Overviewjga001/Chapter 5 Alford.pdf · 2012-03-01 · NE PE NS. Example 5.1 - Probability of...

Overview

5.1 Introducing Probability

5.2 Combining Events

5.3 Conditional Probability

5.4 Counting Methods (OMIT)

5.1 Introducing Probability

Objectives:

By the end of this section, I will be

able to…

1) Understand the meaning of an experiment, an outcome, an event, and a sample space.

2) Describe the classical method of assigning probability.

3) Explain the Law of Large Numbers and the relative frequency method of assigning probability.

Probability

Defined as the long-term proportion of times the outcome occurs

Building Blocks of Probability

Experiment - any activity for which the

outcome is uncertain

Outcome - the result of a single performance of an experiment

Sample space (S) - collection of all possible outcomes

Event - collection of outcomes from the sample space

Example Experiment: A pair of dice are rolled.

List the sample space.

Example The sample space has 36 events:

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

where the pairs represent the numbers rolled on

each dice. Note that here we assume order matters

so that 1,2 and 2,1 are different outcomes (think of

each dice as having a different color).

Example Which elements of the sample space correspond to

the event that the sum of each dice is 4?

Example The sample space has 36 events:

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

})1,3(),2,2(),3,1{(ANSWER:

Rules of Probability

The probability P(E) for any event E is always between 0 and 1, inclusive.

That is, 0 ≤ P(E ) ≤ 1.

Law of Total Probability:

For any experiment, the sum of all the

outcome probabilities in the sample space must equal 1.

Interpreting Probability

If P(E) is near 0, the event is very unlikely to occur.

If P(E) is equal to 0, the event cannot occur.

If P(E) is near 1, the event is very likely to occur.

If P(E) is equal to 1, the event is will occur.

If P(E) is low, the event is unusual

If P(E) is high, the event is not unusual

Classical Method of Assigning Probabilities

Let N(E) and N(S) denote the number of outcomes in event E and the sample space S, respectively.

If the experiment has equally likely outcomes, then the probability of event E is then

number outcomes in E

number of outcomes in sample space

N EP E

N S

Example 5.1 - Probability of drawing an ace

Find the probability of drawing an ace when

drawing a single card at random from a

deck of cards.

Example 5.1 continued

Solution

The sample space for the experiment where a subject chooses a single card at random from a deck of cards is given in Figure 5.1.

FIGURE 5.1 Sample space for drawing a card at random from a deck of cards.


If a card is chosen at random, then each card has the same chance of being drawn.

Since each card is equally likely to be drawn, we can use the classical method to assign probabilities.

There are 52 outcomes in this sample space, so N(S) = 52.


Let E be the event that an ace is drawn.

Event E consists of the four aces

{A♥, A♦, A♣, A♠}, so N(E ) = 4.

Therefore, the probability of drawing

an ace is

4 1

52 13

N EP E

N S

Example

What is the probability of rolling two die and getting a sum of 4?

Assume each number is equally likely to be rolled on the die.

Example

ANSWER:

E = event that sum of the dice is 4

Rolling a sum of 4 can happen in one of three ways (see previous slide) with 36 events so:

0833.036

3

)(

)()(

SN

ENEP

Example

What is the probability of getting no heads when a “fair” coin is tossed three times? (A fair coin has an equal probability of showing heads or tails when tossed.)

Example

ANSWER

Sample Space

TTTTTHTHTTHHHTTHTHHHTHHH ,,,,,,,

E = event that no heads occurs in three tosses

125.08

1

)(

)()(

SN

ENEP

Tree Diagram

Device used to count the number of outcomes of an experiment

Graphical display to visualize a multistage experiment

Helps us to construct the sample space for

a multistage experiment

FIGURE 5.2

Tree diagram for the experiment of tossing a fair coin twice.

Example: Tree Diagrams

A bag contains three different colored marbles:

red, blue, and green. Suppose two marbles are

drawn from the bag and after the first marble is

drawn, it is put back into the bag before the

second marble is drawn. Construct a tree

diagram that depicts all possible outcomes.

