perceived the futureas little more than a matter of luck or the result of randomvariations, and most of their decisions were driven by instinct.

The inappropriateness of the numbering system was importantreason of why Europeans were not induced to explore themastery of risk.

Without numbers, there are no odds and no probabilities;without odds and probabilities, the only way to deal with riskis to appeal to the gods and the fates.

Sebastian M. Saiegh Probability Theory

Page 24: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

It is hard for us to imagine a time without numbers. However,if we were able to bring a well-educated man from the year1000 to the present, he probably would not recognize thenumber zero and would surely flunk third-grade arithmetic;few people from the year 1500 would fare much better.

The story of numbers in the West begins in 1202, when abook titled Liber Abaci, or Book of Abacus, appeared in Italy.The author, Leonardo Pisano, was known for most of his lifeas Fibonacci (yes, the “Fibonacci series”).

It took almost three hundred years, though, for theHindu-Arabic numbering system to be widely adopted in theWestern world.

Sebastian M. Saiegh Probability Theory

Page 25: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

It is hard for us to imagine a time without numbers. However,if we were able to bring a well-educated man from the year1000 to the present, he probably would not recognize thenumber zero and would surely flunk third-grade arithmetic;few people from the year 1500 would fare much better.

The story of numbers in the West begins in 1202, when abook titled Liber Abaci, or Book of Abacus, appeared in Italy.The author, Leonardo Pisano, was known for most of his lifeas Fibonacci (yes, the “Fibonacci series”).

It took almost three hundred years, though, for theHindu-Arabic numbering system to be widely adopted in theWestern world.

Sebastian M. Saiegh Probability Theory

Page 26: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

It is hard for us to imagine a time without numbers. However,if we were able to bring a well-educated man from the year1000 to the present, he probably would not recognize thenumber zero and would surely flunk third-grade arithmetic;few people from the year 1500 would fare much better.

The story of numbers in the West begins in 1202, when abook titled Liber Abaci, or Book of Abacus, appeared in Italy.The author, Leonardo Pisano, was known for most of his lifeas Fibonacci (yes, the “Fibonacci series”).

It took almost three hundred years, though, for theHindu-Arabic numbering system to be widely adopted in theWestern world.

Sebastian M. Saiegh Probability Theory

Page 27: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

In 1494, a Franciscan monk named Luca Paccioli published hisSumma de arithmetic, geometria et proportionalita. He posedthe following problem in the book:

A and B are playing a fair game of balla. They agree tocontinue until one has won six rounds. The game actuallystops when A has won five and B three. How should thestakes be divided?

The puzzle, which came to be known as the problem of thepoints, was more significant than it appears. The resolution ofhow to divide the stakes in an uncompleted game marked thebeginning of a systematic analysis of probability.

Sebastian M. Saiegh Probability Theory

Page 28: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

In 1494, a Franciscan monk named Luca Paccioli published hisSumma de arithmetic, geometria et proportionalita. He posedthe following problem in the book:

A and B are playing a fair game of balla. They agree tocontinue until one has won six rounds. The game actuallystops when A has won five and B three. How should thestakes be divided?

The puzzle, which came to be known as the problem of thepoints, was more significant than it appears. The resolution ofhow to divide the stakes in an uncompleted game marked thebeginning of a systematic analysis of probability.

Sebastian M. Saiegh Probability Theory

Page 29: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Gambling in the Dark (cont.)

In 1494, a Franciscan monk named Luca Paccioli published hisSumma de arithmetic, geometria et proportionalita. He posedthe following problem in the book:

A and B are playing a fair game of balla. They agree tocontinue until one has won six rounds. The game actuallystops when A has won five and B three. How should thestakes be divided?

The puzzle, which came to be known as the problem of thepoints, was more significant than it appears. The resolution ofhow to divide the stakes in an uncompleted game marked thebeginning of a systematic analysis of probability.

Sebastian M. Saiegh Probability Theory

Page 30: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers

In 1654, the Chevalier de Mere, a French nobleman with a taste forboth gambling and mathematics, challenged the famed Frenchmathematician Blaise Pascal to solve the puzzle.

Pascal turned for help to Pierre de Fermat, a lawyer who wasalso a brilliant mathematician.Their collaboration led to the discovery of the theory ofprobability.

Sebastian M. Saiegh Probability Theory

Page 31: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers

In 1654, the Chevalier de Mere, a French nobleman with a taste forboth gambling and mathematics, challenged the famed Frenchmathematician Blaise Pascal to solve the puzzle.

Pascal turned for help to Pierre de Fermat, a lawyer who wasalso a brilliant mathematician.Their collaboration led to the discovery of the theory ofprobability.

Sebastian M. Saiegh Probability Theory

Page 32: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

Given that more things can happen than will happen, Pascaland Fermat established a procedure for determining thelikelihood of each of the possible results – assuming alwaysthat the outcomes can be measured mathematically.

As the years passed, mathematicians transformed probabilitytheory from a gamblers’ toy into a powerful instrument fororganizing, interpreting, and applying information.

In particular, they showed how to infer previously unknownprobabilities from the empirical facts of reality.

Sebastian M. Saiegh Probability Theory

Page 33: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

Given that more things can happen than will happen, Pascaland Fermat established a procedure for determining thelikelihood of each of the possible results – assuming alwaysthat the outcomes can be measured mathematically.

As the years passed, mathematicians transformed probabilitytheory from a gamblers’ toy into a powerful instrument fororganizing, interpreting, and applying information.

In particular, they showed how to infer previously unknownprobabilities from the empirical facts of reality.

Sebastian M. Saiegh Probability Theory

Page 34: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

Given that more things can happen than will happen, Pascaland Fermat established a procedure for determining thelikelihood of each of the possible results – assuming alwaysthat the outcomes can be measured mathematically.

As the years passed, mathematicians transformed probabilitytheory from a gamblers’ toy into a powerful instrument fororganizing, interpreting, and applying information.

In particular, they showed how to infer previously unknownprobabilities from the empirical facts of reality.

Sebastian M. Saiegh Probability Theory

Page 35: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

In 1703, Gottfried von Leibniz commented to the Swiss scientistand mathematician Jacob Bernoulli that “[N]ature has establishedpatterns originating in the return of events, but only for the mostpart,” thereby prompting Bernoulli to invent the Law of LargeNumbers.

In 1730, Abraham de Moivre suggested the structure of thenormal distribution – also known as the bell curve – anddiscovered the concept of standard deviation.

Sebastian M. Saiegh Probability Theory

Page 36: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

In 1703, Gottfried von Leibniz commented to the Swiss scientistand mathematician Jacob Bernoulli that “[N]ature has establishedpatterns originating in the return of events, but only for the mostpart,” thereby prompting Bernoulli to invent the Law of LargeNumbers.

In 1730, Abraham de Moivre suggested the structure of thenormal distribution – also known as the bell curve – anddiscovered the concept of standard deviation.

Sebastian M. Saiegh Probability Theory

Page 37: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

Around 1760, a dissident English minister named ThomasBayes made a striking advance in statistics by demonstratinghow to make better-informed decisions by mathematicallyblending new information into old information.

With some exceptions, all the tools we use today in theanalysis of decisions and choice, from the strict rationality ofgame theory to the challenges of chaos theory, stem most fromthe developments that took place between 1654 and 1760.

Sebastian M. Saiegh Probability Theory

Page 38: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Non-Degenerate Gamblers (cont.)

Around 1760, a dissident English minister named ThomasBayes made a striking advance in statistics by demonstratinghow to make better-informed decisions by mathematicallyblending new information into old information.

With some exceptions, all the tools we use today in theanalysis of decisions and choice, from the strict rationality ofgame theory to the challenges of chaos theory, stem most fromthe developments that took place between 1654 and 1760.

Sebastian M. Saiegh Probability Theory

Page 39: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Today’s Class

In today’s class we will focus the ideas developed by theseremarkable thinkers.

Then, we will apply the calculus of probabilities to a widevariety of situations requiring the use of sophisticatedcounting techniques.

We shall often refer to games of chance. Yet these gameshave applications that extend far beyond the spin of theroulette wheel.

Sebastian M. Saiegh Probability Theory

Page 40: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Today’s Class

In today’s class we will focus the ideas developed by theseremarkable thinkers.

Then, we will apply the calculus of probabilities to a widevariety of situations requiring the use of sophisticatedcounting techniques.

We shall often refer to games of chance. Yet these gameshave applications that extend far beyond the spin of theroulette wheel.

Sebastian M. Saiegh Probability Theory

Page 41: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional ProbabilitySome Background

Today’s Class

In today’s class we will focus the ideas developed by theseremarkable thinkers.

Then, we will apply the calculus of probabilities to a widevariety of situations requiring the use of sophisticatedcounting techniques.

We shall often refer to games of chance. Yet these gameshave applications that extend far beyond the spin of theroulette wheel.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

Page 43: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Introduction

In everyday life, we often talk loosely about chance. What are thechances of getting a job? of meeting someone? of rain tomorrow?But for the purposes of this class, we need to give the word chancea definite, clear interpretation.This turns out to be hard, and, mathematicians have struggledwith the job for centuries. We will now focus on the frequencytheory, which works best for processes which can be repeated overand over again, independently and under the same conditions.Games of chance fail into category. And, as we already know, muchof the frequency theory was developed to solve gambling problems.

Sebastian M. Saiegh Probability Theory

Page 44: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Introduction

In everyday life, we often talk loosely about chance. What are thechances of getting a job? of meeting someone? of rain tomorrow?But for the purposes of this class, we need to give the word chancea definite, clear interpretation.This turns out to be hard, and, mathematicians have struggledwith the job for centuries. We will now focus on the frequencytheory, which works best for processes which can be repeated overand over again, independently and under the same conditions.Games of chance fail into category. And, as we already know, muchof the frequency theory was developed to solve gambling problems.

Sebastian M. Saiegh Probability Theory

Page 45: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Introduction

In everyday life, we often talk loosely about chance. What are thechances of getting a job? of meeting someone? of rain tomorrow?But for the purposes of this class, we need to give the word chancea definite, clear interpretation.This turns out to be hard, and, mathematicians have struggledwith the job for centuries. We will now focus on the frequencytheory, which works best for processes which can be repeated overand over again, independently and under the same conditions.Games of chance fail into category. And, as we already know, muchof the frequency theory was developed to solve gambling problems.

Sebastian M. Saiegh Probability Theory

Page 46: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins

One of the simplest games of chance involves betting on the tossof a coin.

If a coin is tossed in the air, then it is certain that the coinwill come down, but it is not certain that, say, it will come up“head”.

When trying to figure chances, it is usually very helpful to listall the possible ways that a chance process can turn out.

In the case of a coin toss, we ordinarily agree to regard“head” and “tail” as the only possible outcomes.

If we denote these outcomes by H and T respectively, theneach outcome would correspond to exactly one of theelements of the set H,T.

Sebastian M. Saiegh Probability Theory

Page 47: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins

One of the simplest games of chance involves betting on the tossof a coin.

If a coin is tossed in the air, then it is certain that the coinwill come down, but it is not certain that, say, it will come up“head”.

When trying to figure chances, it is usually very helpful to listall the possible ways that a chance process can turn out.

In the case of a coin toss, we ordinarily agree to regard“head” and “tail” as the only possible outcomes.

If we denote these outcomes by H and T respectively, theneach outcome would correspond to exactly one of theelements of the set H,T.

Sebastian M. Saiegh Probability Theory

Page 48: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins

One of the simplest games of chance involves betting on the tossof a coin.

If a coin is tossed in the air, then it is certain that the coinwill come down, but it is not certain that, say, it will come up“head”.

When trying to figure chances, it is usually very helpful to listall the possible ways that a chance process can turn out.

In the case of a coin toss, we ordinarily agree to regard“head” and “tail” as the only possible outcomes.

If we denote these outcomes by H and T respectively, theneach outcome would correspond to exactly one of theelements of the set H,T.

Sebastian M. Saiegh Probability Theory

Page 49: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins

One of the simplest games of chance involves betting on the tossof a coin.

If a coin is tossed in the air, then it is certain that the coinwill come down, but it is not certain that, say, it will come up“head”.

When trying to figure chances, it is usually very helpful to listall the possible ways that a chance process can turn out.

In the case of a coin toss, we ordinarily agree to regard“head” and “tail” as the only possible outcomes.

If we denote these outcomes by H and T respectively, theneach outcome would correspond to exactly one of theelements of the set H,T.

Sebastian M. Saiegh Probability Theory

Page 50: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins

One of the simplest games of chance involves betting on the tossof a coin.

If a coin is tossed in the air, then it is certain that the coinwill come down, but it is not certain that, say, it will come up“head”.

When trying to figure chances, it is usually very helpful to listall the possible ways that a chance process can turn out.

In the case of a coin toss, we ordinarily agree to regard“head” and “tail” as the only possible outcomes.

If we denote these outcomes by H and T respectively, theneach outcome would correspond to exactly one of theelements of the set H,T.

Sebastian M. Saiegh Probability Theory

Page 51: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Sample Space

Suppose now that we repeat this experiment of tossing a coin.

The process of tossing the coin can certainly be repeated overand over again, independently and under the same conditions:the outcomes of this experiment would still correspond toexactly one of the elements of the set H,T.This set, Ω = H,T is called a sample space for theexperiment.

Suppose now that we toss a die and observe the number thatappears on top. Then the sample space consist of the six possiblenumbers: Ω = 1, 2, 3, 4, 5, 6.

Sebastian M. Saiegh Probability Theory

Page 52: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Sample Space

Suppose now that we repeat this experiment of tossing a coin.

The process of tossing the coin can certainly be repeated overand over again, independently and under the same conditions:the outcomes of this experiment would still correspond toexactly one of the elements of the set H,T.This set, Ω = H,T is called a sample space for theexperiment.

Suppose now that we toss a die and observe the number thatappears on top. Then the sample space consist of the six possiblenumbers: Ω = 1, 2, 3, 4, 5, 6.

Sebastian M. Saiegh Probability Theory

Page 53: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Sample Space

Suppose now that we repeat this experiment of tossing a coin.

The process of tossing the coin can certainly be repeated overand over again, independently and under the same conditions:the outcomes of this experiment would still correspond toexactly one of the elements of the set H,T.This set, Ω = H,T is called a sample space for theexperiment.

Suppose now that we toss a die and observe the number thatappears on top. Then the sample space consist of the six possiblenumbers: Ω = 1, 2, 3, 4, 5, 6.

Sebastian M. Saiegh Probability Theory

Page 54: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Sample Space

Suppose now that we repeat this experiment of tossing a coin.

The process of tossing the coin can certainly be repeated overand over again, independently and under the same conditions:the outcomes of this experiment would still correspond toexactly one of the elements of the set H,T.This set, Ω = H,T is called a sample space for theexperiment.

Suppose now that we toss a die and observe the number thatappears on top. Then the sample space consist of the six possiblenumbers: Ω = 1, 2, 3, 4, 5, 6.

Sebastian M. Saiegh Probability Theory

Page 55: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Sample Space (cont.)

Definition

The set Ω of all possible outcomes associated with a real orconceptual experiment is called the sample space. A particularoutcome ω, i.e. an element in Ω, is called a sample point orsample.

Sebastian M. Saiegh Probability Theory

Page 56: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins, Again

Suppose now that we perform the following experiment: we toss acoin three successive times.Let Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT bethe associated sample space.

We may be interested in the event, “the number of headsexceeds the number of tails”.

For any outcome of the experiment we can determine whetherthis event does or does not occur.

We find that HHH, HHT ,HTH and THH are the onlyelements of Ω corresponding to outcomes for which this eventdoes occur.

Sebastian M. Saiegh Probability Theory

Page 57: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins, Again

Suppose now that we perform the following experiment: we toss acoin three successive times.Let Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT bethe associated sample space.

We may be interested in the event, “the number of headsexceeds the number of tails”.

For any outcome of the experiment we can determine whetherthis event does or does not occur.

We find that HHH, HHT ,HTH and THH are the onlyelements of Ω corresponding to outcomes for which this eventdoes occur.

Sebastian M. Saiegh Probability Theory

Page 58: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins, Again

Suppose now that we perform the following experiment: we toss acoin three successive times.Let Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT bethe associated sample space.

We may be interested in the event, “the number of headsexceeds the number of tails”.

For any outcome of the experiment we can determine whetherthis event does or does not occur.

We find that HHH, HHT ,HTH and THH are the onlyelements of Ω corresponding to outcomes for which this eventdoes occur.

Sebastian M. Saiegh Probability Theory

Page 59: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins, Again

Suppose now that we perform the following experiment: we toss acoin three successive times.Let Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT bethe associated sample space.

We may be interested in the event, “the number of headsexceeds the number of tails”.

For any outcome of the experiment we can determine whetherthis event does or does not occur.

We find that HHH, HHT ,HTH and THH are the onlyelements of Ω corresponding to outcomes for which this eventdoes occur.

Sebastian M. Saiegh Probability Theory

Page 60: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Tossing Coins, Again

Suppose now that we perform the following experiment: we toss acoin three successive times.Let Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT bethe associated sample space.

We may be interested in the event, “the number of headsexceeds the number of tails”.

For any outcome of the experiment we can determine whetherthis event does or does not occur.

We find that HHH, HHT ,HTH and THH are the onlyelements of Ω corresponding to outcomes for which this eventdoes occur.

Sebastian M. Saiegh Probability Theory

Page 61: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events

To say that the event “the number of heads exceeds the number oftails” occurs is the same as saying the experiment results in anoutcome corresponding to an element of the setA = HHH,HHT ,HTH,THH.

Notice that A is a subset of the sample space Ω.

Definition

An event A is a set of outcomes or, in other words, a subset ofsome underlying sample space Ω.

Sebastian M. Saiegh Probability Theory

Page 62: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events

To say that the event “the number of heads exceeds the number oftails” occurs is the same as saying the experiment results in anoutcome corresponding to an element of the setA = HHH,HHT ,HTH,THH.

