Introduction to probability - UPR-RPepsilon.uprrp.edu/aniel/Lectures/groebner_bsdma9_ch04.pdf ·...

Copyright ©2014 Pearson Education, Inc.

Introduction to probability

Copyright ©2014 Pearson Education, Inc. 4-2

4.1 The Basics of Probability

• Probability– The chance that a particular event will occur

– The probability value will be in the range 0 to 1

• Experiment– A process that produces a single outcome

whose result cannot be predicted with certainty

• Sample Space– The collection of all outcomes that can result

from a selection, decision, or experiment.


Sample Space Examples

• Example 1: All 6 faces of a die:

• Example 2: All 52 cards of a deck of cards:

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Defining the Sample Space

• Step 1: Define the experiment– For example, the sale. The item of interest is the product sold

• Step 2: Define the outcomes for one trial of the experiment– Outcomes: e1 - Hamburger, e2 – Cheeseburger, e3 – Bacon Burger

• Step 3: Define the sample space (SS)– For single sale: SS = {e1, e2, e3}

– If experiment includes two sales, there will be nine outcomes:SS = {o1, o2, o3, o4, o5, o6, o7, o8, o9}o1={e1, e1}, o2={e1, e2}, o3={e1, e3}, o4={e2, e1}, . . . , o9={e3, e3}

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Tree Diagrams

• A useful way to define the sample space

• Step 1: Define the experiment– Three students were asked if they like statistics

• Step 2: Define the outcomes– Possible outcomes are: ‘No’ and ‘Yes’

• Step 3: Define the sample space for three trials using a tree diagram– For a single trial:

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Tree Diagram

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Event of Interest

• A collection of experimental outcomes

• Step 1: Define the experiment– Two randomly chosen audits

• Step 2: List the outcomes associated with one trial of the experiment– Outcomes: audit done early, on time, or late

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Event of Interest

• Step 3: Define the sample space– There are nine experimental outcomes for two audits

e1={early, early}, e2={early, on time}, e3={early, late},e4={on time, early}, e5={on time, on time}, e6={on time, late},e7={late, early}, e8={late, on time}, e9={late, late}

• Step 4: Define the event of interest– At least one audit is completed late:

E = {e3, e6, e7, e8, e9}

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Types of Events

• Mutually Exclusive Events– Two events are mutually exclusive if the occurrence of

one event precludes the occurrence of the other event.

• Independent Events– Two events are independent if the occurrence of one

event in no way influences the probability of the occurrence of the other event.

• Dependent Events– Two events are dependent if the occurrence of one

event impacts the probability of the other event occurring.

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Events Examples

• Mutually exclusive– The product can be either good of defective

• Independent

– Result of second flip does not depend on the result of the first flip

• Dependent

– Probability of the second event is affected by the occurrence of the first event

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E1 = heads on one flip of fair coin

E2 = heads on second flip of same coin

E1 = heads on one flip of fair coin

E2 = heads on second flip of same coin

E1 = rain forecasted on the news

E2 = take umbrella to work

E1 = rain forecasted on the news

E2 = take umbrella to work


Assigning Probability

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Methods ofAssigning Probability

Methods ofAssigning Probability

ClassicalProbability

Assessment

RelativeFrequency

Assessment

SubjectiveProbability

Assessment

P(Ei) = Probability of event Ei occurringP(Ei) = Probability of event Ei occurring


Classical Probability Assessment

• The method is based on the ratio of the number of ways an outcome or event of interest can occur to the number of ways any outcome or event can occur when the individual outcomes are equally likely.

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Classical Probability Assessment

• Step 1: Define the experiment

• Step 2: Determine whether the possible outcomes are equally likely

• Step 3: Determine the total number of outcomes

• Step 4: Define the event of interest

• Step 5: Determine the number of outcomes associated with the event of interest

• Step 6: Compute the classical probability

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Relative Frequency Probability Assessment

• The method defines probability as the number of times an event occurs divided by the total number of times an experiment is performed in a large number of trials.

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Ei - The event of interestN - Number of trials


Relative Frequency Probability Assessment


• Step 2: Define the events of interest

• Step 3: Determine the total number of occurrences

• Step 4: For the event of interest, determine the number of occurrences

• Step 5: Determine the relative frequency probability assessment.

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Subjective Probability Assessment

• The method defines probability of an event as reflecting a decision maker’s state of mind regarding the chances that the particular event will occur

• Represents a person’s belief that an event will occur

• Only approximate guide to the future

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4.1 The Rules of Probability

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Probability RulesProbability Rules

Possible Values

Summation ofPossible Values

Addition Rule for Any Two Events

Addition Rule for Individual Outcomes

Complement Rule

Addition Rule for Mutually Exclusive

Events

Conditional Probability for Any

Two Events

Multiplication Rule for Any Two Events

Multiplication Rule for Independent

Events

Bayes’ Theorem

Conditional Probability for

Independent Events


Possible Values and Summation Rules

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Probability Rule 1Probability Rule 1 Probability Rule 2Probability Rule 2

For any event Ei

0 ≤ P(Ei) ≤ 1 for all i

k - Number of outcomes in the sampleei - ith outcome

0 = no chance of occurring1 = 100% chance of occurring

Sum of the probabilities of allpossible outcomes is 1.