Example: Tree Diagrams

Example: Computing Probability

withTree Diagrams

Use the previous tree diagram to compute

the probability of drawing a red marble on

the first draw and a blue marble on the

second draw.

Example: Computing Probability

withTree Diagrams

111.0

9

1

)(

)()(

SN

ENEP

E = event of drawing red first then

blue

Probability Model

A probability model is a table or listing of all the possible outcomes of an experiment together with the probability of each outcome.

A probability model must follow the rules of probability stated earlier.

Probability Model Example

A probability model for the number of heads

when a “fair” coin is tossed three times.

Sample Space

Number of

Heads

Probability

0 1/8=0.125

1 3/8=0.375

2 3/8=0.375

3 1/8=0.125



page 213


ANSWER:

Not a probability model since probability is never less than zero.

Law of Large Numbers

As the number of times that an experiment is repeated increases, the relative frequency (proportion) of a particular outcome tends to approach the probability of the outcome.

For quantitative data, as the number of times that an experiment is repeated increases, the mean of the outcomes tends to approach the population mean.

For categorical (qualitative) data, as the number of times that an experiment is repeated increases, the proportion of times a particular outcome occurs tends to approach the population proportion.

Relative Frequency Method

The probability of event E is approximately equal to the relative frequency of event E.

Also known as the empirical method

frequency of Erelative frequency of E

number of trials of experimentP E

Example 5.7 - Teen bloggers

A recent study found that 35% of all online

teen girls are bloggers, compared to 20% of

online teen boys. Suppose that the 35% came

from a random sample of 100 teen girls

who use the Internet, 35 of whom are

bloggers. If we choose one teen girl at

random, find the probability that she is a

blogger.


Solution

Define the event.

B: The online girl is a blogger.

We use the relative frequency method to find the probability of event B:

frequency of Brelative frequency of B

number of trials of experiment

35 0.35

100

P B

Polygraph Data:

Example

Did Not Lie Did Lie TOTALS

Positive Test Result 15

(false positive)

42

(true positive)

57

NegativeTest Result 32

(true negative)

9

(false negative)

41

TOTALS 47 51 98

What is the probability a person lied on a

polygraph test?

Example

If one response is randomly selected, what

is the probability it is a lie?

E = event of a lie

Example

520.098

51

experiment of trialsofnumber

E offrequency )(EP

Example

Page 214, problem 36

Example

ANSWER;

(a) No, the outcomes parked illegally and not parked illegally may not be equally likely.

Example

ANSWER;

(b) Using the frequency method, the probability a car parked in a handicap zone is illegally parked is:

10/50 = 1/5 = 0.20

Subjective Method

Should be used when the event is not (even theoretically) repeatable

The assignment of a probability value to an outcome based on personal judgment

EXAMPLE: the probability of having an accident on the way home is 0.00001 (very low, 1 in 100,000 trips I may have an accident)

Summary

Section 5.1 introduces the building blocks of probability, including the concepts of probability, outcome, experiment, and sample space.

Probabilities always take values between 0 and 1, where 0 means that the outcome cannot occur and 1 means that the outcome is certain.

Summary

The classical method of assigning probability is used if all outcomes are equally likely.

The classical method states that the probability of an event A equals the number of outcomes in A divided by the number of outcomes in the sample space.

Summary

The Law of Large Numbers states that, as an experiment is repeated many times, the relative frequency (proportion) of a particular outcome tends to approach the probability of the outcome.

The relative frequency method of assigning probability uses prior knowledge about the relative frequency of an outcome.

The subjective method of assigning probability is used when the other methods are not applicable.

5.2 Combining Events

Objectives:


able to…

1) Understand how to combine events using complement, union, and intersection.

2) Apply the Addition Rule to events in general and to mutually exclusive events in particular.

Complement of A

Symbolized by AC

Collection of outcomes not in event A

Complement comes from the word “to complete”

Any event and its complement together make up the complete sample space

Example 5.10 - Finding the probability of the complement of an event

If A is the event “observing a sum of 4 when

the two fair dice are rolled,” then your

roommate is interested in the probability of

AC, the event that a 4 is not rolled. Find the

probability that your roommate does not roll

a 4.


Solution

Which outcomes belong to AC?