Notice that A is a subset of the sample space Ω.

Definition

An event A is a set of outcomes or, in other words, a subset ofsome underlying sample space Ω.

Sebastian M. Saiegh Probability Theory

Page 63: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events

To say that the event “the number of heads exceeds the number oftails” occurs is the same as saying the experiment results in anoutcome corresponding to an element of the setA = HHH,HHT ,HTH,THH.

Notice that A is a subset of the sample space Ω.

Definition

An event A is a set of outcomes or, in other words, a subset ofsome underlying sample space Ω.

Sebastian M. Saiegh Probability Theory

Page 64: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

The event ω consisting of a single sample ω ∈ Ω is calledan elementary event.

The certain (or sure) event, which always occurs regardless ofthe outcome of the experiment, is formally identical with thewhole space Ω.

The impossible event is the empty set ∅, containing none ofthe elementary events ω.

Sebastian M. Saiegh Probability Theory

Page 65: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

The event ω consisting of a single sample ω ∈ Ω is calledan elementary event.

The certain (or sure) event, which always occurs regardless ofthe outcome of the experiment, is formally identical with thewhole space Ω.

The impossible event is the empty set ∅, containing none ofthe elementary events ω.

Sebastian M. Saiegh Probability Theory

Page 66: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

The event ω consisting of a single sample ω ∈ Ω is calledan elementary event.

The certain (or sure) event, which always occurs regardless ofthe outcome of the experiment, is formally identical with thewhole space Ω.

The impossible event is the empty set ∅, containing none ofthe elementary events ω.

Sebastian M. Saiegh Probability Theory

Page 67: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

We can combine events to form new events using the various setoperations:

(i) A1 ∪A2 is the event that occurs iff A1 occurs or A2 occurs (orboth);

(ii) A1 ∩ A2 is the event that occurs iff A1 occurs and A2 occurs;

(iii) A ′ = A−Ω, the complement of A, is the event that occurs iffA does not occur.

Sebastian M. Saiegh Probability Theory

Page 68: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

We can combine events to form new events using the various setoperations:

(i) A1 ∪A2 is the event that occurs iff A1 occurs or A2 occurs (orboth);

(ii) A1 ∩ A2 is the event that occurs iff A1 occurs and A2 occurs;

(iii) A ′ = A−Ω, the complement of A, is the event that occurs iffA does not occur.

Sebastian M. Saiegh Probability Theory

Page 69: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

We can combine events to form new events using the various setoperations:

(i) A1 ∪A2 is the event that occurs iff A1 occurs or A2 occurs (orboth);

(ii) A1 ∩ A2 is the event that occurs iff A1 occurs and A2 occurs;

(iii) A ′ = A−Ω, the complement of A, is the event that occurs iffA does not occur.

Sebastian M. Saiegh Probability Theory

Page 70: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

We can combine events to form new events using the various setoperations:

(i) A1 ∪A2 is the event that occurs iff A1 occurs or A2 occurs (orboth);

(ii) A1 ∩ A2 is the event that occurs iff A1 occurs and A2 occurs;

(iii) A ′ = A−Ω, the complement of A, is the event that occurs iffA does not occur.

Sebastian M. Saiegh Probability Theory

Page 71: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

Given two events A1 and A2, suppose A1 occurs if and only if A2

occurs.

Then, A1 and A2 are said to be identical (or equivalent), andwe write A1 = A2.

Two events A1 and A2 are called mutually exclusive if they aredisjoint, i.e. if A1 ∩ A2 = ∅.

In other words, A1 and A2 are mutually exclusive if theycannot occur simultaneously.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

Given two events A1 and A2, suppose A1 occurs if and only if A2

occurs.

Then, A1 and A2 are said to be identical (or equivalent), andwe write A1 = A2.

Two events A1 and A2 are called mutually exclusive if they aredisjoint, i.e. if A1 ∩ A2 = ∅.

In other words, A1 and A2 are mutually exclusive if theycannot occur simultaneously.

Sebastian M. Saiegh Probability Theory

Page 73: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

Given two events A1 and A2, suppose A1 occurs if and only if A2

occurs.

Then, A1 and A2 are said to be identical (or equivalent), andwe write A1 = A2.

Two events A1 and A2 are called mutually exclusive if they aredisjoint, i.e. if A1 ∩ A2 = ∅.

In other words, A1 and A2 are mutually exclusive if theycannot occur simultaneously.

Sebastian M. Saiegh Probability Theory

Page 74: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events (cont.)

Given two events A1 and A2, suppose A1 occurs if and only if A2

occurs.

Then, A1 and A2 are said to be identical (or equivalent), andwe write A1 = A2.

Two events A1 and A2 are called mutually exclusive if they aredisjoint, i.e. if A1 ∩ A2 = ∅.

In other words, A1 and A2 are mutually exclusive if theycannot occur simultaneously.

Sebastian M. Saiegh Probability Theory

Page 75: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example

Example

Suppose that we toss a die and observe the number that appearson top. The sample space consist of the six possible numbers:Ω = 1, 2, 3, 4, 5, 6. Let A1 be the event that an even numberoccurs, A2 that an odd number occurs, and A3 that a numberhigher than 3 occurs.

Then, A1 = 2, 4, 6, A2 = 1, 3, 5, and A3 = 4, 5, 6. And,

A1 ∪ A3 = 2, 4, 5, 6 is the event that an even number or anumber higher than 3 occurs;A1 ∩ A3 = 4, 6 is the event that an even number higherthan 3 occurs;A3′ = 1, 2, 3 is the event that a number higher than 3 does

not occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example

Example

Suppose that we toss a die and observe the number that appearson top. The sample space consist of the six possible numbers:Ω = 1, 2, 3, 4, 5, 6. Let A1 be the event that an even numberoccurs, A2 that an odd number occurs, and A3 that a numberhigher than 3 occurs.

Then, A1 = 2, 4, 6, A2 = 1, 3, 5, and A3 = 4, 5, 6. And,

A1 ∪ A3 = 2, 4, 5, 6 is the event that an even number or anumber higher than 3 occurs;A1 ∩ A3 = 4, 6 is the event that an even number higherthan 3 occurs;A3′ = 1, 2, 3 is the event that a number higher than 3 does

not occur.

Sebastian M. Saiegh Probability Theory

Page 77: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example

Example

Suppose that we toss a die and observe the number that appearson top. The sample space consist of the six possible numbers:Ω = 1, 2, 3, 4, 5, 6. Let A1 be the event that an even numberoccurs, A2 that an odd number occurs, and A3 that a numberhigher than 3 occurs.

Then, A1 = 2, 4, 6, A2 = 1, 3, 5, and A3 = 4, 5, 6. And,

A1 ∪ A3 = 2, 4, 5, 6 is the event that an even number or anumber higher than 3 occurs;A1 ∩ A3 = 4, 6 is the event that an even number higherthan 3 occurs;A3′ = 1, 2, 3 is the event that a number higher than 3 does

not occur.

Sebastian M. Saiegh Probability Theory

Page 78: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example

Example

Suppose that we toss a die and observe the number that appearson top. The sample space consist of the six possible numbers:Ω = 1, 2, 3, 4, 5, 6. Let A1 be the event that an even numberoccurs, A2 that an odd number occurs, and A3 that a numberhigher than 3 occurs.

Then, A1 = 2, 4, 6, A2 = 1, 3, 5, and A3 = 4, 5, 6. And,

A1 ∪ A3 = 2, 4, 5, 6 is the event that an even number or anumber higher than 3 occurs;A1 ∩ A3 = 4, 6 is the event that an even number higherthan 3 occurs;A3′ = 1, 2, 3 is the event that a number higher than 3 does

not occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example

Example

Suppose that we toss a die and observe the number that appearson top. The sample space consist of the six possible numbers:Ω = 1, 2, 3, 4, 5, 6. Let A1 be the event that an even numberoccurs, A2 that an odd number occurs, and A3 that a numberhigher than 3 occurs.

Then, A1 = 2, 4, 6, A2 = 1, 3, 5, and A3 = 4, 5, 6. And,

A1 ∪ A3 = 2, 4, 5, 6 is the event that an even number or anumber higher than 3 occurs;A1 ∩ A3 = 4, 6 is the event that an even number higherthan 3 occurs;A3′ = 1, 2, 3 is the event that a number higher than 3 does

not occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example (cont.)

Note that A1 and A2 are mutually exclusive: A1 ∩ A2 = ∅.

In other words, an even number and an odd number cannotoccur simultaneously.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Events: Example (cont.)

Note that A1 and A2 are mutually exclusive: A1 ∩ A2 = ∅.

In other words, an even number and an odd number cannotoccur simultaneously.

Sebastian M. Saiegh Probability Theory

Page 82: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Definition

Let Ω be a finite sample space; say, Ω = ω1, ω2, ..., ωn.

A finite probability space is obtained by assigning to eachsample point ωi ∈ Ω a real number pi , called the probabilityof ωi , satisfying the following properties:

(i) each pi is non-negative, pi ≥ 0

(ii) the sum of the pi is one,n∑

i=1

pi = 1

The probability P(A) of an event A, is then defined to be the sumof the probabilities of the sample points in A. For notationalconvenience we write P(ωi ) for P(ωi).

Sebastian M. Saiegh Probability Theory

Page 84: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Definition

Let Ω be a finite sample space; say, Ω = ω1, ω2, ..., ωn.

A finite probability space is obtained by assigning to eachsample point ωi ∈ Ω a real number pi , called the probabilityof ωi , satisfying the following properties:

(i) each pi is non-negative, pi ≥ 0

(ii) the sum of the pi is one,n∑

i=1

pi = 1

The probability P(A) of an event A, is then defined to be the sumof the probabilities of the sample points in A. For notationalconvenience we write P(ωi ) for P(ωi).

Sebastian M. Saiegh Probability Theory

Page 85: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Definition

Let Ω be a finite sample space; say, Ω = ω1, ω2, ..., ωn.

A finite probability space is obtained by assigning to eachsample point ωi ∈ Ω a real number pi , called the probabilityof ωi , satisfying the following properties:

(i) each pi is non-negative, pi ≥ 0

(ii) the sum of the pi is one,n∑

i=1

pi = 1

The probability P(A) of an event A, is then defined to be the sumof the probabilities of the sample points in A. For notationalconvenience we write P(ωi ) for P(ωi).

Sebastian M. Saiegh Probability Theory

Page 86: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Definition

Let Ω be a finite sample space; say, Ω = ω1, ω2, ..., ωn.

A finite probability space is obtained by assigning to eachsample point ωi ∈ Ω a real number pi , called the probabilityof ωi , satisfying the following properties:

(i) each pi is non-negative, pi ≥ 0

(ii) the sum of the pi is one,n∑

i=1

pi = 1

The probability P(A) of an event A, is then defined to be the sumof the probabilities of the sample points in A. For notationalconvenience we write P(ωi ) for P(ωi).

Sebastian M. Saiegh Probability Theory

Page 87: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Definition

Let Ω be a finite sample space; say, Ω = ω1, ω2, ..., ωn.

A finite probability space is obtained by assigning to eachsample point ωi ∈ Ω a real number pi , called the probabilityof ωi , satisfying the following properties:

(i) each pi is non-negative, pi ≥ 0

(ii) the sum of the pi is one,n∑

i=1

pi = 1

The probability P(A) of an event A, is then defined to be the sumof the probabilities of the sample points in A. For notationalconvenience we write P(ωi ) for P(ωi).

Sebastian M. Saiegh Probability Theory

Page 88: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Example

Example

Let three coins be tossed and the number of heads observed; thenΩ = 0, 1, 2, 3 is a sample space associated with this experiment.

We obtain a probability space by the following assignment,P(0) = 1

8 , P(1) = 38 , P(2) = 3

8 , and P(3) = 18 since each

probability is nonnegative and the sum of the probabilities is 1.Let now A1 be the event that at least one head appears and let A2

be the event that all heads or all tails appear: A1 = 1, 2, 3 andA2 = 0, 3.Then, by definition, P(A1) = P(1) + P(2) + P(3) = 3

8 + 38 + 1

8 = 78

and P(A2) = P(0) + P(3) = 18 + 1

8 = 14 .

Sebastian M. Saiegh Probability Theory

Page 89: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Example

Example

Let three coins be tossed and the number of heads observed; thenΩ = 0, 1, 2, 3 is a sample space associated with this experiment.

We obtain a probability space by the following assignment,P(0) = 1

8 , P(1) = 38 , P(2) = 3

8 , and P(3) = 18 since each

probability is nonnegative and the sum of the probabilities is 1.Let now A1 be the event that at least one head appears and let A2

be the event that all heads or all tails appear: A1 = 1, 2, 3 andA2 = 0, 3.Then, by definition, P(A1) = P(1) + P(2) + P(3) = 3

8 + 38 + 1

8 = 78

and P(A2) = P(0) + P(3) = 18 + 1

8 = 14 .

Sebastian M. Saiegh Probability Theory

Page 90: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Example

Example

Let three coins be tossed and the number of heads observed; thenΩ = 0, 1, 2, 3 is a sample space associated with this experiment.

We obtain a probability space by the following assignment,P(0) = 1

8 , P(1) = 38 , P(2) = 3

8 , and P(3) = 18 since each

probability is nonnegative and the sum of the probabilities is 1.Let now A1 be the event that at least one head appears and let A2

be the event that all heads or all tails appear: A1 = 1, 2, 3 andA2 = 0, 3.Then, by definition, P(A1) = P(1) + P(2) + P(3) = 3

8 + 38 + 1

8 = 78

and P(A2) = P(0) + P(3) = 18 + 1

8 = 14 .

Sebastian M. Saiegh Probability Theory

Page 91: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability of an Event: Example

Example

Let three coins be tossed and the number of heads observed; thenΩ = 0, 1, 2, 3 is a sample space associated with this experiment.

We obtain a probability space by the following assignment,P(0) = 1

8 , P(1) = 38 , P(2) = 3

8 , and P(3) = 18 since each

probability is nonnegative and the sum of the probabilities is 1.Let now A1 be the event that at least one head appears and let A2

be the event that all heads or all tails appear: A1 = 1, 2, 3 andA2 = 0, 3.Then, by definition, P(A1) = P(1) + P(2) + P(3) = 3

8 + 38 + 1

8 = 78

and P(A2) = P(0) + P(3) = 18 + 1

8 = 14 .

Sebastian M. Saiegh Probability Theory

Page 92: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Assignment of Probabilities and Probability of an Event

Note that the probability of an event depends on the previousassignment of probabilities to the sample points.

The following question then arises: which assignment ofprobabilities to sample points should be made?

The answer to this question is not a mathematical one.Rather, it depends upon our assessment of the real-worldsituation to which the theory is to be applied.

Sebastian M. Saiegh Probability Theory

Page 93: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Assignment of Probabilities and Probability of an Event

Note that the probability of an event depends on the previousassignment of probabilities to the sample points.

The following question then arises: which assignment ofprobabilities to sample points should be made?

The answer to this question is not a mathematical one.Rather, it depends upon our assessment of the real-worldsituation to which the theory is to be applied.

Sebastian M. Saiegh Probability Theory

Page 94: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Assignment of Probabilities and Probability of an Event

Note that the probability of an event depends on the previousassignment of probabilities to the sample points.

The following question then arises: which assignment ofprobabilities to sample points should be made?

The answer to this question is not a mathematical one.Rather, it depends upon our assessment of the real-worldsituation to which the theory is to be applied.

Sebastian M. Saiegh Probability Theory

Page 95: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability

Let Ω be a sample space, let C be the class of all events, and let Pbe a real-valued function defined on C.Then P is called a probability function, and P(A) is called theprobability of the event A if the following axioms hold:

1. The probability of a certain event is 1, P(Ω) = 1.

2. If A is any event, then 0 ≤ P(A) ≤ 1.

3. If A1 and A2 are mutually exclusive events, then

P(A1 ∪ A2) = P(A1) + P(A2).

Sebastian M. Saiegh Probability Theory

Page 96: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability

Let Ω be a sample space, let C be the class of all events, and let Pbe a real-valued function defined on C.Then P is called a probability function, and P(A) is called theprobability of the event A if the following axioms hold:

1. The probability of a certain event is 1, P(Ω) = 1.

2. If A is any event, then 0 ≤ P(A) ≤ 1.

3. If A1 and A2 are mutually exclusive events, then

P(A1 ∪ A2) = P(A1) + P(A2).

Sebastian M. Saiegh Probability Theory

Page 97: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability

Let Ω be a sample space, let C be the class of all events, and let Pbe a real-valued function defined on C.Then P is called a probability function, and P(A) is called theprobability of the event A if the following axioms hold:

1. The probability of a certain event is 1, P(Ω) = 1.

2. If A is any event, then 0 ≤ P(A) ≤ 1.

3. If A1 and A2 are mutually exclusive events, then

P(A1 ∪ A2) = P(A1) + P(A2).

Sebastian M. Saiegh Probability Theory

Page 98: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability

Let Ω be a sample space, let C be the class of all events, and let Pbe a real-valued function defined on C.Then P is called a probability function, and P(A) is called theprobability of the event A if the following axioms hold:

1. The probability of a certain event is 1, P(Ω) = 1.

2. If A is any event, then 0 ≤ P(A) ≤ 1.

3. If A1 and A2 are mutually exclusive events, then

P(A1 ∪ A2) = P(A1) + P(A2).

Sebastian M. Saiegh Probability Theory

Page 99: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability

Let Ω be a sample space, let C be the class of all events, and let Pbe a real-valued function defined on C.Then P is called a probability function, and P(A) is called theprobability of the event A if the following axioms hold:

1. The probability of a certain event is 1, P(Ω) = 1.

2. If A is any event, then 0 ≤ P(A) ≤ 1.

3. If A1 and A2 are mutually exclusive events, then

P(A1 ∪ A2) = P(A1) + P(A2).