Probability of event occurringis always between 0 and 1.



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Probability Rule 3Probability Rule 3

The probability of an event Ei is equal to the sumof the probabilities of the individual outcomes

forming Ei

For example, if Ei = {e1, e2, e3} thenP(Ei) = P(e1) + P(e2) + P(e3)




• Step 2: Define the possible outcomes

• Step 3: Determine the probability of each possible outcome

• Step 4: Define the event of interest

• Step 5: Apply Probability Rule 3 to compute the desired probability

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Complement Rule

• The complement of an event E is the collection of all possible outcomes not contained in event E

• This rule is corollary to Probability Rules 1 and 2

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The probability of the complement of event E is 1 minus the probability of event E

P(E) = 1 - P(E)



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P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)



• Step 1: Define the Experiment

• Step 2: Define the events of interest

• Step 3: Determine the probability for each event

• Step 4: Determine whether the two events overlap, and if so, compute the joint probability

• Step 5: Apply Probability Rule 4 to compute the desired probability

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Addition Rule for Any Two Events - Example

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P(Red or Ace) = P(Red) + P(Ace) - P(Red and Ace)

P(Red or Ace) = 26/52 + 4/52 - 2/52 = 28/52

Don’t count the two red aces twice!

Don’t count the two red aces twice!Black

ColorType Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52


Addition Rule for Mutually Exclusive Events

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P(E1 or E2) = P(E1) + P(E2)

Both events cannot occur at the same time: P(E1 and E2) = 0


Conditional Probability for Any Two Events

• Conditional Probability– The probability that an event will occur given

that some other event has already happened.

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Given thatGiven that


Conditional Probability Example

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No CDCD Total

AC 0.2 0.5 0.7

No AC 0.2 0.1 0.3

Total 0.4 0.6 1.0


Conditional Probability for Independent Events

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For independent events E1, E2

P(E1|E2) = P(E1) P(E2) > 0P(E2|E1) = P(E2) P(E1) > 0

The conditional probability of one event occurring,given a second independent event has already occurred, is simply the probability of the first event occurring.


Multiplication Rule for Any Two Events

• If we do not know the joint relative frequencies the multiplication rule for two events can be used

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For two events E1 and E2

P(E1 and E2) = P(E1) P(E2|E1)


Multiplication Rule – Tree Diagram

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DieselP(E1) = 0.2

Gasoline P(E1) = 0.8

Truck: P(E3|E1

) = 0.2

Car: P(E4|E1) = 0.5

SUV: P(E5|E

1) = 0.3

P(E1 and E3) = 0.8 x 0.2 = 0.16

P(E1 and E4) = 0.8 x 0.5 = 0.40

P(E1 and E5) = 0.8 x 0.3 = 0.24

P(E2 and E3) = 0.2 x 0.6 = 0.12

P(E2 and E4) = 0.2 x 0.1 = 0.02

P(E3 and E4) = 0.2 x 0.3 = 0.06

Truck: P(E3|E2

) = 0.6

Car: P(E4|E2) = 0.1

SUV: P(E5|E

2) = 0.3


Multiplication Rule for Independent Events

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For independent events E1, E2

P(E1 and E2) = P(E1) P(E2)

• The joint probability of two independent events is simply the product of the probabilities of the two events


Bayes’ Theorem

• A special application of conditional probability

• A way to formally incorporate the new information

• Probability assessment for events of interest may be based on relative frequency or subjectivity

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Bayes’ Theorem Equation

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Bayes’ Theorem Tabular Approach

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Event PriorProbabilities

ConditionalProbabilities

JointProbabilities

RevisedProbabilities

E1 P(E1) P(B | E1) P(E1) P(B | E1) P(E1 | B)

E2 P(E2) P(B | E2) P(E2) P(B | E2) P(E2 |B)

• Step 1: Define the events• Step 2: Determine the prior probability for the events• Step 3: Define an event that if it occurs could alter

the prior probabilities• Step 4: Determine the conditional probabilities• Step 5: Determine the revised probabilities


Bayes’ Theorem Example

• A drilling company has estimated a 40% chance of striking oil for their new well.

• A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests.

• Given that this well has been scheduled for a detailed test, what is the probability

that the well will be successful?

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Bayes’ Theorem Example

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• Let S = successful well and U = unsuccessful well• P(S) = 0.4 , P(U) = 0.6 (prior probabilities)• Define the detailed test event as D• Conditional probabilities: P(D|S) = 0.6 P(D|U) = 0.2• Revised probabilities

Event PriorProbabilities

ConditionalProbabilities

JointProbabilities

RevisedProbabilities

S 0.4 0.6 0.4 x 0.6 = 0.24 0.24/0.36 = 0.67

U 0.6 0.2 0.6 x 0.2 = 0.12 0.12/0.36 = 0.33

Sum = 0.36

Solution: Given the detailed test, the revised probability of a successful well has risen to 0.67 from the original estimate of 0.4Solution: Given the detailed test, the revised probability of a successful well has risen to 0.67 from the original estimate of 0.4

Introduction to probability - UPR-RPepsilon.uprrp.edu/aniel/Lectures/groebner_bsdma9_ch04.pdf ·...

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