By the definition, AC is all the outcomes in the sample space that do not belong in A.

There are the following outcomes in A:

{(3,1)(2,2)(1,3)}.


Figure 5.10 shows all the outcomes except the outcomes from A in the two-dice sample space.


There are 33 outcomes in AC and 36 outcomes in the sample space.

The classical probability method then gives the probability of not rolling a 4 to be

The probability is high that, on this roll at least, your roommate will not land on Boardwalk.

33 11

36 12

C

CN A

P AN S

Example

If a fair coin is tossed three times.

A = event that exactly one heads occurs

Find the complement of A.

Example

Sample space:

Event A corresponds to HTT, THT, TTH

Therefore, the complement of A are the events:

HHH, HHT, HTH, THH, TTT


Probabilities for Complements

For any event A and its complement AC,

P(A) + P(AC) = 1.

P(A) = 1 - P(AC)

P(AC ) = 1 - P(A)

Example

Page 222, problem 20

Consider the experiment of drawing a

card at random from a shuffled deck of 52 cards. Find the probability of drawing a card that is not a face card.

NOTE: a face card is a king, queen, or jack

Solution

The sample space for the experiment where a subject chooses a single card at random from a deck of cards is depicted here.

FIGURE 5.1 Sample space for drawing a card at random from a deck of cards.

Example

Example

ANSWER:

There are 52 cards, 12 of which are face

cards.

77.013

10

52

40

52

121

)(

)(1)(1)(

SN

ENEPEP C

card face a drawyou event E

card face a drawnot doyou event CE

Union of Events- Venn Diagram

The event representing all the outcomes that belong to A or B or both

Denoted as A B

Associated with “or”

FIGURE 5.11 Union of events A, B.

Intersection of Events- Venn Diagram

The event representing all the outcomes that belong to both A and B

Denoted as A B

Associated with “and”

FIGURE 5.12 Intersection of Events A, B.

Example 5.11 - Union and intersection

Let our experiment be to draw a single card at

random from a deck of cards. Define

the following events:

A: The card drawn is an ace.

H: The card drawn is a heart.

a. Find A H.


Solution

a. A H is event containing all outcomes are either aces or hearts or both (the ace of hearts).

A H is all cards shown in Figure 5.13.

Figure 5.13

Example 5.11 - Union and intersection

Let our experiment be to draw a single card at

random from a deck of cards. Define

the following events:

A: The card drawn is an ace.

H: The card drawn is a heart.

b. Find A H.


Solution

b. The intersection of A and H is the event containing the outcomes that are common to both A and H.

There is only one such outcome: the ace of hearts (see Figure 5.13).

Experiment: A pair of dice are rolled. Define the

following events:

Find both

5 equals dice two theof sum event the A

4 a is dice theof oneleast at event B

BA BA

Example

The sample space has 36 events:

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

where the pairs represent the numbers rolled on

each dice. Note that here we assume order matters

so that 1,2 and 2,1 are different outcomes (think of

each dice as having a different color).

Example

(3,2)}(2,3),

),6,4(),5,4(),3,4(),2,4(),1,4(

),4,6(),4,5(),4,4(),4,3(),4,2(),4,1{(BA

Example

The event that the sum of the two dice equals five

or at least one of the dice is a 4.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

})1,4(),4,1{(BA

Example


and one of the dice is a 4.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6


following events

Find the probability that the sum of the two dice

equals five and at least one of the dice is a 4.


)( BAP

Example


ANSWER:

06.018

1

36

2

)(

)()(

SN

BANBAP

Example


following events


equals five or at least one of the dice is a 4.


)( BAP

Example


ANSWER:

36.036

13

)(

)()(

SN

BANBAP

Example

Addition Rule

The probability that either one event or another event may occur

Count the probabilities of the outcomes in A

Add the probabilities of the outcomes in B

Subtract the probability of the intersection (overlap)

)()()()( BAPBPAPBAP


following events


equals five or at least one of the dice is a 4.