Sebastian M. Saiegh Probability Theory

Page 100: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

More generally, given n mutually exclusive events A1,A2, ...,An, wehave the formula

P( n∪

k=1Ak

)=

n∑k=1

P(Ak).

This equation is called the addition rule for probabilities.

We can now prove a few theorems which follow directly fromour axioms.

Sebastian M. Saiegh Probability Theory

Page 101: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

More generally, given n mutually exclusive events A1,A2, ...,An, wehave the formula

P( n∪

k=1Ak

)=

n∑k=1

P(Ak).

This equation is called the addition rule for probabilities.

We can now prove a few theorems which follow directly fromour axioms.

Sebastian M. Saiegh Probability Theory

Page 102: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If ∅ is the empty set, then P(∅) = 0.

Proof. Let A be any set; then A and ∅ are disjoint and A ∪ ∅ = A.By axiom 3,

P(A) = P(A ∪ ∅) = P(A) + P(∅)

Subtracting P(A) from both sides gives our result.

Sebastian M. Saiegh Probability Theory

Page 103: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If ∅ is the empty set, then P(∅) = 0.

Proof. Let A be any set; then A and ∅ are disjoint and A ∪ ∅ = A.By axiom 3,

P(A) = P(A ∪ ∅) = P(A) + P(∅)

Subtracting P(A) from both sides gives our result.

Sebastian M. Saiegh Probability Theory

Page 104: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A ′ is the complement of an event A, then P(A ′) = 1− P(A).

In words, the probability that A does not occur is obtained bysubtracting from 1 the probability that A does occur.

Proof. The sample space Ω can be decomposed into mutuallyexclusive events A and A ′; that is, Ω = A ∪ A ′. By axioms 1 and3 we obtain

1 = P(Ω) = P(A ∪ A ′) = P(A) + P(A ′)

from which our result follows.

Sebastian M. Saiegh Probability Theory

Page 105: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A ′ is the complement of an event A, then P(A ′) = 1− P(A).

In words, the probability that A does not occur is obtained bysubtracting from 1 the probability that A does occur.

Proof. The sample space Ω can be decomposed into mutuallyexclusive events A and A ′; that is, Ω = A ∪ A ′. By axioms 1 and3 we obtain

1 = P(Ω) = P(A ∪ A ′) = P(A) + P(A ′)

from which our result follows.

Sebastian M. Saiegh Probability Theory

Page 106: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A ′ is the complement of an event A, then P(A ′) = 1− P(A).

In words, the probability that A does not occur is obtained bysubtracting from 1 the probability that A does occur.

Proof. The sample space Ω can be decomposed into mutuallyexclusive events A and A ′; that is, Ω = A ∪ A ′. By axioms 1 and3 we obtain

1 = P(Ω) = P(A ∪ A ′) = P(A) + P(A ′)

from which our result follows.

Sebastian M. Saiegh Probability Theory

Page 107: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A1 ⊆ A2, then P(A1) ≤ P(A2).

In words, if A1 implies A2, then the probability of A1 cannotexceed the probability of A2.

Proof. If A1 ⊆ A2, then A2 can be decomposed into the mutuallyexclusive events A1 and A2 \ A1.Therefore,

P(A2) = P(A1) + P(A2 \ A1)

The result now follows from the fact that P(A2 \ A1) ≥ 1.

Sebastian M. Saiegh Probability Theory

Page 108: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A1 ⊆ A2, then P(A1) ≤ P(A2).

In words, if A1 implies A2, then the probability of A1 cannotexceed the probability of A2.

Proof. If A1 ⊆ A2, then A2 can be decomposed into the mutuallyexclusive events A1 and A2 \ A1.Therefore,

P(A2) = P(A1) + P(A2 \ A1)

The result now follows from the fact that P(A2 \ A1) ≥ 1.

Sebastian M. Saiegh Probability Theory

Page 109: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A1 ⊆ A2, then P(A1) ≤ P(A2).

In words, if A1 implies A2, then the probability of A1 cannotexceed the probability of A2.

Proof. If A1 ⊆ A2, then A2 can be decomposed into the mutuallyexclusive events A1 and A2 \ A1.Therefore,

P(A2) = P(A1) + P(A2 \ A1)

The result now follows from the fact that P(A2 \ A1) ≥ 1.

Sebastian M. Saiegh Probability Theory

Page 110: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A1 ⊆ A2, then P(A1) ≤ P(A2).

In words, if A1 implies A2, then the probability of A1 cannotexceed the probability of A2.

Proof. If A1 ⊆ A2, then A2 can be decomposed into the mutuallyexclusive events A1 and A2 \ A1.Therefore,

P(A2) = P(A1) + P(A2 \ A1)

The result now follows from the fact that P(A2 \ A1) ≥ 1.

Sebastian M. Saiegh Probability Theory

Page 111: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A1 and A2 are any two events, then

P(A1 \ A2) = P(A1)− P(A1 ∩ A2).

Proof. Now A1 can be decomposed into the mutually exclusiveevents A1 \ A2 and A1 ∩ A2; that is A1 = (A1 \ A2) ∪ (A1 ∩ A2).Thus by axiom 3,

P(A1) = P(A1 \ A2) + P(A1 ∩ A2)

from which our result follows.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Theorem

If A1 and A2 are any two events, then

P(A1 \ A2) = P(A1)− P(A1 ∩ A2).

Proof. Now A1 can be decomposed into the mutually exclusiveevents A1 \ A2 and A1 ∩ A2; that is A1 = (A1 \ A2) ∪ (A1 ∩ A2).Thus by axiom 3,

P(A1) = P(A1 \ A2) + P(A1 ∩ A2)

from which our result follows.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

The addition rule (axiom 3) becomes a little more complicated ifwe drop the requirement that the events be mutually exclusive.The following theorem, called the general addition law is similar tothe inclusion-exclusion principle for sets.

Theorem

If A1 and A2 are any two events, then

P(A1 ∪ A2) = P(A1) + P(A2)− P(A1 ∩ A2).

In words, the probability that at leat one of the events A1 and A2

occurs is obtained by adding the probability that A1 occurs and theprobability that A2 occurs, and then subtracting the probabilitythat both A1 and A2 occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

The addition rule (axiom 3) becomes a little more complicated ifwe drop the requirement that the events be mutually exclusive.The following theorem, called the general addition law is similar tothe inclusion-exclusion principle for sets.

Theorem

If A1 and A2 are any two events, then

P(A1 ∪ A2) = P(A1) + P(A2)− P(A1 ∩ A2).

In words, the probability that at leat one of the events A1 and A2

occurs is obtained by adding the probability that A1 occurs and theprobability that A2 occurs, and then subtracting the probabilitythat both A1 and A2 occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

The addition rule (axiom 3) becomes a little more complicated ifwe drop the requirement that the events be mutually exclusive.The following theorem, called the general addition law is similar tothe inclusion-exclusion principle for sets.

Theorem

If A1 and A2 are any two events, then

P(A1 ∪ A2) = P(A1) + P(A2)− P(A1 ∩ A2).

In words, the probability that at leat one of the events A1 and A2

occurs is obtained by adding the probability that A1 occurs and theprobability that A2 occurs, and then subtracting the probabilitythat both A1 and A2 occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Proof. Note that A1 ∪ A2 can be decomposed into the mutuallyexclusive events A1 \ A2 and A2; that is A1 ∪ A2 = (A1 \ A2) ∪ A2.Thus by axiom 3 and our previous theorem,

P(A1 ∪ A2) = P(A1 \ A2) + P(A2)

= P(A1)− P(A1 ∩ A2) + P(A2)

= P(A1) + P(A2)− P(A1 ∩ A2)

which is the desired result.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Applying the above theorem twice we obtain

Corollary

For any events A1, A2, A3,

P(A1 ∪ A2 ∪ A3) = P(A1) + P(A2) + P(A3)− P(A1 ∩ A2)− P(A1 ∩ A3)− P(A2 ∩ A3)

+ P(A1 ∩ A2 ∩ A3).

More generally, given any n events A1,A2, ...An,

P( n∪

k=1Ak

)= P1 − P2 + P3 − P4 + ...± Pn.

(see notes)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Applying the above theorem twice we obtain

Corollary

For any events A1, A2, A3,

P(A1 ∪ A2 ∪ A3) = P(A1) + P(A2) + P(A3)− P(A1 ∩ A2)− P(A1 ∩ A3)− P(A2 ∩ A3)

+ P(A1 ∩ A2 ∩ A3).

More generally, given any n events A1,A2, ...An,

P( n∪

k=1Ak

)= P1 − P2 + P3 − P4 + ...± Pn.

(see notes)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability (cont.)

Applying the above theorem twice we obtain

Corollary

For any events A1, A2, A3,

P(A1 ∪ A2 ∪ A3) = P(A1) + P(A2) + P(A3)− P(A1 ∩ A2)− P(A1 ∩ A3)− P(A2 ∩ A3)

+ P(A1 ∩ A2 ∩ A3).

More generally, given any n events A1,A2, ...An,

P( n∪

k=1Ak

)= P1 − P2 + P3 − P4 + ...± Pn.

(see notes)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples

The following examples illustrate how our formulas can be used tocompute probabilities:

Example

A card is dealt off the top of a well-shuffled ordinary deck of 52cards. There is 1 chance in 4 for it to be a heart. There is 1chance in 4 for it to be a spade.

What is the chance for it to be in a major suit (hearts orspades)?

The question asks for the chance that one of the following twothings will happen:

the card is a heart;

the card is a spade.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples

The following examples illustrate how our formulas can be used tocompute probabilities:

Example

A card is dealt off the top of a well-shuffled ordinary deck of 52cards. There is 1 chance in 4 for it to be a heart. There is 1chance in 4 for it to be a spade.

What is the chance for it to be in a major suit (hearts orspades)?

The question asks for the chance that one of the following twothings will happen:

the card is a heart;

the card is a spade.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples

The following examples illustrate how our formulas can be used tocompute probabilities:

Example

A card is dealt off the top of a well-shuffled ordinary deck of 52cards. There is 1 chance in 4 for it to be a heart. There is 1chance in 4 for it to be a spade.

What is the chance for it to be in a major suit (hearts orspades)?

The question asks for the chance that one of the following twothings will happen:

the card is a heart;

the card is a spade.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples

The following examples illustrate how our formulas can be used tocompute probabilities:

Example

A card is dealt off the top of a well-shuffled ordinary deck of 52cards. There is 1 chance in 4 for it to be a heart. There is 1chance in 4 for it to be a spade.

What is the chance for it to be in a major suit (hearts orspades)?

The question asks for the chance that one of the following twothings will happen:

the card is a heart;

the card is a spade.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Well, if the card is a heart then it cannot be a spade: these aremutually exclusive events.

So it is legitimate to add the chances.

The chance of getting a card in a major suit is 14 + 1

4 = 12 .

(A check on the reasoning: there are 13 hearts and 13 spades, so2652 = 1

2 of the cards in the deck are in a major suit.)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Well, if the card is a heart then it cannot be a spade: these aremutually exclusive events.

So it is legitimate to add the chances.

The chance of getting a card in a major suit is 14 + 1

4 = 12 .

(A check on the reasoning: there are 13 hearts and 13 spades, so2652 = 1

2 of the cards in the deck are in a major suit.)

Sebastian M. Saiegh Probability Theory

Page 126: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Well, if the card is a heart then it cannot be a spade: these aremutually exclusive events.

So it is legitimate to add the chances.

The chance of getting a card in a major suit is 14 + 1

4 = 12 .

(A check on the reasoning: there are 13 hearts and 13 spades, so2652 = 1

2 of the cards in the deck are in a major suit.)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Well, if the card is a heart then it cannot be a spade: these aremutually exclusive events.

So it is legitimate to add the chances.

The chance of getting a card in a major suit is 14 + 1

4 = 12 .

(A check on the reasoning: there are 13 hearts and 13 spades, so2652 = 1

2 of the cards in the deck are in a major suit.)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Three coins are tossed. We assign equal probabilities to the eightsample points in the sample space,Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT.

What is the chance of getting at least one head?

If A is the event “at least one head,” then the complementaryevent A ′ is “no heads.” By theorem 2,

P(A) = 1− P(A ′)

= 1− P(TTT)

= 1− 1

8=

7

8.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Three coins are tossed. We assign equal probabilities to the eightsample points in the sample space,Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT.

What is the chance of getting at least one head?

If A is the event “at least one head,” then the complementaryevent A ′ is “no heads.” By theorem 2,

P(A) = 1− P(A ′)

= 1− P(TTT)

= 1− 1

8=

7

8.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Three coins are tossed. We assign equal probabilities to the eightsample points in the sample space,Ω = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT.

What is the chance of getting at least one head?

If A is the event “at least one head,” then the complementaryevent A ′ is “no heads.” By theorem 2,

P(A) = 1− P(A ′)

= 1− P(TTT)

= 1− 1

8=

7

8.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Someone throws a pair of dice. True or false: The chance ofgetting at least one ace is 1

6 + 16 = 1

3 .

This is false. Imagine one of the dice is white, the other red.

The question asks for the chance that one of the twofollowing things will happen:

the white die lands ace;the red die lands ace.

A white ace does not prevent a red ace. These two events are notmutually exclusive, so the simple addition rule does not apply.Adding the chances gives the wrong answer.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Someone throws a pair of dice. True or false: The chance ofgetting at least one ace is 1

6 + 16 = 1

3 .

This is false. Imagine one of the dice is white, the other red.

The question asks for the chance that one of the twofollowing things will happen:

the white die lands ace;the red die lands ace.

A white ace does not prevent a red ace. These two events are notmutually exclusive, so the simple addition rule does not apply.Adding the chances gives the wrong answer.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Someone throws a pair of dice. True or false: The chance ofgetting at least one ace is 1

6 + 16 = 1

3 .

This is false. Imagine one of the dice is white, the other red.

The question asks for the chance that one of the twofollowing things will happen:

the white die lands ace;the red die lands ace.

A white ace does not prevent a red ace. These two events are notmutually exclusive, so the simple addition rule does not apply.Adding the chances gives the wrong answer.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Example

Someone throws a pair of dice. True or false: The chance ofgetting at least one ace is 1

6 + 16 = 1

3 .

This is false. Imagine one of the dice is white, the other red.

The question asks for the chance that one of the twofollowing things will happen:

the white die lands ace;the red die lands ace.

A white ace does not prevent a red ace. These two events are notmutually exclusive, so the simple addition rule does not apply.Adding the chances gives the wrong answer.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Look at the following figure: there are six ways for the white die toshow an ace; there are six ways for the red die to show an ace; butthe number of ways to get at least one ace is not 6 + 6.

Addition double counts the outcome “snake eyes” at the topleft corner.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

Look at the following figure: there are six ways for the white die toshow an ace; there are six ways for the red die to show an ace; butthe number of ways to get at least one ace is not 6 + 6.

Addition double counts the outcome “snake eyes” at the topleft corner.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

The chance of getting at least one ace is: 6+6−136 = 11

36 , or30.5% (and not 33.3%).

If you want to find the probability that at least one eventoccurs, and the events are not mutually exclusive, do not addthe probabilities; the sum will be too big.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Axioms of Probability: Examples (cont.)

The chance of getting at least one ace is: 6+6−136 = 11

36 , or30.5% (and not 33.3%).

If you want to find the probability that at least one eventoccurs, and the events are not mutually exclusive, do not addthe probabilities; the sum will be too big.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces

As some of our examples show, many times the physicalcharacteristics of an experiment suggest that the various outcomesof the sample space be assigned equal probabilities.

Such a finite probability space Ω, where each sample pointhas the same probability, is called an equiprobable or uniformspace.

In particular, if Ω contains n points then the probability ofeach point is 1

n .

Furthermore, if an event A contains r points then itsprobability is r · 1

n = rn .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces

As some of our examples show, many times the physicalcharacteristics of an experiment suggest that the various outcomesof the sample space be assigned equal probabilities.

Such a finite probability space Ω, where each sample pointhas the same probability, is called an equiprobable or uniformspace.

In particular, if Ω contains n points then the probability ofeach point is 1

n .

Furthermore, if an event A contains r points then itsprobability is r · 1

n = rn .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces

As some of our examples show, many times the physicalcharacteristics of an experiment suggest that the various outcomesof the sample space be assigned equal probabilities.

Such a finite probability space Ω, where each sample pointhas the same probability, is called an equiprobable or uniformspace.

In particular, if Ω contains n points then the probability ofeach point is 1

n .

Furthermore, if an event A contains r points then itsprobability is r · 1

n = rn .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces

As some of our examples show, many times the physicalcharacteristics of an experiment suggest that the various outcomesof the sample space be assigned equal probabilities.

Such a finite probability space Ω, where each sample pointhas the same probability, is called an equiprobable or uniformspace.

In particular, if Ω contains n points then the probability ofeach point is 1

n .

Furthermore, if an event A contains r points then itsprobability is r · 1

n = rn .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces (cont.)

In other words, the probability P(A) of the event A is defined asthe fraction of the outcomes in which A occurs:

P(A) =number of elements in A

number of elements in Ω=

n(A)

n(Ω).

For example, in tossing a die an even number can occur in 3 waysout of 6 “equally likely” ways; hence p = 3

6 = 12 .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces (cont.)

In other words, the probability P(A) of the event A is defined asthe fraction of the outcomes in which A occurs:

P(A) =number of elements in A

number of elements in Ω=

n(A)

n(Ω).

For example, in tossing a die an even number can occur in 3 waysout of 6 “equally likely” ways; hence p = 3

6 = 12 .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces (cont.)

Theorem

Given a finite sample space Ω, for any subset A of Ω, letP(A) = n(A)

n(Ω) . Then, P satisfies the probability axioms presentedabove.

The expression “at random” will be used only with respect to anequiprobable space; formally, the statement “choose a point atrandom from a set Ω” shall mean that Ω is an equiprobable space,i.e. that each sample point in Ω has the same probability.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces (cont.)

