Slides Prepared by JOHN S. LOUCKS St. Edward’s University

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© 2006 Thomson/South-Western© 2006 Thomson/South-Western

Slides Prepared bySlides Prepared by

JOHN S. LOUCKSJOHN S. LOUCKSSt. Edward’s UniversitySt. Edward’s University

Slides Prepared bySlides Prepared by

JOHN S. LOUCKSJOHN S. LOUCKSSt. Edward’s UniversitySt. Edward’s University

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Chapter 4Chapter 4 Introduction to Probability Introduction to Probability

Experiments, Counting Rules, Experiments, Counting Rules,

and Assigning Probabilitiesand Assigning Probabilities Events and Their ProbabilityEvents and Their Probability Some Basic RelationshipsSome Basic Relationships

of Probabilityof Probability Conditional ProbabilityConditional Probability Bayes’ TheoremBayes’ Theorem

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Probability as a Numerical MeasureProbability as a Numerical Measureof the Likelihood of Occurrenceof the Likelihood of Occurrence

00 11..55

Increasing Likelihood of OccurrenceIncreasing Likelihood of Occurrence

ProbabilitProbability:y:

The eventThe eventis veryis veryunlikelyunlikelyto occur.to occur.

The occurrenceThe occurrenceof the event isof the event is

just as likely asjust as likely asit is unlikely.it is unlikely.

The eventThe eventis almostis almostcertaincertain

to occur.to occur.

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An Experiment and Its Sample SpaceAn Experiment and Its Sample Space

An An experimentexperiment is any process that generatesis any process that generates well-defined outcomes.well-defined outcomes. An An experimentexperiment is any process that generatesis any process that generates well-defined outcomes.well-defined outcomes.

The The sample spacesample space for an experiment is the set of for an experiment is the set of all experimental outcomes.all experimental outcomes. The The sample spacesample space for an experiment is the set of for an experiment is the set of all experimental outcomes.all experimental outcomes.

An experimental outcome is also called a An experimental outcome is also called a samplesample pointpoint.. An experimental outcome is also called a An experimental outcome is also called a samplesample pointpoint..

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Example: Bradley InvestmentsExample: Bradley Investments

Bradley has invested in two stocks, Markley Oil and Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that theCollins Mining. Bradley has determined that thepossible outcomes of these investments three possible outcomes of these investments three

monthsmonthsfrom now are as follows.from now are as follows.

Investment Gain or LossInvestment Gain or Loss in 3 Months (in $000)in 3 Months (in $000)

Markley OilMarkley Oil Collins MiningCollins Mining

1010 55 002020

8822

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A Counting Rule for A Counting Rule for Multiple-Step ExperimentsMultiple-Step Experiments

If an experiment consists of a sequence of If an experiment consists of a sequence of kk steps steps in which there are in which there are nn11 possible results for the first step, possible results for the first step,

nn22 possible results for the second step, and so on, possible results for the second step, and so on,

then the total number of experimental outcomes isthen the total number of experimental outcomes is given by (given by (nn11)()(nn22) . . . () . . . (nnkk).).

A helpful graphical representation of a multiple-stepA helpful graphical representation of a multiple-step

experiment is a experiment is a tree diagramtree diagram..

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Bradley Investments can be viewed as aBradley Investments can be viewed as atwo-step experiment. It involves two stocks, two-step experiment. It involves two stocks, eacheachwith a set of experimental outcomes.with a set of experimental outcomes.

Markley Oil:Markley Oil: nn11 = 4 = 4

Collins Mining:Collins Mining: nn22 = 2 = 2Total Number of Total Number of

Experimental Outcomes:Experimental Outcomes: nn11nn22 = (4)(2) = 8 = (4)(2) = 8

A Counting Rule for A Counting Rule for Multiple-Step ExperimentsMultiple-Step Experiments

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Tree DiagramTree Diagram

Gain 5Gain 5

Gain 8Gain 8

Gain 8Gain 8

Gain 10Gain 10

Gain 8Gain 8

Gain 8Gain 8

Lose 20Lose 20

Lose 2Lose 2

Lose 2Lose 2

Lose 2Lose 2

Lose 2Lose 2

EvenEven

Markley OilMarkley Oil(Stage 1)(Stage 1)

Collins MiningCollins Mining(Stage 2)(Stage 2)

ExperimentalExperimentalOutcomesOutcomes

(10, 8) (10, 8) Gain $18,000 Gain $18,000

(10, -2) (10, -2) Gain $8,000 Gain $8,000

(5, 8) (5, 8) Gain $13,000 Gain $13,000

(5, -2) (5, -2) Gain $3,000 Gain $3,000

(0, 8) (0, 8) Gain $8,000 Gain $8,000

(0, -2) (0, -2) Lose Lose $2,000$2,000

(-20, 8) (-20, 8) Lose Lose $12,000$12,000

(-20, -2)(-20, -2) Lose Lose $22,000$22,000

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A second useful counting rule enables us to count theA second useful counting rule enables us to count thenumber of experimental outcomes when number of experimental outcomes when nn objects are to objects are tobe selected from a set of be selected from a set of NN objects. objects.