)( BAP

Example


36

4

)(

)()(

)}1,4(),2,3(),3,2(),4,1{(

SN

ANAP

A

Example


1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

36

11

)(

)()(

})6,4(),5,4(),3,4(),2,4(),1,4(

),4,6(),4,5(),4,4(),4,3(),4,2(),4,1{(

SN

BNBP

B

Example

The event that at least one of the dice is a 4.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

36

2)(

})1,4(),4,1{(

BAP

BA

Example


and at least one of the dice is a 4.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

ANSWER:

)()()()( BAPBPAPBAP

Example

36

2

36

11

36

4

36.036

13

Example 5.12 - Addition Rule applied to a deck of cards

Suppose you pay $1 to play the following

game. You choose one card at random from

a deck of 52 cards, and you will win $3 if the

card is either an ace or a heart. Find the

probability of winning this game.


Solution

Using the same events defined in Example 5.11, we find P(A or H) = P(A H).

By the Addition Rule, we know that

P(A H) = P(A) + P(H ) - P(A H)

There are 4 aces in a deck of 52 cards, so by the classical method (equally likely outcomes), P(A) = 4/52.


There are 13 hearts in a deck of 52 cards, so P(H ) = 13/52.

From Example 5.11, we know that A H represents the ace of hearts.

Since each card is equally likely to be drawn, then P(ace of hearts) = P(A H) = 1/52.

P(A H) = P(A) + P(H ) - P(A H)

4 13 1 16 4

52 52 52 52 13

Mutually Exclusive Events

Also known as disjoint events

Events having no outcomes in common

FIGURE 5.14 Mutually exclusive events.

symbol) set"empty " (the BA

0)()( PBAP

Mutually Exclusive Events

Addition rule for disjoint (mutually exclusive) events A and B

)()()( BPAPBAP


following events


equals five or the sum of the two dice equals 2.


2 is dice two theof sum event the B

)( BAP

Example

Example

The event that the sum of the two dice equals five.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

36

4

)(

)()(

)}1,4(),2,3(),3,2(),4,1{(

SN

ANAP

A

Example

The event that the sum of the two dice equals two.

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

36

1

)(

)()(

)}1,1{(

SN

BNBP

B

Example


and event that the sum of the two dice equals two

are mutually exclusive and

BA

)()()( BPAPBAP

14.036

5

36

1

36

4

Summary

Combinations of events may be formed using the concepts of complement, union, and intersection.

The Addition Rule provides the probability of Event A or Event B to be the sum of their two probabilities minus the probability of their intersection.

Mutually exclusive events have no outcomes in common.

5.3 Conditional Probability

Objectives:


able to…

1) Calculate conditional probabilities.

2) Explain independent and dependent events.

3) Solve problems using the Multiplication Rule.

4) Recognize the difference between sampling with replacement and sampling without replacement.

Conditional probability

For two related events A and B, the probability of B given A is denoted

P(B|A)

FIGURE 5.16 How conditional probability works.

Calculating Conditional Probability

The conditional probability that B will occur, given that event A has already taken place:

The conditional probability that A will occur, given that event B has already taken place:

)(

)(

)(

)()|(

AN

BAN

AP

BAPABP

)(

)(

)(

)()|(

BN

BAN

BP

BAPBAP

Polygraph data:

Find the probability a subject had a negative

test result, given that the subject did not lie.

Example



(false positive)

42

(true positive)

57


(true negative)

9

(false negative)

41

TOTALS 47 51 98

There are 47 subjects who did not lie,

32 of which had a negative test result.

681.047

32

)(

)() |(

BN

BANBAP

lienot didsubject event the B

result test negative a hadsubject event the A

Example

Previous example using probabilities

instead of counts:

Example

47

32

98/47

98/32

)(

)()|(

BP

BAPBAP

Example Do problem 44(c)

Example

ANSWER:

6

1

180

30

300/180

300/30

)(

)()|(

FP

FOPFOP

Careful

The order of conditional probability may

matter! In general,

)|()|( BAPABP

Use the polygraph example to show

that:

where

is the probability a subject did not lie

(event B) given that he/she had a

negative test result (event A).

Example

)|()|( BAPABP

)|( ABP

780.041

32

98/41

98/32

A)(

)A( )|(

P

BPABP

Example

which is not the same as our previous result

681.047

32) | ( BAP

Independent Events

If the occurrence of an event does not affect the probability of a second event, then the two events are independent

Events A and B are independent means

P(A| B) = P(A) or P(B| A) = P(B)

Otherwise the events are said to be dependent.