Theorem

Given a finite sample space Ω, for any subset A of Ω, letP(A) = n(A)

n(Ω) . Then, P satisfies the probability axioms presentedabove.

The expression “at random” will be used only with respect to anequiprobable space; formally, the statement “choose a point atrandom from a set Ω” shall mean that Ω is an equiprobable space,i.e. that each sample point in Ω has the same probability.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces: Example

Example

A card is dealt off the top of a well-shuffled deck. Let A1 = thecard is a spade, and A2 = the card is a face card (i.e. a jack,queen or king). What is the chance for it to be a face card ofspades?

We compute P(A1), P(A2), and P(A1 ∩ A2). Since we have anequiprobable space,

P(A1) =number of spades

number of cards=

13

52=

1

4,

P(A2) =number of face cards

number of cards=

12

52=

3

13,

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces: Example

Example

A card is dealt off the top of a well-shuffled deck. Let A1 = thecard is a spade, and A2 = the card is a face card (i.e. a jack,queen or king). What is the chance for it to be a face card ofspades?

We compute P(A1), P(A2), and P(A1 ∩ A2). Since we have anequiprobable space,

P(A1) =number of spades

number of cards=

13

52=

1

4,

P(A2) =number of face cards

number of cards=

12

52=

3

13,

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Finite Equiprobable Spaces: Example (cont.)

P(A1 ∩ A2) =number of spade face cards

number of cards=

3

52.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Relative Frequency

Sometimes we do not face situations like the ones described above.However, the accumulated experience of innumerable observationsmay reveal a remarkable regularity of behavior, allowing us toassign a precise meaning to the concept of probability not only inthe case of experiments with equiprobable outcomes, but also inthe most general case.

Suppose the experiment under consideration can be repeated anynumber of times, so that, in principle at leat, we can produce awhole series of “independent trials under identical conditions,” ineach of which, depending on chance, a particular event A ofinterest either occurs or does not occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Relative Frequency

Sometimes we do not face situations like the ones described above.However, the accumulated experience of innumerable observationsmay reveal a remarkable regularity of behavior, allowing us toassign a precise meaning to the concept of probability not only inthe case of experiments with equiprobable outcomes, but also inthe most general case.

Suppose the experiment under consideration can be repeated anynumber of times, so that, in principle at leat, we can produce awhole series of “independent trials under identical conditions,” ineach of which, depending on chance, a particular event A ofinterest either occurs or does not occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Relative Frequency (cont.)

Let n be the total number of experiments in the whole series oftrials, and let n(A) be the number of experiments in which Aoccurs. Then the ratio

n(A)

n

is called the relative frequency of the event A (in the given seriesof trials).

It turns out that the relative frequencies n(A)n observed in different

series of trials are virtually the same for a large n, clustering aboutsome constant

P(A) ∼ n(A)

n,

called the probability of the event A.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Relative Frequency (cont.)

Let n be the total number of experiments in the whole series oftrials, and let n(A) be the number of experiments in which Aoccurs. Then the ratio

n(A)

n

is called the relative frequency of the event A (in the given seriesof trials).

It turns out that the relative frequencies n(A)n observed in different

series of trials are virtually the same for a large n, clustering aboutsome constant

P(A) ∼ n(A)

n,

called the probability of the event A.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Relative Frequency (cont.)

More exactly, the previous equation means that

P(A) = limn→∞

n(A)

n.

Roughly speaking, the probability P(A) of the event A equals thefraction of experiments leading to the occurrence of A in a largeseries of trials.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Relative Frequency (cont.)

More exactly, the previous equation means that

P(A) = limn→∞

n(A)

n.

Roughly speaking, the probability P(A) of the event A equals thefraction of experiments leading to the occurrence of A in a largeseries of trials.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Introduction

The above definition of probability is the “classical” one proposedby Laplace (1749-1827).

What we have to keep in mind is that the rule for computingprobabilities delineated above is applicable only when allsample points have been assigned the same probability.

Thus, the formula does not apply to a wide variety ofimportant problems where it is not reasonable to make thisspecial assignment of probabilities to sample points.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Introduction

The above definition of probability is the “classical” one proposedby Laplace (1749-1827).

What we have to keep in mind is that the rule for computingprobabilities delineated above is applicable only when allsample points have been assigned the same probability.

Thus, the formula does not apply to a wide variety ofimportant problems where it is not reasonable to make thisspecial assignment of probabilities to sample points.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere

In the seventeenth century, French gamblers used to bet on theevent that with 4 rolls of a die, at least one ace would turn up.

In another game, they bet on the event that with 24 rolls of apair of dice, at least one double-ace would turn up.

The Chavalier de Mere, a French nobleman of the period,thought the two events were equally likely.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere

In the seventeenth century, French gamblers used to bet on theevent that with 4 rolls of a die, at least one ace would turn up.

In another game, they bet on the event that with 24 rolls of apair of dice, at least one double-ace would turn up.

The Chavalier de Mere, a French nobleman of the period,thought the two events were equally likely.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere

In the seventeenth century, French gamblers used to bet on theevent that with 4 rolls of a die, at least one ace would turn up.

In another game, they bet on the event that with 24 rolls of apair of dice, at least one double-ace would turn up.

The Chavalier de Mere, a French nobleman of the period,thought the two events were equally likely.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

He reasoned this way about the first game:

In one roll of a die, I have 16 of a chance to get an ace.

So in 4 rolls, I have 4× 16 = 2

3 of a chance to get at least oneace.

His reasoning for the second game was similar:

In one roll of a pair of dice, I have 136 of a chance to get a

double-ace.

So in 24 rolls, I must have 24× 136 = 2

3 of a chance to get atleast one double-ace.

By this argument, both chances were the same, namely 23 . But

experience showed the first event to be a bit more likely than thesecond.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

He reasoned this way about the first game:

In one roll of a die, I have 16 of a chance to get an ace.

So in 4 rolls, I have 4× 16 = 2

3 of a chance to get at least oneace.

His reasoning for the second game was similar:

In one roll of a pair of dice, I have 136 of a chance to get a

double-ace.

So in 24 rolls, I must have 24× 136 = 2

3 of a chance to get atleast one double-ace.

By this argument, both chances were the same, namely 23 . But

experience showed the first event to be a bit more likely than thesecond.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

He reasoned this way about the first game:

In one roll of a die, I have 16 of a chance to get an ace.

So in 4 rolls, I have 4× 16 = 2

3 of a chance to get at least oneace.

His reasoning for the second game was similar:

In one roll of a pair of dice, I have 136 of a chance to get a

double-ace.

So in 24 rolls, I must have 24× 136 = 2

3 of a chance to get atleast one double-ace.

By this argument, both chances were the same, namely 23 . But

experience showed the first event to be a bit more likely than thesecond.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

He reasoned this way about the first game:

In one roll of a die, I have 16 of a chance to get an ace.

So in 4 rolls, I have 4× 16 = 2

3 of a chance to get at least oneace.

His reasoning for the second game was similar:

In one roll of a pair of dice, I have 136 of a chance to get a

double-ace.

So in 24 rolls, I must have 24× 136 = 2

3 of a chance to get atleast one double-ace.

By this argument, both chances were the same, namely 23 . But

experience showed the first event to be a bit more likely than thesecond.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

He reasoned this way about the first game:

In one roll of a die, I have 16 of a chance to get an ace.

So in 4 rolls, I have 4× 16 = 2

3 of a chance to get at least oneace.

His reasoning for the second game was similar:

In one roll of a pair of dice, I have 136 of a chance to get a

double-ace.

So in 24 rolls, I must have 24× 136 = 2

3 of a chance to get atleast one double-ace.

By this argument, both chances were the same, namely 23 . But

experience showed the first event to be a bit more likely than thesecond.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

He reasoned this way about the first game:

In one roll of a die, I have 16 of a chance to get an ace.

So in 4 rolls, I have 4× 16 = 2

3 of a chance to get at least oneace.

His reasoning for the second game was similar:

In one roll of a pair of dice, I have 136 of a chance to get a

double-ace.

So in 24 rolls, I must have 24× 136 = 2

3 of a chance to get atleast one double-ace.

By this argument, both chances were the same, namely 23 . But

experience showed the first event to be a bit more likely than thesecond.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

This contradiction became to be known as the Paradox of theChavalier de Mere.

De Mere asked the philosopher Blaise Pascal about theproblem, and Pascal solved it with the help of his friend,Pierre de Fermat.

The solution rested on the fact that the outcomes listed by deMere are not actually equiprobable.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

This contradiction became to be known as the Paradox of theChavalier de Mere.

De Mere asked the philosopher Blaise Pascal about theproblem, and Pascal solved it with the help of his friend,Pierre de Fermat.

The solution rested on the fact that the outcomes listed by deMere are not actually equiprobable.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

This contradiction became to be known as the Paradox of theChavalier de Mere.

De Mere asked the philosopher Blaise Pascal about theproblem, and Pascal solved it with the help of his friend,Pierre de Fermat.

The solution rested on the fact that the outcomes listed by deMere are not actually equiprobable.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Let’s look at the first game first.

The chance of winning is hard to compute, so let’s work outthe chance of the opposite event –losing.

The gambler loses when none of the four rolls shows an ace.With one roll, the chance is 5

6 .

With two rolls, the chance that the first roll doesn’t give anace and the second doesn’t give an ace equals 5

6 ×56 = ( 5

6 )2.

By the same logic, the chance of losing when the die is rolled fortimes is ( 5

6 )4, or 48.2 percent. So the chance of winning is100%-48.2%=51.8%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Let’s look at the first game first.

The chance of winning is hard to compute, so let’s work outthe chance of the opposite event –losing.

The gambler loses when none of the four rolls shows an ace.With one roll, the chance is 5

6 .

With two rolls, the chance that the first roll doesn’t give anace and the second doesn’t give an ace equals 5

6 ×56 = ( 5

6 )2.

By the same logic, the chance of losing when the die is rolled fortimes is ( 5

6 )4, or 48.2 percent. So the chance of winning is100%-48.2%=51.8%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Let’s look at the first game first.

The chance of winning is hard to compute, so let’s work outthe chance of the opposite event –losing.

The gambler loses when none of the four rolls shows an ace.With one roll, the chance is 5

6 .

With two rolls, the chance that the first roll doesn’t give anace and the second doesn’t give an ace equals 5

6 ×56 = ( 5

6 )2.

By the same logic, the chance of losing when the die is rolled fortimes is ( 5

6 )4, or 48.2 percent. So the chance of winning is100%-48.2%=51.8%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Let’s look at the first game first.

The chance of winning is hard to compute, so let’s work outthe chance of the opposite event –losing.

The gambler loses when none of the four rolls shows an ace.With one roll, the chance is 5

6 .

With two rolls, the chance that the first roll doesn’t give anace and the second doesn’t give an ace equals 5

6 ×56 = ( 5

6 )2.

By the same logic, the chance of losing when the die is rolled fortimes is ( 5

6 )4, or 48.2 percent. So the chance of winning is100%-48.2%=51.8%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Let’s look at the first game first.

The chance of winning is hard to compute, so let’s work outthe chance of the opposite event –losing.

The gambler loses when none of the four rolls shows an ace.With one roll, the chance is 5

6 .

With two rolls, the chance that the first roll doesn’t give anace and the second doesn’t give an ace equals 5

6 ×56 = ( 5

6 )2.

By the same logic, the chance of losing when the die is rolled fortimes is ( 5

6 )4, or 48.2 percent. So the chance of winning is100%-48.2%=51.8%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Now what about the second game?

In one roll of a pair of dice, there is 1 chance in 36 of getting adouble-ace, and 35 chances in 36 of not getting a double-ace.

In 24 rolls of a pair of dice, the chance of getting nodouble-aces must be ( 35

36 )24.

That is about 50.9%. So the chance of winning is a little bit lessthan 50%, 100%-50.9%=49.1%. That’s why you win the secondgame a bit less frequently than the first.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Now what about the second game?

In one roll of a pair of dice, there is 1 chance in 36 of getting adouble-ace, and 35 chances in 36 of not getting a double-ace.

In 24 rolls of a pair of dice, the chance of getting nodouble-aces must be ( 35

36 )24.

That is about 50.9%. So the chance of winning is a little bit lessthan 50%, 100%-50.9%=49.1%. That’s why you win the secondgame a bit less frequently than the first.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Now what about the second game?

In one roll of a pair of dice, there is 1 chance in 36 of getting adouble-ace, and 35 chances in 36 of not getting a double-ace.

In 24 rolls of a pair of dice, the chance of getting nodouble-aces must be ( 35

36 )24.

That is about 50.9%. So the chance of winning is a little bit lessthan 50%, 100%-50.9%=49.1%. That’s why you win the secondgame a bit less frequently than the first.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

Now what about the second game?

In one roll of a pair of dice, there is 1 chance in 36 of getting adouble-ace, and 35 chances in 36 of not getting a double-ace.

In 24 rolls of a pair of dice, the chance of getting nodouble-aces must be ( 35

36 )24.

That is about 50.9%. So the chance of winning is a little bit lessthan 50%, 100%-50.9%=49.1%. That’s why you win the secondgame a bit less frequently than the first.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

As this example shows, in order to compute the probability of anevent P(A) we must often determine first which elements of thesample space are in A, and then add the probabilities of thecorresponding simple events.

Moreover, let Ω be a finite sample space; say,Ω = ω1, ω2, ..., ωn. Then, there are 2n different subsets of Ωand since each subset is an event, there are 2n different events.

It is therefore extremely useful to learn some effectivetechniques for counting the elements in sets specified bydefining properties.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

As this example shows, in order to compute the probability of anevent P(A) we must often determine first which elements of thesample space are in A, and then add the probabilities of thecorresponding simple events.

Moreover, let Ω be a finite sample space; say,Ω = ω1, ω2, ..., ωn. Then, there are 2n different subsets of Ωand since each subset is an event, there are 2n different events.

It is therefore extremely useful to learn some effectivetechniques for counting the elements in sets specified bydefining properties.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Example: The Paradox of the Chevalier de Mere (cont.)

As this example shows, in order to compute the probability of anevent P(A) we must often determine first which elements of thesample space are in A, and then add the probabilities of thecorresponding simple events.

Moreover, let Ω be a finite sample space; say,Ω = ω1, ω2, ..., ωn. Then, there are 2n different subsets of Ωand since each subset is an event, there are 2n different events.

It is therefore extremely useful to learn some effectivetechniques for counting the elements in sets specified bydefining properties.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Fundamental Principle of Counting

Suppose a license plate contains two distinct letter followed bythree digits with the first digit not zero.

How many different license plates can be printed?

The first letter can be printed in 26 different ways, the secondletter in 25 different ways (since the letter printed first cannotbe chosen for the second letter), the first digit in 9 ways andeach of the other two digits in 10 ways.

Hence

26× 25× 9× 10× 10 = 585, 000

different plates can be printed.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Fundamental Principle of Counting

Suppose a license plate contains two distinct letter followed bythree digits with the first digit not zero.

How many different license plates can be printed?

The first letter can be printed in 26 different ways, the secondletter in 25 different ways (since the letter printed first cannotbe chosen for the second letter), the first digit in 9 ways andeach of the other two digits in 10 ways.

Hence

26× 25× 9× 10× 10 = 585, 000

different plates can be printed.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Fundamental Principle of Counting

Suppose a license plate contains two distinct letter followed bythree digits with the first digit not zero.

How many different license plates can be printed?

The first letter can be printed in 26 different ways, the secondletter in 25 different ways (since the letter printed first cannotbe chosen for the second letter), the first digit in 9 ways andeach of the other two digits in 10 ways.

Hence

26× 25× 9× 10× 10 = 585, 000

different plates can be printed.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Fundamental Principle of Counting

Suppose a license plate contains two distinct letter followed bythree digits with the first digit not zero.

How many different license plates can be printed?

The first letter can be printed in 26 different ways, the secondletter in 25 different ways (since the letter printed first cannotbe chosen for the second letter), the first digit in 9 ways andeach of the other two digits in 10 ways.

Hence

26× 25× 9× 10× 10 = 585, 000

different plates can be printed.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Fundamental Principle of Counting (cont.)

If some procedure can be performed in n1 different ways, and if,following this procedure, a second procedure can be performed inn2 different ways, and if, following this second procedure, a thirdprocedure can be performed in n3 different ways, and so forth;then the number of ways the procedures can be performed in theorder indicated is the product n1 × n2 × n3....

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Factorial Notation

The product of the positive integers from 1 to n inclusive occursvery often in mathematics and hence is denoted by the specialsymbol n! (read “n factorial”):

n! = 1× 2× 3× ...× (n − 2)(n − 1)n

In words, we use the exclamation mark (!) to indicate the result ofmultiplying together a number and all the numbers which comebefore it. So, for example: 4! = 4× 3× 2× 1 = 24 ( and is read“four-factorial equal twenty-four.”)It is also convenient to define 1! = 1, and 0! = 1 (it is aconvention of mathematics that zero-factorial equals 1).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Factorial Notation

The product of the positive integers from 1 to n inclusive occursvery often in mathematics and hence is denoted by the specialsymbol n! (read “n factorial”):

n! = 1× 2× 3× ...× (n − 2)(n − 1)n

In words, we use the exclamation mark (!) to indicate the result ofmultiplying together a number and all the numbers which comebefore it. So, for example: 4! = 4× 3× 2× 1 = 24 ( and is read“four-factorial equal twenty-four.”)It is also convenient to define 1! = 1, and 0! = 1 (it is aconvention of mathematics that zero-factorial equals 1).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Factorial Notation

The product of the positive integers from 1 to n inclusive occursvery often in mathematics and hence is denoted by the specialsymbol n! (read “n factorial”):

n! = 1× 2× 3× ...× (n − 2)(n − 1)n

In words, we use the exclamation mark (!) to indicate the result ofmultiplying together a number and all the numbers which comebefore it. So, for example: 4! = 4× 3× 2× 1 = 24 ( and is read“four-factorial equal twenty-four.”)It is also convenient to define 1! = 1, and 0! = 1 (it is aconvention of mathematics that zero-factorial equals 1).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations

An arrangement of a set of n objects in a given order is called apermutation of the objects (taken all at a time).

An arrangement of any r ≤ n of these objects in a given orderis called an r-permutation or a permutation of the n objectstaken r at a time.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations

An arrangement of a set of n objects in a given order is called apermutation of the objects (taken all at a time).

An arrangement of any r ≤ n of these objects in a given orderis called an r-permutation or a permutation of the n objectstaken r at a time.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Example

Consider the set of letters a,b,c, and d. Then:

(i) bdca, dcba, and acdb are permutations of the 4 letters (takenall at a time);

(ii) bad, adb, cbd, and bca are permutations of the 4 letters taken3 at a time;

(iii) ad,cb,da, and bd are permutations of the 4 letters taken 2 ata time.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

The number of permutations of n objects taken r at a time will bedenoted by

P(n, r)

Before we derive the general formula for P(n, r) we consider aspecial case:

Suppose we want to find the number of permutations of 6objects, say a,b,c,d,e,f, taken three at a time. In other words,we want to find the number of “three letter words” withdistinct letters that can be formed from the above six letters.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

The number of permutations of n objects taken r at a time will bedenoted by

P(n, r)

Before we derive the general formula for P(n, r) we consider aspecial case:

Suppose we want to find the number of permutations of 6objects, say a,b,c,d,e,f, taken three at a time. In other words,we want to find the number of “three letter words” withdistinct letters that can be formed from the above six letters.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

The number of permutations of n objects taken r at a time will bedenoted by

P(n, r)

Before we derive the general formula for P(n, r) we consider aspecial case:

Suppose we want to find the number of permutations of 6objects, say a,b,c,d,e,f, taken three at a time. In other words,we want to find the number of “three letter words” withdistinct letters that can be formed from the above six letters.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Let the general three letter word be represented by three boxes:

Now the first letter can be chosen in 6 different ways; followingthis, the second letter can be chosen in 5 different ways; and,following this, the last letter can be chosen in 4 different ways.

If we write each number in its appropriate box, they shouldlook like this:

6 5 4

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Let the general three letter word be represented by three boxes:

Now the first letter can be chosen in 6 different ways; followingthis, the second letter can be chosen in 5 different ways; and,following this, the last letter can be chosen in 4 different ways.

If we write each number in its appropriate box, they shouldlook like this:

6 5 4

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Let the general three letter word be represented by three boxes:

Now the first letter can be chosen in 6 different ways; followingthis, the second letter can be chosen in 5 different ways; and,following this, the last letter can be chosen in 4 different ways.

If we write each number in its appropriate box, they shouldlook like this:

6 5 4

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

By the fundamental principle of counting there are 6× 5× 4 = 120possible three letter words without repetitions from the six letters,or there are 120 permutations of 6 objects taken 3 at a time. Thatis,

P(6, 3) = 120

The derivation of the formula for P(n, r) follows the procedure inthe preceding example.

The first element in an r -permutation of n-objects can bechosen in n different ways; following this, the second elementin the permutation can be chosen in n− 1 ways; and followingthis, the third element in the permutation can be chosen inn − 2 ways.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

By the fundamental principle of counting there are 6× 5× 4 = 120possible three letter words without repetitions from the six letters,or there are 120 permutations of 6 objects taken 3 at a time. Thatis,

P(6, 3) = 120

The derivation of the formula for P(n, r) follows the procedure inthe preceding example.

The first element in an r -permutation of n-objects can bechosen in n different ways; following this, the second elementin the permutation can be chosen in n− 1 ways; and followingthis, the third element in the permutation can be chosen inn − 2 ways.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

By the fundamental principle of counting there are 6× 5× 4 = 120possible three letter words without repetitions from the six letters,or there are 120 permutations of 6 objects taken 3 at a time. Thatis,

P(6, 3) = 120

The derivation of the formula for P(n, r) follows the procedure inthe preceding example.

The first element in an r -permutation of n-objects can bechosen in n different ways; following this, the second elementin the permutation can be chosen in n− 1 ways; and followingthis, the third element in the permutation can be chosen inn − 2 ways.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Continuing in this manner, we have that the r th (last) element inthe permutation can be chosen in n − (r − 1) = n − r + 1 ways.

Theorem

P(n, r) = n(n − 1)(n − 2) · · · (n − r + 1) = n!(n−r)!

The second part of the formula follows from the fact that

P(n, r) =

=n(n − 1)(n − 2) · · · (n − r + 1)× (n − r)!

(n − r)!

=n!

(n − r)!

So, P(6, 3) = 6!3! = 720

6 = 120.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Continuing in this manner, we have that the r th (last) element inthe permutation can be chosen in n − (r − 1) = n − r + 1 ways.

Theorem

P(n, r) = n(n − 1)(n − 2) · · · (n − r + 1) = n!(n−r)!

The second part of the formula follows from the fact that

P(n, r) =

=n(n − 1)(n − 2) · · · (n − r + 1)× (n − r)!

(n − r)!

=n!

(n − r)!

So, P(6, 3) = 6!3! = 720

6 = 120.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

Continuing in this manner, we have that the r th (last) element inthe permutation can be chosen in n − (r − 1) = n − r + 1 ways.

Theorem

P(n, r) = n(n − 1)(n − 2) · · · (n − r + 1) = n!(n−r)!

The second part of the formula follows from the fact that

P(n, r) =

=n(n − 1)(n − 2) · · · (n − r + 1)× (n − r)!

(n − r)!

=n!

(n − r)!

So, P(6, 3) = 6!3! = 720

6 = 120.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

In the special case that r = n, we haveP(n, n) = n(n − 1)(n − 2) · · · 3× 2× 1 = n!

Namely, there are n! permutations of n objects (taken all at atime).

For example, if we have 3 objects, say a, b, c , there are 3!permutations: abc, acb, bac, bca, cab, cba.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations (cont.)

In the special case that r = n, we haveP(n, n) = n(n − 1)(n − 2) · · · 3× 2× 1 = n!

Namely, there are n! permutations of n objects (taken all at atime).

For example, if we have 3 objects, say a, b, c , there are 3!permutations: abc, acb, bac, bca, cab, cba.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions

Suppose we want to know the number of permutations of objectssome of which are alike.

For example, say an urn contains one red ball and nine greenones. Five draws are made at random. Each time a ball isdrawn, we put it back into the box.

We interested in the situation where two red balls (no more and noless) are drawn. One way this can happen is that the first twodraws are red an the final three are green.

With R for red and G for green, this possibility can be written

R R G G G

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions

Suppose we want to know the number of permutations of objectssome of which are alike.

For example, say an urn contains one red ball and nine greenones. Five draws are made at random. Each time a ball isdrawn, we put it back into the box.

We interested in the situation where two red balls (no more and noless) are drawn. One way this can happen is that the first twodraws are red an the final three are green.

With R for red and G for green, this possibility can be written

R R G G G

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions

Suppose we want to know the number of permutations of objectssome of which are alike.

For example, say an urn contains one red ball and nine greenones. Five draws are made at random. Each time a ball isdrawn, we put it back into the box.

We interested in the situation where two red balls (no more and noless) are drawn. One way this can happen is that the first twodraws are red an the final three are green.

With R for red and G for green, this possibility can be written

R R G G G

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions

Suppose we want to know the number of permutations of objectssome of which are alike.

For example, say an urn contains one red ball and nine greenones. Five draws are made at random. Each time a ball isdrawn, we put it back into the box.

We interested in the situation where two red balls (no more and noless) are drawn. One way this can happen is that the first twodraws are red an the final three are green.

With R for red and G for green, this possibility can be written

R R G G G

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

Of course, there are many other ways to get two reds. Forexample, the second and the fifth draws might be red, while all therest are green:

G R G G R

To figure out all the possible ways in which two red balls aredrawn, we must first calculate the chance of each possible way,and then use the addition rule to add up the chances.

The chance of the pattern R R G G G is:

1

10× 1

10× 9

10× 9

10× 9

10=( 1

10

)2( 9

10

)3

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

Of course, there are many other ways to get two reds. Forexample, the second and the fifth draws might be red, while all therest are green:

G R G G R

To figure out all the possible ways in which two red balls aredrawn, we must first calculate the chance of each possible way,and then use the addition rule to add up the chances.

The chance of the pattern R R G G G is:

1

10× 1

10× 9

10× 9

10× 9

10=( 1

10

)2( 9

10

)3

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

Of course, there are many other ways to get two reds. Forexample, the second and the fifth draws might be red, while all therest are green:

G R G G R

To figure out all the possible ways in which two red balls aredrawn, we must first calculate the chance of each possible way,and then use the addition rule to add up the chances.

The chance of the pattern R R G G G is:

1

10× 1

10× 9

10× 9

10× 9

10=( 1

10

)2( 9

10

)3

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

Similarly, the chance of the pattern G R G G R equals

9

10× 1

10× 9

10× 9

10× 1

10=( 1

10

)2( 9

10

)3

The pattern G R G G R has the same chance as the pattern R R GG G. In fact, each pattern with 2 reds and 3 greens has the same

chance,(

110

)2(9

10

)3, since the 2 reds will contribute

(1

10

)2to the

product and the 3 greens will contribute(

910

)3.

The sum of the chances of all the patterns, therefore, equalsthe number of patterns times the common chance.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

Similarly, the chance of the pattern G R G G R equals

9

10× 1

10× 9

10× 9

10× 1

10=( 1

10

)2( 9

10

)3

The pattern G R G G R has the same chance as the pattern R R GG G. In fact, each pattern with 2 reds and 3 greens has the same

chance,(

110

)2(9

10

)3, since the 2 reds will contribute

(1

10

)2to the

product and the 3 greens will contribute(

910

)3.

The sum of the chances of all the patterns, therefore, equalsthe number of patterns times the common chance.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

Similarly, the chance of the pattern G R G G R equals

9

10× 1

10× 9

10× 9

10× 1

10=( 1

10

)2( 9

10

)3

The pattern G R G G R has the same chance as the pattern R R GG G. In fact, each pattern with 2 reds and 3 greens has the same

chance,(

110

)2(9

10

)3, since the 2 reds will contribute

(1

10

)2to the

product and the 3 greens will contribute(

910

)3.

The sum of the chances of all the patterns, therefore, equalsthe number of patterns times the common chance.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

How many patterns are there? Each pattern is specified by writingdown in a row 2 R’s and 3 G’s, in some order. The number ofpatterns is given by:

5!

2!3!= 10

In words, this formula gives us the number of different ways ofarranging 2 R’s and 3 G’s in a row. The 5 in the numerator of theformula is the sum of 2 and 3 in the denominator.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

How many patterns are there? Each pattern is specified by writingdown in a row 2 R’s and 3 G’s, in some order. The number ofpatterns is given by:

5!

2!3!= 10

In words, this formula gives us the number of different ways ofarranging 2 R’s and 3 G’s in a row. The 5 in the numerator of theformula is the sum of 2 and 3 in the denominator.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

For example, the number of ways to arrange four R’s and one G ina row is:

5!

4!1!= 5

The patterns are:

R R R R G, R R R G R, R R G R R, R G R R R, G R R R R

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

For example, the number of ways to arrange four R’s and one G ina row is:

5!

4!1!= 5

The patterns are:

R R R R G, R R R G R, R R G R R, R G R R R, G R R R R

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

For example, the number of ways to arrange four R’s and one G ina row is:

5!

4!1!= 5

The patterns are:

R R R R G, R R R G R, R R G R R, R G R R R, G R R R R

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

The general formula follows:

Theorem

The number of permutations of n objects of which n1 are alike, n2

are alike, ..., nr are alike is

n!

n1!n2! · · · nr !

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Permutations with Repetitions (cont.)

The general formula follows:

Theorem

The number of permutations of n objects of which n1 are alike, n2

are alike, ..., nr are alike is

n!

n1!n2! · · · nr !

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Ordered Samples

Our last example is representative of many other problems incombinatorial analysis.

In particular, a lot of problems in probability are concernedwith choosing a ball from an urn containing n balls (or a cardfrom a deck, or a person from a population).

When we choose one ball after another from the urn, say rtimes, we call the choice an ordered sample of size r .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Ordered Samples

Our last example is representative of many other problems incombinatorial analysis.

In particular, a lot of problems in probability are concernedwith choosing a ball from an urn containing n balls (or a cardfrom a deck, or a person from a population).

When we choose one ball after another from the urn, say rtimes, we call the choice an ordered sample of size r .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Ordered Samples

Our last example is representative of many other problems incombinatorial analysis.

In particular, a lot of problems in probability are concernedwith choosing a ball from an urn containing n balls (or a cardfrom a deck, or a person from a population).

When we choose one ball after another from the urn, say rtimes, we call the choice an ordered sample of size r .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Ordered Samples (cont.)

We consider two cases:

(i) Sampling with replacement. Here the ball is replaced in theurn before the next ball is chosen. Now since there are ndifferent ways to choose each ball, there are by thefundamental principle of counting

r times︷ ︸︸ ︷n × n × n...n = nr

different ordered samples with replacement of size r .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Ordered Samples (cont.)

We consider two cases:

(i) Sampling with replacement. Here the ball is replaced in theurn before the next ball is chosen. Now since there are ndifferent ways to choose each ball, there are by thefundamental principle of counting

r times︷ ︸︸ ︷n × n × n...n = nr

different ordered samples with replacement of size r .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Ordered Samples (cont.)

(ii) Sampling without replacement. Here the ball is not replacedin the urn before the next ball is chosen. In other words, anordered sample of size r without replacement is simply anr -permutation of the objects in the urn. Thus there are

P(n, r) = n(n − 1)(n − 2) · · · (n − r + 1) =n!

(n − r)!

different ordered samples of size r without replacement from apopulation of n objects.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula

Let’s go back to the example of the urn containing one red balland nine green ones. Five draws were made at random withreplacement.

We learned that there are 10 different patterns in which 2draws are red:

5!

2!3!= 10

An expression of this form is called a binomial coefficient.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula

Let’s go back to the example of the urn containing one red balland nine green ones. Five draws were made at random withreplacement.

We learned that there are 10 different patterns in which 2draws are red:

5!

2!3!= 10

An expression of this form is called a binomial coefficient.

Sebastian M. Saiegh Probability Theory

Page 233: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula

Let’s go back to the example of the urn containing one red balland nine green ones. Five draws were made at random withreplacement.

We learned that there are 10 different patterns in which 2draws are red:

5!

2!3!= 10

An expression of this form is called a binomial coefficient.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We will usually write the binomial coefficient as follows:(n

r

)=

n!

r !(n − r)!

The symbol(nr

)reads as “n choose r ,” the idea being that the

formula gives the number of ways to choose r things out of n.

Some books write the binomial coefficient as “nCr” or Cnr ,

the “number of combinations of n things taken r at a time.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We will usually write the binomial coefficient as follows:(n

r

)=

n!

r !(n − r)!

The symbol(nr

)reads as “n choose r ,” the idea being that the

formula gives the number of ways to choose r things out of n.

Some books write the binomial coefficient as “nCr” or Cnr ,

the “number of combinations of n things taken r at a time.”

Sebastian M. Saiegh Probability Theory

Page 236: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We will usually write the binomial coefficient as follows:(n

r

)=

n!

r !(n − r)!

The symbol(nr

)reads as “n choose r ,” the idea being that the

formula gives the number of ways to choose r things out of n.

Some books write the binomial coefficient as “nCr” or Cnr ,

the “number of combinations of n things taken r at a time.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We are finally ready to answer the following question.

Given our example of the red and green balls: What is theprobability that exactly two draws will be red?

We know that the chance of red is 110 , and the chance of

green is 910 .

We also know that there are 10 different patterns with 2 R’sand 3 G’s.

So the probability of drawing exactly 2 reds is:

10×( 1

10

)2( 9

10

)3≈ 7%

Sebastian M. Saiegh Probability Theory

Page 238: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We are finally ready to answer the following question.

Given our example of the red and green balls: What is theprobability that exactly two draws will be red?

We know that the chance of red is 110 , and the chance of

green is 910 .

We also know that there are 10 different patterns with 2 R’sand 3 G’s.

So the probability of drawing exactly 2 reds is:

10×( 1

10

)2( 9

10

)3≈ 7%

Sebastian M. Saiegh Probability Theory

Page 239: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We are finally ready to answer the following question.

Given our example of the red and green balls: What is theprobability that exactly two draws will be red?

We know that the chance of red is 110 , and the chance of

green is 910 .

We also know that there are 10 different patterns with 2 R’sand 3 G’s.

So the probability of drawing exactly 2 reds is:

10×( 1

10

)2( 9

10

)3≈ 7%

Sebastian M. Saiegh Probability Theory

Page 240: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We are finally ready to answer the following question.

Given our example of the red and green balls: What is theprobability that exactly two draws will be red?

We know that the chance of red is 110 , and the chance of

green is 910 .

We also know that there are 10 different patterns with 2 R’sand 3 G’s.

So the probability of drawing exactly 2 reds is:

10×( 1

10

)2( 9

10

)3≈ 7%

Sebastian M. Saiegh Probability Theory

Page 241: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

We are finally ready to answer the following question.

Given our example of the red and green balls: What is theprobability that exactly two draws will be red?

We know that the chance of red is 110 , and the chance of

green is 910 .

We also know that there are 10 different patterns with 2 R’sand 3 G’s.

So the probability of drawing exactly 2 reds is:

10×( 1

10

)2( 9

10

)3≈ 7%

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The reasoning of this example is summarized in the binomialformula.

Suppose a chance process is carried out as a sequence oftrails; an example would be rolling a die 10 times, where eachroll counts as a trial.

There is an event of interest which may or may not occur ateach trial; the die may or may not land ace.

The problem is to calculate the probability that the event willoccur a specified number of times.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The reasoning of this example is summarized in the binomialformula.