, problem 38

A single fair dice is rolled twice in

succession. Find the probability that you

observe an even number on the second

roll, given that you observe an even

number on the first roll.

Example

ANSWER:

These are “obviously” independent events since

the probability of the second roll being an even

number is not affected by getting an even

number on the first roll.

Example

rollfirst on number even an ofevent 1E

roll secondon number even an ofevent 2E

2

1

6

3)()|( 212 EPEEP

Is it “obvious” when events are

independent?

Independent or Not?

Systematic Strategy for Determining Whether Two Events Are Independent

1) Find P(B)

2) Find P(B|A)

3) Compare the two probabilities.

◦ If they are equal, then A and B are independent events.

◦ Otherwise, A and B are dependent events.

A bag contains three different colored

marbles: red, blue, and green. Suppose

two marbles are drawn from the bag

without replacing the first marble after it is

drawn. Determine whether the following

events are independent.

Example

first red drawyou event R1

second blue drawyou event B2

Example

Sample Space:

{RB, RG, BR, BG, GR, GB}

Example

333.03

1

6

2)( 2BP

500.06/2

6/1

)(

)()|(

1

2112

RP

BRPRBP

)()|( 212 BPRBPSince:

eventst independennot are and 12 RB





second marble is drawn. Determine whether the

following events are independent.



Example

Sample Space:

{RR,RB,RG,BR,BB,BG,GR,GB,GG}

Example

Example

3

1

9

3)( 2BP

3

1

9/3

9/1

)(

)()|(

1

2112

RP

BRPRBP

)()|( 212 BPRBPSince:

eventst independen are and 12 RB

Sampling

The previous examples illustrate a difference between sampling with and without replacement.

Sampling

Sampling with replacement the randomly selected unit is returned to the population after being selected

It is possible for the same unit to be sampled more than once

Successive draws can be considered independent

Sampling

Sampling without replacement the randomly selected unit is not returned to the population after being selected

It is not possible for the same unit to be sampled more than once

Successive draws should be considered dependent (not independent)

Multiplication Rule

Used to find probabilities of intersections of events

or equivalently

Recall: intersection is associated with the word “and”

)|()()( ABPAPBAP

)|()()( BAPBPBAP

Polygraph data:

Find the probability a subject had a negative

test result and the subject did not lie.

Example



(false positive)

42

(true positive)

57


(true negative)

9

(false negative)

41

TOTALS 47 51 98

ANSWER:

327.098

32

98

47

47

32

)P() |()( BBAPBAP

Example

lienot didsubject event the B

result test negative a hadsubject event the A

ANSWER (direct method using relative

frequency formula):

327.098

32

experiment of trialsofnumber

offrequency )(

BABAP

Example

Example Do problem 43(c)

ANSWER:

6

1

300

50

300

150

150

50

)P() |()( CCMPCMP

Example

A bag contains three different colored

marbles: red, blue, and green. Suppose

two marbles are drawn from the bag

without replacing the first marble after it is

drawn. Determine the probability that you

draw a red on the first draw and a blue on

the second draw.

Example

ANSWER:

Must find:

Example



)|()()( 12121 RBPRPBRP

Example

3

1

6

2)( 2BP

2

1)|( 12 RBP

ANSWER:

6

1

2

1

3

1

)|()()( 12121 RBPRPBRP

Multiplication Rule for Two Independent Events

If A and B are any two independent events we have that:

The multiplication rule:

Becomes:

)()|( BPABP

)()()( BPAPBAP

)|()()( ABPAPBAP





second marble is drawn. Determine the

probability that you draw a red on the first draw

and a blue on the second draw.

Example

ANSWER:

Example

3

1)( 1RP

3

1)()|( 212 BPRBP

111.09

1

3

1

3

1

)()()( 2121 BPRPBRP

If two subjects are randomly selected without

replacement, what is probability they both had false

positive test results?

Example



(false positive)

42

(true positive)

57


(true negative)

9

(false negative)

41

TOTALS 47 51 98

Use the following polygraph data to answer the

question below.