Suppose a chance process is carried out as a sequence oftrails; an example would be rolling a die 10 times, where eachroll counts as a trial.

There is an event of interest which may or may not occur ateach trial; the die may or may not land ace.

The problem is to calculate the probability that the event willoccur a specified number of times.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The reasoning of this example is summarized in the binomialformula.

Suppose a chance process is carried out as a sequence oftrails; an example would be rolling a die 10 times, where eachroll counts as a trial.

There is an event of interest which may or may not occur ateach trial; the die may or may not land ace.

The problem is to calculate the probability that the event willoccur a specified number of times.

Sebastian M. Saiegh Probability Theory

Page 245: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The reasoning of this example is summarized in the binomialformula.

Suppose a chance process is carried out as a sequence oftrails; an example would be rolling a die 10 times, where eachroll counts as a trial.

There is an event of interest which may or may not occur ateach trial; the die may or may not land ace.

The problem is to calculate the probability that the event willoccur a specified number of times.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The Binomial Formula. The chance that an event will occurexactly r times out of n is given by the binomial formula

n!

r !(n − r)!pr (1− p)n−r .

In this formula, n is the number of trials, r is the number of timesthe event is to occur, and p is the probability that the event willoccur on any particular trial. The assumptions:

The value of n must be fixed in advance.

p must be the same from trial to trial.

The trials must be independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The Binomial Formula. The chance that an event will occurexactly r times out of n is given by the binomial formula

n!

r !(n − r)!pr (1− p)n−r .

In this formula, n is the number of trials, r is the number of timesthe event is to occur, and p is the probability that the event willoccur on any particular trial. The assumptions:

The value of n must be fixed in advance.

p must be the same from trial to trial.

The trials must be independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

The Binomial Formula. The chance that an event will occurexactly r times out of n is given by the binomial formula

n!

r !(n − r)!pr (1− p)n−r .

In this formula, n is the number of trials, r is the number of timesthe event is to occur, and p is the probability that the event willoccur on any particular trial. The assumptions:

The value of n must be fixed in advance.

p must be the same from trial to trial.

The trials must be independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

Example

A die is rolled 10 times. What is the chance of getting exactly 2aces?

The number of trials is fixed in advance. It is 10. So n = 10. Theevent of interest is rolling an ace. The probability of rolling an aceis the same from trial to trial. It is 1

6 . So, p = 16 . The trials are

independent.

The binomial formula can be used, and the answer is:

10!

2!8!

(1

6

)2(5

6

)8≈ 29%.

Sebastian M. Saiegh Probability Theory

Page 250: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

Example

A die is rolled 10 times. What is the chance of getting exactly 2aces?

The number of trials is fixed in advance. It is 10. So n = 10. Theevent of interest is rolling an ace. The probability of rolling an aceis the same from trial to trial. It is 1

6 . So, p = 16 . The trials are

independent.

The binomial formula can be used, and the answer is:

10!

2!8!

(1

6

)2(5

6

)8≈ 29%.

Sebastian M. Saiegh Probability Theory

Page 251: Probability Theory - POLI 270 - Mathematical and ...pages.ucsd.edu/~ssaiegh/Slides7.pdf · Probability Theory POLI 270 - Mathematical and Statistical Foundations Sebastian M. Saiegh

ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

The Binomial Coefficient and Formula (cont.)

Example

A die is rolled 10 times. What is the chance of getting exactly 2aces?

The number of trials is fixed in advance. It is 10. So n = 10. Theevent of interest is rolling an ace. The probability of rolling an aceis the same from trial to trial. It is 1

6 . So, p = 16 . The trials are

independent.

The binomial formula can be used, and the answer is:

10!

2!8!

(1

6

)2(5

6

)8≈ 29%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem

We seek the probability p that n people have distinct birthdays. Insolving this problem we ignore leap years and assume that aperson’s birthday can fall on any day with the same probability.

Since there are n people and 365 different days, there are365n ways in which the n people can have their birthdays. Onthe other hand, if the n persons are to have distinct birthdays,then the first person can be born on any of the 365 days, thesecond person can be born on the remaining 364 days, thethird person can be born on the remaining 363 days, etc.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem

We seek the probability p that n people have distinct birthdays. Insolving this problem we ignore leap years and assume that aperson’s birthday can fall on any day with the same probability.

Since there are n people and 365 different days, there are365n ways in which the n people can have their birthdays. Onthe other hand, if the n persons are to have distinct birthdays,then the first person can be born on any of the 365 days, thesecond person can be born on the remaining 364 days, thethird person can be born on the remaining 363 days, etc.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem (cont.)

Thus there are 365× 364× 363 · · · (365− n + 1) ways in which then persons can have distinct birthdays.

Accordingly, p = 365×364×363···(365−n+1)365n

p =365

365× 364

365× 363

365· · · 365− n + 1

365

And the probability that two or more people have the samebirthday is:

1− 365× 364× 363 · · · (365− n + 1)

365n

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem (cont.)

Thus there are 365× 364× 363 · · · (365− n + 1) ways in which then persons can have distinct birthdays.

Accordingly, p = 365×364×363···(365−n+1)365n

p =365

365× 364

365× 363

365· · · 365− n + 1

365

And the probability that two or more people have the samebirthday is:

1− 365× 364× 363 · · · (365− n + 1)

365n

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem (cont.)

Thus there are 365× 364× 363 · · · (365− n + 1) ways in which then persons can have distinct birthdays.

Accordingly, p = 365×364×363···(365−n+1)365n

p =365

365× 364

365× 363

365· · · 365− n + 1

365

And the probability that two or more people have the samebirthday is:

1− 365× 364× 363 · · · (365− n + 1)

365n

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem (cont.)

The following table shows the value of p for different values of n:

When n = 22 the probability p = 0.524, and when n = 23 theprobability is p = 0.493.

Thus, amongst 23 or more people it is more likely that atleast two of them have the same birthday than that they allhave distinct birthdays.

In a group of 60 or more people, the probability that two ormore of them have the same birthday exceeds 99%.

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem (cont.)

The following table shows the value of p for different values of n:

When n = 22 the probability p = 0.524, and when n = 23 theprobability is p = 0.493.

Thus, amongst 23 or more people it is more likely that atleast two of them have the same birthday than that they allhave distinct birthdays.

In a group of 60 or more people, the probability that two ormore of them have the same birthday exceeds 99%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Sample Space and EventsProbability of an EventCounting Techniques

Bonus Track: Classical Birthday Problem (cont.)

The following table shows the value of p for different values of n:

When n = 22 the probability p = 0.524, and when n = 23 theprobability is p = 0.493.

Thus, amongst 23 or more people it is more likely that atleast two of them have the same birthday than that they allhave distinct birthdays.

In a group of 60 or more people, the probability that two ormore of them have the same birthday exceeds 99%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Introduction

Sometimes when we observe two outcomes we may beinterested in how the outcome of one event E is influenced bythat of another event F .

For example, in one extreme case the relation between E andF may be such that E always occur if F does, while in theother case, E never occurs if F does.

To characterize the relation between E and F , I will introduce theconditional probability of E on the hypothesis F , i.e., the“probability of E occurring under the condition that F is known tohave occurred.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Introduction

Sometimes when we observe two outcomes we may beinterested in how the outcome of one event E is influenced bythat of another event F .

For example, in one extreme case the relation between E andF may be such that E always occur if F does, while in theother case, E never occurs if F does.

To characterize the relation between E and F , I will introduce theconditional probability of E on the hypothesis F , i.e., the“probability of E occurring under the condition that F is known tohave occurred.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Introduction

Sometimes when we observe two outcomes we may beinterested in how the outcome of one event E is influenced bythat of another event F .

For example, in one extreme case the relation between E andF may be such that E always occur if F does, while in theother case, E never occurs if F does.

To characterize the relation between E and F , I will introduce theconditional probability of E on the hypothesis F , i.e., the“probability of E occurring under the condition that F is known tohave occurred.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

It will be helpful first to take a look at an example in which we cansee how this works on intuitive grounds.

Example

A deck of cards is shuffled and the top two cards are put on a table,face down. You win $1 if the second card is the queen of hearts.

(a) What is your chance of winning the dollar?

(b) You turn over the first card. It is the seven of clubs. Nowwhat is your chance of winning?

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

It will be helpful first to take a look at an example in which we cansee how this works on intuitive grounds.

Example

A deck of cards is shuffled and the top two cards are put on a table,face down. You win $1 if the second card is the queen of hearts.

(a) What is your chance of winning the dollar?

(b) You turn over the first card. It is the seven of clubs. Nowwhat is your chance of winning?

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Solution. Part (a). The bet is about the second card, not the first.For example,

If the first card is the two of spades and the second is thequeen of hearts, you win.

If the first card is the seven of clubs and the second is theking of hearts, you lose.

The bet can be settled without even looking at the first card: thesecond card is all you need to know. The chance of winning is 1

52 .A well-shuffled deck brings the cards into random order. Thequeen of hearts has to wind up somewhere. There are 52 possiblepositions, and they are all equally likely.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Solution. Part (a). The bet is about the second card, not the first.For example,

If the first card is the two of spades and the second is thequeen of hearts, you win.

If the first card is the seven of clubs and the second is theking of hearts, you lose.

The bet can be settled without even looking at the first card: thesecond card is all you need to know. The chance of winning is 1

52 .A well-shuffled deck brings the cards into random order. Thequeen of hearts has to wind up somewhere. There are 52 possiblepositions, and they are all equally likely.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Solution. Part (a). The bet is about the second card, not the first.For example,

If the first card is the two of spades and the second is thequeen of hearts, you win.

If the first card is the seven of clubs and the second is theking of hearts, you lose.

The bet can be settled without even looking at the first card: thesecond card is all you need to know. The chance of winning is 1

52 .A well-shuffled deck brings the cards into random order. Thequeen of hearts has to wind up somewhere. There are 52 possiblepositions, and they are all equally likely.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Part (b). There are 51 cards left. They are in random order, andthe queen of hearts is one of them. She has 1 chance in 51 to be onthe table. Your chance goes up a little, to 1

51 . That is the answer.

The 151 in part (b) is called a conditional chance. The

problem puts a condition on the first card: it has to be theseven of clubs.

We will refer to this as the conditional probability that thesecond card is the queen of hearts given the first card is theseven of clubs.

To emphasize the contrast, the 152 in part (a) is called an

unconditional chance: the problem puts no conditions on thefirst card.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Part (b). There are 51 cards left. They are in random order, andthe queen of hearts is one of them. She has 1 chance in 51 to be onthe table. Your chance goes up a little, to 1

51 . That is the answer.

The 151 in part (b) is called a conditional chance. The

problem puts a condition on the first card: it has to be theseven of clubs.

We will refer to this as the conditional probability that thesecond card is the queen of hearts given the first card is theseven of clubs.

To emphasize the contrast, the 152 in part (a) is called an

unconditional chance: the problem puts no conditions on thefirst card.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Part (b). There are 51 cards left. They are in random order, andthe queen of hearts is one of them. She has 1 chance in 51 to be onthe table. Your chance goes up a little, to 1

51 . That is the answer.

The 151 in part (b) is called a conditional chance. The

problem puts a condition on the first card: it has to be theseven of clubs.

We will refer to this as the conditional probability that thesecond card is the queen of hearts given the first card is theseven of clubs.

To emphasize the contrast, the 152 in part (a) is called an

unconditional chance: the problem puts no conditions on thefirst card.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example (cont.)

Part (b). There are 51 cards left. They are in random order, andthe queen of hearts is one of them. She has 1 chance in 51 to be onthe table. Your chance goes up a little, to 1

51 . That is the answer.

The 151 in part (b) is called a conditional chance. The

problem puts a condition on the first card: it has to be theseven of clubs.

We will refer to this as the conditional probability that thesecond card is the queen of hearts given the first card is theseven of clubs.

To emphasize the contrast, the 152 in part (a) is called an

unconditional chance: the problem puts no conditions on thefirst card.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition

This informal and intuitive reasoning can be described in anotherway. In general, given a sample space Ω and an acceptableassignment of probabilities to the sample points in Ω, we computethe probability of an event E by adding the probabilities of thesample points whose union is E .

Since P(Ω) = 1 and E ∩ Ω = E , we can write the identity

P(E ) =P(E ∩ Ω)

P(Ω),

which shows that P(E ) is the ratio of the probability of thatpart of E included in Ω (which happens to be all of E ) to theprobability of Ω itself (which happens to be 1).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition

This informal and intuitive reasoning can be described in anotherway. In general, given a sample space Ω and an acceptableassignment of probabilities to the sample points in Ω, we computethe probability of an event E by adding the probabilities of thesample points whose union is E .

Since P(Ω) = 1 and E ∩ Ω = E , we can write the identity

P(E ) =P(E ∩ Ω)

P(Ω),

which shows that P(E ) is the ratio of the probability of thatpart of E included in Ω (which happens to be all of E ) to theprobability of Ω itself (which happens to be 1).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition (cont.)

But if we are told that event F has occurred, then the outcomescorresponding to elements of F ′, the complement of F , are nolonger possible.

Therefore, in light of our added information, the event Freplaces the sample space Ω as the set whose elementscorrespond to all possible outcomes.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition (cont.)

But if we are told that event F has occurred, then the outcomescorresponding to elements of F ′, the complement of F , are nolonger possible.

Therefore, in light of our added information, the event Freplaces the sample space Ω as the set whose elementscorrespond to all possible outcomes.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition (cont.)

With this in mind, observe how reasonable it appears to write

P(E |F ) =P(E ∩ F )

P(F ),

which says that P(E |F ), the conditional probability of E given F(the vertical bar is read “given”), is the ratio of the probability ofthat part of E included in F (which is E ∩ F ) to the probability ofF itself.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition (cont.)

Definition

Let F be an arbitrary event in a sample space Ω with P(F ) > 0.The probability that an event E occurs once F has occurred or, inother words, the conditional probability of E given F , writtenP(E |F ), is defined as follows:

P(E |F ) =P(E ∩ F )

P(F ).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition (cont.)

If Ω is a finite equiprobable space and |E | denotes the number ofelements in an event E , then

P(E ∩ F ) =|E ∩ F ||Ω|

, P(F ) =|F ||Ω|

and so

P(E |F ) =P(E ∩ F )

P(F )=|E ∩ F ||F |

.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Definition (cont.)

If Ω is a finite equiprobable space and |E | denotes the number ofelements in an event E , then

P(E ∩ F ) =|E ∩ F ||Ω|

, P(F ) =|F ||Ω|

and so

P(E |F ) =P(E ∩ F )

P(F )=|E ∩ F ||F |

.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Three fair coins are tossed, one after the other. Let E be the event“at least two heads” and F be the event “first coin falls heads.”

The sample space for this experiment isΩ = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT.Thus, the event E is the union of four sample points. Theevent F is also the union of four sample points. And, E ∩ F isthe union of three sample points.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Three fair coins are tossed, one after the other. Let E be the event“at least two heads” and F be the event “first coin falls heads.”

The sample space for this experiment isΩ = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT.Thus, the event E is the union of four sample points. Theevent F is also the union of four sample points. And, E ∩ F isthe union of three sample points.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Three fair coins are tossed, one after the other. Let E be the event“at least two heads” and F be the event “first coin falls heads.”

The sample space for this experiment isΩ = HHH,HHT ,HTH,THH,HTT ,THT ,TTH,TTT.Thus, the event E is the union of four sample points. Theevent F is also the union of four sample points. And, E ∩ F isthe union of three sample points.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Therefore, the conditional probability of E given F is

P(E |F ) =P(E ∩ F )

P(F )=

3848

=3

4.

Note that the added knowledge that the first coin falls headsincreases the probability of getting at least two heads. Before thisadditional information is revealed, P(E ) = 1

2 . Afterwards,P(E |F ) = 3

4 .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Therefore, the conditional probability of E given F is

P(E |F ) =P(E ∩ F )

P(F )=

3848

=3

4.

Note that the added knowledge that the first coin falls headsincreases the probability of getting at least two heads. Before thisadditional information is revealed, P(E ) = 1

2 . Afterwards,P(E |F ) = 3

4 .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability

Theorem

Let Ω be a finite equiprobable space with events E and F . Then

P(E |F ) =number of elements in E ∩ F

number of elements in F.

or

P(E |F ) =number of ways E and F can occur

number of ways F can occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Let a pair of dice be tossed. If the sum is 6, find the probabilitythat one of the dice is a 2.

We can restate this problem as F = sum is6=(1, 5), (2, 4), (3, 3), (4, 2), (5, 1). And, E = a 2 appearson at least one die, find P(E |F ).

Now F consists of five elements and two of them, (2, 4) and(4, 2) belong to E : E ∩ F = (2, 4), (4, 2). ThenP(E |F ) = 2

5 ≈ 0.4.

As before, the added knowledge that the sum of the dice is 6increases the probability that one of the dice is a 2. Beforethis additional information is revealed, P(E ) = 11

36 ≈ 0.3.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Let a pair of dice be tossed. If the sum is 6, find the probabilitythat one of the dice is a 2.

We can restate this problem as F = sum is6=(1, 5), (2, 4), (3, 3), (4, 2), (5, 1). And, E = a 2 appearson at least one die, find P(E |F ).

Now F consists of five elements and two of them, (2, 4) and(4, 2) belong to E : E ∩ F = (2, 4), (4, 2). ThenP(E |F ) = 2

5 ≈ 0.4.

As before, the added knowledge that the sum of the dice is 6increases the probability that one of the dice is a 2. Beforethis additional information is revealed, P(E ) = 11

36 ≈ 0.3.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Let a pair of dice be tossed. If the sum is 6, find the probabilitythat one of the dice is a 2.

We can restate this problem as F = sum is6=(1, 5), (2, 4), (3, 3), (4, 2), (5, 1). And, E = a 2 appearson at least one die, find P(E |F ).

Now F consists of five elements and two of them, (2, 4) and(4, 2) belong to E : E ∩ F = (2, 4), (4, 2). ThenP(E |F ) = 2

5 ≈ 0.4.

As before, the added knowledge that the sum of the dice is 6increases the probability that one of the dice is a 2. Beforethis additional information is revealed, P(E ) = 11

36 ≈ 0.3.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Conditional Probability: Example

Example

Let a pair of dice be tossed. If the sum is 6, find the probabilitythat one of the dice is a 2.

We can restate this problem as F = sum is6=(1, 5), (2, 4), (3, 3), (4, 2), (5, 1). And, E = a 2 appearson at least one die, find P(E |F ).

Now F consists of five elements and two of them, (2, 4) and(4, 2) belong to E : E ∩ F = (2, 4), (4, 2). ThenP(E |F ) = 2

5 ≈ 0.4.

As before, the added knowledge that the sum of the dice is 6increases the probability that one of the dice is a 2. Beforethis additional information is revealed, P(E ) = 11

36 ≈ 0.3.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities

Given two events E and F , we already learned how to calculateP(E |F ).

Notice now that if P(E ) > 0, then the roles of E and F canbe interchanged.

The conditional probability of F given E is

P(F |E ) =P(F ∩ E )

P(E )=

P(E ∩ F )

P(E ),

the last equality following from the commutative law for theintersection of two sets.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities

Given two events E and F , we already learned how to calculateP(E |F ).

Notice now that if P(E ) > 0, then the roles of E and F canbe interchanged.

The conditional probability of F given E is

P(F |E ) =P(F ∩ E )

P(E )=

P(E ∩ F )

P(E ),

the last equality following from the commutative law for theintersection of two sets.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities

Given two events E and F , we already learned how to calculateP(E |F ).

Notice now that if P(E ) > 0, then the roles of E and F canbe interchanged.

The conditional probability of F given E is

P(F |E ) =P(F ∩ E )

P(E )=

P(E ∩ F )

P(E ),

the last equality following from the commutative law for theintersection of two sets.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities (cont.)

If we cross-multiply the above equation defining conditionalprobability, we obtain the following useful formula:

P(E ∩ F ) = P(E )P(F |E ) = P(F )P(E |F ).

In words, the chance that two things will both happen equals thechance that the first will happen, multiplied by the chance that thesecond will happen given that the first has happened.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities (cont.)

If we cross-multiply the above equation defining conditionalprobability, we obtain the following useful formula:

P(E ∩ F ) = P(E )P(F |E ) = P(F )P(E |F ).

In words, the chance that two things will both happen equals thechance that the first will happen, multiplied by the chance that thesecond will happen given that the first has happened.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities (cont.)

The last result can be extended by induction as follows:

Theorem

If n is any integer (n ≥ 2) and E1,E2, ...,En are any n events forwhich P(E1 ∩ E2... ∩ En−1) 6= 0, then P(E1 ∩ E2... ∩ En) can bewritten as

P(E1)P(E2|E1)P(E3|E1 ∩ E2)...P(En|E1 ∩ E2... ∩ En−1).

This theorem is sometimes referred to as the theorem oncompound probabilities.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities (cont.)

The last result can be extended by induction as follows:

Theorem

If n is any integer (n ≥ 2) and E1,E2, ...,En are any n events forwhich P(E1 ∩ E2... ∩ En−1) 6= 0, then P(E1 ∩ E2... ∩ En) can bewritten as

P(E1)P(E2|E1)P(E3|E1 ∩ E2)...P(En|E1 ∩ E2... ∩ En−1).

This theorem is sometimes referred to as the theorem oncompound probabilities.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities (cont.)

The last result can be extended by induction as follows:

Theorem

If n is any integer (n ≥ 2) and E1,E2, ...,En are any n events forwhich P(E1 ∩ E2... ∩ En−1) 6= 0, then P(E1 ∩ E2... ∩ En) can bewritten as

P(E1)P(E2|E1)P(E3|E1 ∩ E2)...P(En|E1 ∩ E2... ∩ En−1).

This theorem is sometimes referred to as the theorem oncompound probabilities.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples

Example

Two cards will be dealt off the top of a well-shuffled deck. What isthe chance that the first card will be the seven of clubs and thesecond card will be the queen of hearts?

This is very similar to the first example. The chance that the firstcard will be the seven of clubs is 1

52 . Given that the first card wasthe seven of clubs, the chance that the second card will be thequeen of hearts is 1

51 . The chance of getting both cards is,

1

52× 1

51=

1

2652.

This is a very small chance: about 4 in 10,000, or 0.04 of 1%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples

Example

Two cards will be dealt off the top of a well-shuffled deck. What isthe chance that the first card will be the seven of clubs and thesecond card will be the queen of hearts?

This is very similar to the first example. The chance that the firstcard will be the seven of clubs is 1

52 . Given that the first card wasthe seven of clubs, the chance that the second card will be thequeen of hearts is 1

51 . The chance of getting both cards is,

1

52× 1

51=

1

2652.

This is a very small chance: about 4 in 10,000, or 0.04 of 1%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples

Example

Two cards will be dealt off the top of a well-shuffled deck. What isthe chance that the first card will be the seven of clubs and thesecond card will be the queen of hearts?

This is very similar to the first example. The chance that the firstcard will be the seven of clubs is 1

52 . Given that the first card wasthe seven of clubs, the chance that the second card will be thequeen of hearts is 1

51 . The chance of getting both cards is,

1

52× 1

51=

1

2652.

This is a very small chance: about 4 in 10,000, or 0.04 of 1%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples (cont.)

Example

A deck of cards is shuffled, and two cards are dealt. What is thechance that both are aces?

The chance that the first card is an ace equals 452 . Given that the

first card is an ace, there are 3 aces among the 51 remaining cards:so the chance of a second ace equals 3

51 . The chance that bothcards are aces equals,

4

52× 3

51=

12

2652.

This is about 1 in 200, or 0.5 of 1%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples (cont.)

Example

A deck of cards is shuffled, and two cards are dealt. What is thechance that both are aces?

The chance that the first card is an ace equals 452 . Given that the

first card is an ace, there are 3 aces among the 51 remaining cards:so the chance of a second ace equals 3

51 . The chance that bothcards are aces equals,

4

52× 3

51=

12

2652.

This is about 1 in 200, or 0.5 of 1%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples (cont.)

Example

A deck of cards is shuffled, and two cards are dealt. What is thechance that both are aces?

The chance that the first card is an ace equals 452 . Given that the

first card is an ace, there are 3 aces among the 51 remaining cards:so the chance of a second ace equals 3

51 . The chance that bothcards are aces equals,

4

52× 3

51=

12

2652.

This is about 1 in 200, or 0.5 of 1%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples (cont.)

Example

A lot contains 12 items of which 4 are defective. Three items aredrawn at random from the lot one after the other. Find theprobability p that all three are non-defective.

The probability that the first item is non-defective is 812 . If the first

item is non-defective, then the probability that the next item isnon-defective is 7

11 since only 7 of the remaining 11 items are nondefective. If the first two items are non-defective, then theprobability that the last items is non-defective is 6

10 since only 6 ofthe remaining 10 items are now non-defective.

Thus by the theorem on compound probabilities,

p =8

12× 7

11× 6

10=

14

55≈ .25.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples (cont.)

Example

A lot contains 12 items of which 4 are defective. Three items aredrawn at random from the lot one after the other. Find theprobability p that all three are non-defective.

The probability that the first item is non-defective is 812 . If the first

item is non-defective, then the probability that the next item isnon-defective is 7

11 since only 7 of the remaining 11 items are nondefective. If the first two items are non-defective, then theprobability that the last items is non-defective is 6

10 since only 6 ofthe remaining 10 items are now non-defective.

Thus by the theorem on compound probabilities,

p =8

12× 7

11× 6

10=

14

55≈ .25.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Compound Probabilities: Examples (cont.)

Example

A lot contains 12 items of which 4 are defective. Three items aredrawn at random from the lot one after the other. Find theprobability p that all three are non-defective.

The probability that the first item is non-defective is 812 . If the first

item is non-defective, then the probability that the next item isnon-defective is 7

11 since only 7 of the remaining 11 items are nondefective. If the first two items are non-defective, then theprobability that the last items is non-defective is 6

10 since only 6 ofthe remaining 10 items are now non-defective.

Thus by the theorem on compound probabilities,

p =8

12× 7

11× 6

10=

14

55≈ .25.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams

A (finite) sequence of experiments in which each experiment has afinite number of outcomes with given probabilities is called a(finite) stochastic process.

A convenient way of describing such a process and computingthe probability of any event is by a tree diagram.

Then, the theorem on compound probabilities is used tocompute the probability that the result represented by anygiven path of the tree does occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams

A (finite) sequence of experiments in which each experiment has afinite number of outcomes with given probabilities is called a(finite) stochastic process.

A convenient way of describing such a process and computingthe probability of any event is by a tree diagram.

Then, the theorem on compound probabilities is used tocompute the probability that the result represented by anygiven path of the tree does occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams

A (finite) sequence of experiments in which each experiment has afinite number of outcomes with given probabilities is called a(finite) stochastic process.

A convenient way of describing such a process and computingthe probability of any event is by a tree diagram.

Then, the theorem on compound probabilities is used tocompute the probability that the result represented by anygiven path of the tree does occur.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams (cont.)

Example

We are given three boxes as follow:

Box I has 10 light bulbs of which 4 are defective.

Box II has 6 light bulbs of which 1 is defective.

Box III has 8 light bulbs of which 3 are defective.

We select a box at random and then draw a bulb at random. Whatis the probability p that the bulb is defective?

Here we perform a sequence of two experiments:

(i) select one of the three boxes;(ii) select a bulb which is either defective (D) or non-defective (N).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams (cont.)

Example

We are given three boxes as follow:

Box I has 10 light bulbs of which 4 are defective.

Box II has 6 light bulbs of which 1 is defective.

Box III has 8 light bulbs of which 3 are defective.

We select a box at random and then draw a bulb at random. Whatis the probability p that the bulb is defective?

Here we perform a sequence of two experiments:

(i) select one of the three boxes;(ii) select a bulb which is either defective (D) or non-defective (N).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams (cont.)

Example

We are given three boxes as follow:

Box I has 10 light bulbs of which 4 are defective.

Box II has 6 light bulbs of which 1 is defective.

Box III has 8 light bulbs of which 3 are defective.

We select a box at random and then draw a bulb at random. Whatis the probability p that the bulb is defective?

Here we perform a sequence of two experiments:

(i) select one of the three boxes;(ii) select a bulb which is either defective (D) or non-defective (N).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams (cont.)

The following tree diagram describes this process and gives theprobability of each branch of the tree:

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams (cont.)

The probability that any particular path of the tree occurs is, bythe theorem of compound probabilities, the product of theprobabilities of each branch of the path, e.g., the probability ofselecting box I and then a defective bulb is 1

3 ×21 = 2

15 .

Now since there are three mutually exclusive paths which leadto a defective bulb, the sum of the probabilities of these pathsis the required probability:

p =1

3× 2

5+

1

3× 1

6+

1

3× 3

8=

113

360≈ .31.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Finite Stochastic Processes and Tree Diagrams (cont.)

The probability that any particular path of the tree occurs is, bythe theorem of compound probabilities, the product of theprobabilities of each branch of the path, e.g., the probability ofselecting box I and then a defective bulb is 1

3 ×21 = 2

15 .

Now since there are three mutually exclusive paths which leadto a defective bulb, the sum of the probabilities of these pathsis the required probability:

p =1

3× 2

5+

1

3× 1

6+

1

3× 3

8=

113

360≈ .31.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions

Suppose a set Ω is the union of mutually disjoint subsetsE1,E2, ...,En. Furthermore, suppose E is any other subset of Ω.Then, as illustrated in the following figure for the case n = 3,

E = Ω ∩ E = (E1 ∪ E2 ∪ ... ∪ En) ∩ E

= (E1 ∩ E ) ∪ (E2 ∩ E ) ∪ ... ∪ (En ∩ E )

where the n subsets in the second line of the equation are alsomutually disjoint.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions

Suppose a set Ω is the union of mutually disjoint subsetsE1,E2, ...,En. Furthermore, suppose E is any other subset of Ω.Then, as illustrated in the following figure for the case n = 3,

E = Ω ∩ E = (E1 ∪ E2 ∪ ... ∪ En) ∩ E

= (E1 ∩ E ) ∪ (E2 ∩ E ) ∪ ... ∪ (En ∩ E )

where the n subsets in the second line of the equation are alsomutually disjoint.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions (cont.)

Accordingly, the probability of the event E ,

P(E ) = P(E1 ∩ E ) + P(E2 ∩ E ) + ...+ P(En ∩ E ).

And, by the theorem of compound probabilities,

P(E ) = P(E1)P(E |E1) + P(E2)P(E |E2) + ...+ P(En)P(E |En). (1)

(This is called the Law of Total Probability)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions (cont.)

Accordingly, the probability of the event E ,

P(E ) = P(E1 ∩ E ) + P(E2 ∩ E ) + ...+ P(En ∩ E ).

And, by the theorem of compound probabilities,

P(E ) = P(E1)P(E |E1) + P(E2)P(E |E2) + ...+ P(En)P(E |En). (1)

(This is called the Law of Total Probability)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions (cont.)

Accordingly, the probability of the event E ,

P(E ) = P(E1 ∩ E ) + P(E2 ∩ E ) + ...+ P(En ∩ E ).

And, by the theorem of compound probabilities,

P(E ) = P(E1)P(E |E1) + P(E2)P(E |E2) + ...+ P(En)P(E |En). (1)

(This is called the Law of Total Probability)

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions: Example

Example

Find the probability that in a well-shuffled deck of cards, the ace ofspades is next to the king of spades.

We can solve this problem in the following way. Consider thefollowing three events:

the ace of spades is the top card of the deck;

the ace of spades is the bottom card of the deck;

the ace of spades is somewhere within the deck.

Let E1,E2, and E3 denote these events in the order stated. Wechoose as sample space Ω the set of ordered 52-tuples denoting allpossible orderings of the 52-card deck.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions: Example

Example

Find the probability that in a well-shuffled deck of cards, the ace ofspades is next to the king of spades.

We can solve this problem in the following way. Consider thefollowing three events:

the ace of spades is the top card of the deck;

the ace of spades is the bottom card of the deck;

the ace of spades is somewhere within the deck.

Let E1,E2, and E3 denote these events in the order stated. Wechoose as sample space Ω the set of ordered 52-tuples denoting allpossible orderings of the 52-card deck.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions: Example

Example

Find the probability that in a well-shuffled deck of cards, the ace ofspades is next to the king of spades.

We can solve this problem in the following way. Consider thefollowing three events:

the ace of spades is the top card of the deck;

the ace of spades is the bottom card of the deck;

the ace of spades is somewhere within the deck.

Let E1,E2, and E3 denote these events in the order stated. Wechoose as sample space Ω the set of ordered 52-tuples denoting allpossible orderings of the 52-card deck.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions: Example (cont.)

Then E1,E2,E3 is a partition of Ω. If E denotes the event thatthe ace and king of spades are neighboring cards, then noting thatonly one card is next to the top or bottom cards but two cards arenext to a card within the deck, we find

P(E1) = P(E2) =1

52, P(E3) =

50

52=

25

26,

P(E |E1) = P(E |E2) =1

51, P(E |E3) =

2

51.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions: Example (cont.)

Using the preceding formula,

P(E ) =1

52× 1

51+

1

52× 1

51+

25

26× 2

51=

1

26≈ .038.

This is a small chance: about 4 in 100, or 0.04.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Partitions: Example (cont.)

Using the preceding formula,

P(E ) =1

52× 1

51+

1

52× 1

51+

25

26× 2

51=

1

26≈ .038.

This is a small chance: about 4 in 100, or 0.04.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem

From the preceding formula we are also a short step to Bayes’theorem.

Note that for any i , the conditional probability of Ei given Eis defined by

P(Ei |E ) =P(Ei ∩ E )

P(E ).

The final step is to use equation (1) to replace P(E ) and useP(Ei ∩ E ) = P(Ei )P(E |Ei ) to replace P(Ei ∩ E )

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem

From the preceding formula we are also a short step to Bayes’theorem.

Note that for any i , the conditional probability of Ei given Eis defined by

P(Ei |E ) =P(Ei ∩ E )

P(E ).

The final step is to use equation (1) to replace P(E ) and useP(Ei ∩ E ) = P(Ei )P(E |Ei ) to replace P(Ei ∩ E )

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem

From the preceding formula we are also a short step to Bayes’theorem.

Note that for any i , the conditional probability of Ei given Eis defined by

P(Ei |E ) =P(Ei ∩ E )

P(E ).

The final step is to use equation (1) to replace P(E ) and useP(Ei ∩ E ) = P(Ei )P(E |Ei ) to replace P(Ei ∩ E )

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem (cont.)

Theorem

Bayes’ Theorem. Suppose E1,E2, ...,En is a partition of Ω and Eis any event. Then for any i ,

P(Ei |E ) =P(Ei )P(E |Ei )

P(E1)P(E |E1) + P(E2)P(E |E2) + ...+ P(En)P(E |En).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Example

Example

Three machines A, B and C produce respectively 50%, 30% and20% of the total number of items of a factory. The percentages ofdefective output of these machines are 3%, 4% and 5%. Supposean item is selected at random and is found to be defective. Whatis the probability that the item was produced by machine A?

Let X be the event that an item is defective. Then, we want tofind P(A|X ).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Example

Example

Three machines A, B and C produce respectively 50%, 30% and20% of the total number of items of a factory. The percentages ofdefective output of these machines are 3%, 4% and 5%. Supposean item is selected at random and is found to be defective. Whatis the probability that the item was produced by machine A?

Let X be the event that an item is defective. Then, we want tofind P(A|X ).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Example (cont.)