ANSWER:

These events are not independent

(sample without replacement). We

must find:

Example

selectionfirst on positive false a is thereevent that 1F

selection secondon positive false a is thereevent that 2F

)|()() ( 12121 FFPFPFFP

ANSWER:

On first selection there are 98 subjects,

15 of which are false positive:

Example

98

15)( 1FP

On second selection there are 97

subjects (without replacement)

If the first selection was false positive,

there are 14 false positives left:

Example

97

14)|( 12 FFP

Example

0221.0

9506

210

97

14

98

15

)|()() ( 12121 FFPFPFFP

If two subjects are randomly selected with

replacement, what is probability they both

had false positive test results?

Example

ANSWER:

On first selection there are 98 subjects,

15 of which are false positive:

Example

98

15)( 1FP

On second selection there are 98

subjects (with replacement), 15 of

which are false positives.

Example

98

15)()|( 212 FPFFP

Example

0234.0

9604

225

98

15

98

15

)()(

)|()() (

21

12121

FPFP

FFPFPFFP

12:30-2:00

Start Here On Feb. 28

For exercises 13-16, let A and B be two

independent events, with P(A)=0.5 and

P(B)=0.2. Find the indicated probabilities.

(13)

Example

)( BAP

ANSWER:

Example

10.0

)2.0()5.0(

)()()( BPAPBAP

(14)

Example

)|( BAP

ANSWER:

Example

5.0)()|( APBAP

(15)

Example

)|( ABP

ANSWER:

Example

2.0)()|( BPABP

(16)

Example

)( BAP

ANSWER:

(see chapter 5.2) we have that

Example

0.6

0.1 0.2 0.5

)()()()( BAPBPAPBAP

Mutually exclusive (disjoint) events:

Independent events:

These are not the same thing. (see example

5.22 in the book and assigned problems 29-

32 on page 238)

Common Confusion

0)()( PBAP

)()()( BPAPBAP

Multiplication Rule for n Independent Events

If A, B, C, . . .

are independent events, then

P(A B C . . .) = P(A) P(B) P(C ) . . .

Treating Dependent Events as

Independent

Some calculations are cumbersome, but

they can be made manageable by using the

common practice of treating events as

independent when small samples are drawn

from large populations. In such cases, it is

rare to select the same item twice (sample

with replacement).

The 5% Guideline for

Cumbersome Calculations

If a sample size is no more than 5% of the

size of the population, we may treat the

selections as being independent (even if the

selections are made without replacement, so

they are technically dependent).

, problem 40

The Federal Interagency Forum on Child

and Family Statistics reported that the

teenage birth rate in 2000 was 0.0453.

(a) Find the probability that two randomly

selected births are due to teenagers.

Example

ANSWER:

We will assume sample size is relatively

small compared to the population and that

these events are independent.

Example

teenagerafor isbirth selectedfirst event 1T

teenagerafor isbirth selected secondevent 2T

)()()( 2121 TPTPTTP

ANSWER:

Example

0.00205

)0453.0(

)0453.0()0453.0(

)()()(

2

2121 TPTPTTP

, problem 40

The Federal Interagency Forum on Child

and Family Statistics reported that the

teenage birth rate in 2000 was 0.0453.

(b) Find the probability that five randomly

selected births are due to teenagers.

Example

ANSWER:

Example

7-

5

54321

54321

101.9110.00000019

)0453.0(

)0453.0()0453.0()0453.0()0453.0()0453.0(

)()()()()(

)(

TPTPTPTPTP

TTTTTP

Summary

Section 5.3 discusses conditional probability P(B | A), the probability of an event B given that an event A has occurred.

We can compare P(B | A) to P(B) to determine whether the events A and B are independent.

Events are independent if the occurrence of one event does not affect the probability that the other event will occur.

Summary

The Multiplication Rule for Independent Events is the product of the individual probabilities.

Sampling with replacement is associated with independence, while sampling without replacement means that the events are not independent.

Overviewjga001/Chapter 5 Alford.pdf · 2012-03-01 · NE PE NS. Example 5.1 - Probability of...

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