By Bayes’ theorem,

P(A|X ) =P(A)P(X |A)

P(X )P(X |A) + P(B)P(X |B) + P(C )P(X |C )

=(.5)(.03)

(.5)(.03) + (.3)(.04) + (.2)(.05)

=15

37≈ .4

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Example (cont.)

By Bayes’ theorem,

P(A|X ) =P(A)P(X |A)

P(X )P(X |A) + P(B)P(X |B) + P(C )P(X |C )

=(.5)(.03)

(.5)(.03) + (.3)(.04) + (.2)(.05)

=15

37≈ .4

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Terminology

A few words on the terminology often used when Bayes’ theorem isapplied:

The events E1,E2, ...,En are called hypothesis, and they areassumed to be disjoint and exhaustive.

The probability P(Ek) is called the a priori probability ofhypothesis Ek .

The conditional probability P(Ek |E ) is called the a posterioriprobability of the hypothesis Ek given the observed event E .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Terminology

A few words on the terminology often used when Bayes’ theorem isapplied:

The events E1,E2, ...,En are called hypothesis, and they areassumed to be disjoint and exhaustive.

The probability P(Ek) is called the a priori probability ofhypothesis Ek .

The conditional probability P(Ek |E ) is called the a posterioriprobability of the hypothesis Ek given the observed event E .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Terminology

A few words on the terminology often used when Bayes’ theorem isapplied:

The events E1,E2, ...,En are called hypothesis, and they areassumed to be disjoint and exhaustive.

The probability P(Ek) is called the a priori probability ofhypothesis Ek .

The conditional probability P(Ek |E ) is called the a posterioriprobability of the hypothesis Ek given the observed event E .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Terminology

A few words on the terminology often used when Bayes’ theorem isapplied:

The events E1,E2, ...,En are called hypothesis, and they areassumed to be disjoint and exhaustive.

The probability P(Ek) is called the a priori probability ofhypothesis Ek .

The conditional probability P(Ek |E ) is called the a posterioriprobability of the hypothesis Ek given the observed event E .

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Relevance

This is Thomas Bayes’ great contribution to “modern” thinking:

We can make better-informed decisions by mathematicallyblending new information into old information.

In particular, suppose that in a given situation the data aregiven and the probability is the unknown.

Questions put in this manner form the subject matter of whatis known as inverse probability.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Relevance

This is Thomas Bayes’ great contribution to “modern” thinking:

We can make better-informed decisions by mathematicallyblending new information into old information.

In particular, suppose that in a given situation the data aregiven and the probability is the unknown.

Questions put in this manner form the subject matter of whatis known as inverse probability.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Relevance

This is Thomas Bayes’ great contribution to “modern” thinking:

We can make better-informed decisions by mathematicallyblending new information into old information.

In particular, suppose that in a given situation the data aregiven and the probability is the unknown.

Questions put in this manner form the subject matter of whatis known as inverse probability.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Bayes’ Theorem: Relevance

This is Thomas Bayes’ great contribution to “modern” thinking:

We can make better-informed decisions by mathematicallyblending new information into old information.

In particular, suppose that in a given situation the data aregiven and the probability is the unknown.

Questions put in this manner form the subject matter of whatis known as inverse probability.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Using Bayes’ Rule to calculate Conditional Probability

Let E and F be two events that may be related; assume thatP(F ) > 0.

Suppose that the only information we have is P(E ), P(F |E ),and P(F |E ′).

In this case, we need to calculate P(E ∩ F ), and P(F ), whichwere not given directly as data to us.

We can use Bayes’ theorem to calculate the conditionalprobability P(E |F ) in terms of P(E ), P(F |E ), and P(F |E ′):

P(E |F ) =P(E )P(F |E )

P(E )P(F |E ) + P(E ′)Pr(F |E ′).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Using Bayes’ Rule to calculate Conditional Probability

Let E and F be two events that may be related; assume thatP(F ) > 0.

Suppose that the only information we have is P(E ), P(F |E ),and P(F |E ′).

In this case, we need to calculate P(E ∩ F ), and P(F ), whichwere not given directly as data to us.

We can use Bayes’ theorem to calculate the conditionalprobability P(E |F ) in terms of P(E ), P(F |E ), and P(F |E ′):

P(E |F ) =P(E )P(F |E )

P(E )P(F |E ) + P(E ′)Pr(F |E ′).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Using Bayes’ Rule to calculate Conditional Probability

Let E and F be two events that may be related; assume thatP(F ) > 0.

Suppose that the only information we have is P(E ), P(F |E ),and P(F |E ′).

In this case, we need to calculate P(E ∩ F ), and P(F ), whichwere not given directly as data to us.

We can use Bayes’ theorem to calculate the conditionalprobability P(E |F ) in terms of P(E ), P(F |E ), and P(F |E ′):

P(E |F ) =P(E )P(F |E )

P(E )P(F |E ) + P(E ′)Pr(F |E ′).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Using Bayes’ Rule to calculate Conditional Probability

Let E and F be two events that may be related; assume thatP(F ) > 0.

Suppose that the only information we have is P(E ), P(F |E ),and P(F |E ′).

In this case, we need to calculate P(E ∩ F ), and P(F ), whichwere not given directly as data to us.

We can use Bayes’ theorem to calculate the conditionalprobability P(E |F ) in terms of P(E ), P(F |E ), and P(F |E ′):

P(E |F ) =P(E )P(F |E )

P(E )P(F |E ) + P(E ′)Pr(F |E ′).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0.1, and Pr(Sunny) = 0.5.

But, before leaving your house you hear on the news thefollowing forecast: “there will be some precipitation.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0.1, and Pr(Sunny) = 0.5.

But, before leaving your house you hear on the news thefollowing forecast: “there will be some precipitation.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0.1, and Pr(Sunny) = 0.5.

But, before leaving your house you hear on the news thefollowing forecast: “there will be some precipitation.”

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I (cont.)

You can now update your priors via Bayes’ rule:

Pr(R|P) =Pr(Rain)Pr(Precip.|Rain)

Pr(Rain)Pr(Precip.|Rain) + Pr(Snow)Pr(Precip.|Snow).

which yields,

Pr(Rain|Precip.) =0.4

0.5= 0.8.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I (cont.)

You can now update your priors via Bayes’ rule:

Pr(R|P) =Pr(Rain)Pr(Precip.|Rain)

Pr(Rain)Pr(Precip.|Rain) + Pr(Snow)Pr(Precip.|Snow).

which yields,

Pr(Rain|Precip.) =0.4

0.5= 0.8.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I (cont.)

You can now update your priors via Bayes’ rule:

Pr(R|P) =Pr(Rain)Pr(Precip.|Rain)

Pr(Rain)Pr(Precip.|Rain) + Pr(Snow)Pr(Precip.|Snow).

which yields,

Pr(Rain|Precip.) =0.4

0.5= 0.8.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I (cont.)

And, the updated probability of Snow is calculated as:

Pr(Snow |Precip.) =0.1

0.5= 0.2.

Notice that Bayes’ rule is just a re-normalization, so that theupdated probabilities add up to 1.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I (cont.)

And, the updated probability of Snow is calculated as:

Pr(Snow |Precip.) =0.1

0.5= 0.2.

Notice that Bayes’ rule is just a re-normalization, so that theupdated probabilities add up to 1.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather I (cont.)

And, the updated probability of Snow is calculated as:

Pr(Snow |Precip.) =0.1

0.5= 0.2.

Notice that Bayes’ rule is just a re-normalization, so that theupdated probabilities add up to 1.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather II

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0, and Pr(Sunny) = 0.6.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) =0.4

0.4= 1.

Pr(Snow |Precip.) =0

0.4= 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather II

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0, and Pr(Sunny) = 0.6.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) =0.4

0.4= 1.

Pr(Snow |Precip.) =0

0.4= 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather II

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0, and Pr(Sunny) = 0.6.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) =0.4

0.4= 1.

Pr(Snow |Precip.) =0

0.4= 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather II

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0, and Pr(Sunny) = 0.6.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) =0.4

0.4= 1.

Pr(Snow |Precip.) =0

0.4= 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather II

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0.4, Pr(Snow) = 0, and Pr(Sunny) = 0.6.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) =0.4

0.4= 1.

Pr(Snow |Precip.) =0

0.4= 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather III

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0, Pr(Snow) = 0, and Pr(Sunny) = 1.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) 6= defined and Pr(Snow |Precip.) 6= defined .

Bayes’ rule does not work when the conditioning event hasprobability 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather III

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0, Pr(Snow) = 0, and Pr(Sunny) = 1.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) 6= defined and Pr(Snow |Precip.) 6= defined .

Bayes’ rule does not work when the conditioning event hasprobability 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather III

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0, Pr(Snow) = 0, and Pr(Sunny) = 1.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) 6= defined and Pr(Snow |Precip.) 6= defined .

Bayes’ rule does not work when the conditioning event hasprobability 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather III

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0, Pr(Snow) = 0, and Pr(Sunny) = 1.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) 6= defined and Pr(Snow |Precip.) 6= defined .

Bayes’ rule does not work when the conditioning event hasprobability 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Example: Deducing the Weather III

Example

Deducing the weather from a rather uninformative weather report.

Suppose Ω = Rain, Snow, SunnyYou hold the following priors:

Pr(Rain) = 0, Pr(Snow) = 0, and Pr(Sunny) = 1.

As before, the forecast is “there will be some precipitation.”

Pr(Rain|Precip.) 6= defined and Pr(Snow |Precip.) 6= defined .

Bayes’ rule does not work when the conditioning event hasprobability 0.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Probability Theory

1 ProbabilitySome Background

2 Probability in Simple Sample SpacesSample Space and EventsProbability of an EventCounting Techniques

3 Conditional ProbabilityConditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Definition

An event F is said to be independent of an event E if theprobability that F occurs is not influenced by whether E has or hasnot occurred.

In other words, if the probability of F equals the conditionalprobability of F given E : P(F ) = P(F |E ).

Now substituting P(F ) for P(F |E ) in the theorem ofcompound probabilities P(E ∩ F ) = P(E )P(F |E ), we obtain

P(E ∩ F ) = P(E )P(F ).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Definition

An event F is said to be independent of an event E if theprobability that F occurs is not influenced by whether E has or hasnot occurred.

In other words, if the probability of F equals the conditionalprobability of F given E : P(F ) = P(F |E ).

Now substituting P(F ) for P(F |E ) in the theorem ofcompound probabilities P(E ∩ F ) = P(E )P(F |E ), we obtain

P(E ∩ F ) = P(E )P(F ).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Definition

An event F is said to be independent of an event E if theprobability that F occurs is not influenced by whether E has or hasnot occurred.

In other words, if the probability of F equals the conditionalprobability of F given E : P(F ) = P(F |E ).

Now substituting P(F ) for P(F |E ) in the theorem ofcompound probabilities P(E ∩ F ) = P(E )P(F |E ), we obtain

P(E ∩ F ) = P(E )P(F ).

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Definition (cont.)

We will use the above equation as our definition of independence.

Definition

Events F and E are independent if P(E ∩ F ) = P(E )P(F );otherwise they are dependent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Definition (cont.)

We will use the above equation as our definition of independence.

Definition

Events F and E are independent if P(E ∩ F ) = P(E )P(F );otherwise they are dependent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples

Example

A coin is tossed twice. If the coin lands heads on the second toss,you win a dollar.

(a) If the first toss is heads, what is your chance of winning thedollar?

(b) If the first toss is tails, what is your chance of winning thedollar?

If the first toss is heads, the probability to get heads the secondtime is 1

2 . If the first toss is tails, the probability is still 12 . The

chances for the second toss stay the same, however the first tossturns out.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples

Example

A coin is tossed twice. If the coin lands heads on the second toss,you win a dollar.

(a) If the first toss is heads, what is your chance of winning thedollar?

(b) If the first toss is tails, what is your chance of winning thedollar?

If the first toss is heads, the probability to get heads the secondtime is 1

2 . If the first toss is tails, the probability is still 12 . The

chances for the second toss stay the same, however the first tossturns out.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

Example

Two draws will be made at random with replacement from the box.

1 1 2 2 3

(a) Suppose the first draw is 1 . What is the chance of getting a

2 in the second draw?

(b) Suppose the first draw is 2 . What is the chance of getting a

2 in the second draw?

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

Whether the first draw is 1 or 2 or anything else, the

chance of getting 2 in the second draw stays the same – twoin five, or 40%.

The draws are independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

Whether the first draw is 1 or 2 or anything else, the

chance of getting 2 in the second draw stays the same – twoin five, or 40%.

The draws are independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

Example

As in the previous example, but the draws are made withoutreplacement.

If the first draw turns out to be 1 then the second draw is fromthe box

1 2 2 3 .

The chance for the second to be 2 is 50%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

Example

As in the previous example, but the draws are made withoutreplacement.

If the first draw turns out to be 1 then the second draw is fromthe box

1 2 2 3 .

The chance for the second to be 2 is 50%.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

On the other hand, if the first draw turns out to be 2 , then thesecond draw is from the box

1 1 2 3 .

Now there is only a 25% chance for the second to be 2 . Thedraws are dependent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence: Examples (cont.)

On the other hand, if the first draw turns out to be 2 , then thesecond draw is from the box

1 1 2 3 .

Now there is only a 25% chance for the second to be 2 . Thedraws are dependent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events

When drawing at random with replacement, the draws areindependent.

Without replacement, the draws are dependent.

Three events A, B, and C are independent if:

(i) P(A ∩ B) = P(A)P(B), P(A ∩ C ) = P(A)P(C ) andP(C ∩ B) = P(B)P(C ) (i.e. the events are pairwiseindependent), and

(ii) P(A ∩ B ∩ C ) = P(A)P(B)P(C ).

The next example shows that condition (ii) does not follow fromcondition (i); in other words, three events may be pairwiseindependent but not independent themselves.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events

When drawing at random with replacement, the draws areindependent.

Without replacement, the draws are dependent.

Three events A, B, and C are independent if:

(i) P(A ∩ B) = P(A)P(B), P(A ∩ C ) = P(A)P(C ) andP(C ∩ B) = P(B)P(C ) (i.e. the events are pairwiseindependent), and

(ii) P(A ∩ B ∩ C ) = P(A)P(B)P(C ).

The next example shows that condition (ii) does not follow fromcondition (i); in other words, three events may be pairwiseindependent but not independent themselves.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events

When drawing at random with replacement, the draws areindependent.

Without replacement, the draws are dependent.

Three events A, B, and C are independent if:

(i) P(A ∩ B) = P(A)P(B), P(A ∩ C ) = P(A)P(C ) andP(C ∩ B) = P(B)P(C ) (i.e. the events are pairwiseindependent), and

(ii) P(A ∩ B ∩ C ) = P(A)P(B)P(C ).

The next example shows that condition (ii) does not follow fromcondition (i); in other words, three events may be pairwiseindependent but not independent themselves.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events

When drawing at random with replacement, the draws areindependent.

Without replacement, the draws are dependent.

Three events A, B, and C are independent if:

(i) P(A ∩ B) = P(A)P(B), P(A ∩ C ) = P(A)P(C ) andP(C ∩ B) = P(B)P(C ) (i.e. the events are pairwiseindependent), and

(ii) P(A ∩ B ∩ C ) = P(A)P(B)P(C ).

The next example shows that condition (ii) does not follow fromcondition (i); in other words, three events may be pairwiseindependent but not independent themselves.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events: Example

Example

Let a pair of fair coins be tossed.

Let Ω = HH,HT ,TH,TT be the equiprobable finite samplespace associated with this experiment. Consider the events:

A = heads on the first coin = HH,HTB = heads on the second coin = HH,THC = heads on exactly one coin = HT ,TH

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events: Example

Example

Let a pair of fair coins be tossed.

Let Ω = HH,HT ,TH,TT be the equiprobable finite samplespace associated with this experiment. Consider the events:

A = heads on the first coin = HH,HTB = heads on the second coin = HH,THC = heads on exactly one coin = HT ,TH

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events: Example

Example

Let a pair of fair coins be tossed.

Let Ω = HH,HT ,TH,TT be the equiprobable finite samplespace associated with this experiment. Consider the events:

A = heads on the first coin = HH,HTB = heads on the second coin = HH,THC = heads on exactly one coin = HT ,TH

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events: Example (cont.)

Then P(A) = P(B) = P(C ) = 24 = 1

2 andP(A ∩ B) = P(HH) = 1

4 , P(A ∩ C ) = P(HT) = 14 ,

P(B ∩ C ) = P(TH) = 14 .

Thus condition (i) is satisfied. However, A∩B ∩C = ∅ and soP(A ∩ B ∩ C ) = P(∅) = 0 6= P(A)P(B)P(C ).

In other words, condition (ii) is not satisfied and so the threeevents are not independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events: Example (cont.)

Then P(A) = P(B) = P(C ) = 24 = 1

2 andP(A ∩ B) = P(HH) = 1

4 , P(A ∩ C ) = P(HT) = 14 ,

P(B ∩ C ) = P(TH) = 14 .

Thus condition (i) is satisfied. However, A∩B ∩C = ∅ and soP(A ∩ B ∩ C ) = P(∅) = 0 6= P(A)P(B)P(C ).

In other words, condition (ii) is not satisfied and so the threeevents are not independent.

Sebastian M. Saiegh Probability Theory

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ProbabilityProbability in Simple Sample Spaces

Conditional Probability

Conditional and Compound ProbabilitiesBayes’ TheoremIndependent Events

Independence of Events: Example (cont.)

Then P(A) = P(B) = P(C ) = 24 = 1

2 andP(A ∩ B) = P(HH) = 1

4 , P(A ∩ C ) = P(HT) = 14 ,

P(B ∩ C ) = P(TH) = 14 .

Thus condition (i) is satisfied. However, A∩B ∩C = ∅ and soP(A ∩ B ∩ C ) = P(∅) = 0 6= P(A)P(B)P(C ).

In other words, condition (ii) is not satisfied and so the threeevents are not independent.

Sebastian M. Saiegh Probability Theory