Counting Rule for CombinationsCounting Rule for Combinations

CN

nN

n N nnN

!

!( )!C

N

nN

n N nnN

!

!( )!

Number of Number of CombinationsCombinations of of NN Objects Taken Objects Taken nn at a Time at a Time

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

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Number of Number of PermutationsPermutations of of NN Objects Taken Objects Taken nn at a Time at a Time

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

P nN

nN

N nnN

!!

( )!P n

N

nN

N nnN

!!

( )!

Counting Rule for PermutationsCounting Rule for Permutations

A third useful counting rule enables us to count A third useful counting rule enables us to count thethe

number of experimental outcomes when number of experimental outcomes when nn objects are toobjects are to

be selected from a set of be selected from a set of NN objects, where the objects, where the order oforder of

selection is important.selection is important.

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Assigning ProbabilitiesAssigning Probabilities

Classical MethodClassical Method

Relative Frequency MethodRelative Frequency Method

Subjective MethodSubjective Method

Assigning probabilities based on the assumptionAssigning probabilities based on the assumption of of equally likely outcomesequally likely outcomes

Assigning probabilities based on Assigning probabilities based on experimentationexperimentation or historical dataor historical data

Assigning probabilities based on Assigning probabilities based on judgmentjudgment

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Classical MethodClassical Method

If an experiment has If an experiment has nn possible outcomes, this method possible outcomes, this method

would assign a probability of 1/would assign a probability of 1/nn to each outcome. to each outcome.

Experiment: Rolling a dieExperiment: Rolling a die

Sample Space: Sample Space: SS = {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has aProbabilities: Each sample point has a 1/6 chance of occurring1/6 chance of occurring

ExampleExample

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Number ofNumber ofPolishers RentedPolishers Rented

NumberNumberof Daysof Days

0011223344

44 6618181010 22

Lucas Tool Rental would like to assignLucas Tool Rental would like to assign

probabilities to the number of car polishersprobabilities to the number of car polishers

it rents each day. Office records show the it rents each day. Office records show the followingfollowing

frequencies of daily rentals for the last 40 days.frequencies of daily rentals for the last 40 days.

Example: Lucas Tool RentalExample: Lucas Tool Rental

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Each probability assignment is given byEach probability assignment is given bydividing the frequency (number of days) bydividing the frequency (number of days) bythe total frequency (total number of days).the total frequency (total number of days).


4/404/404/404/40

ProbabilityProbabilityNumber ofNumber of

Polishers RentedPolishers RentedNumberNumberof Daysof Days

0011223344

44 6618181010 224040

.10.10 .15.15 .45.45 .25.25 .05.051.001.00

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When economic conditions and a company’sWhen economic conditions and a company’s circumstances change rapidly it might becircumstances change rapidly it might be inappropriate to assign probabilities based solely oninappropriate to assign probabilities based solely on historical data.historical data. We can use any data available as well as ourWe can use any data available as well as our experience and intuition, but ultimately a probabilityexperience and intuition, but ultimately a probability value should express our value should express our degree of beliefdegree of belief that the that the experimental outcome will occur.experimental outcome will occur.

The best probability estimates often are obtained byThe best probability estimates often are obtained by combining the estimates from the classical or relativecombining the estimates from the classical or relative frequency approach with the subjective estimate.frequency approach with the subjective estimate.

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Applying the subjective method, an analyst Applying the subjective method, an analyst made the following probability assignments.made the following probability assignments.

Exper. OutcomeExper. OutcomeNet Gain Net Gain oror Loss Loss ProbabilityProbability(10, 8)(10, 8)(10, (10, 2)2)(5, 8)(5, 8)(5, (5, 2)2)(0, 8)(0, 8)(0, (0, 2)2)((20, 8)20, 8)((20, 20, 2)2)

$18,000 Gain$18,000 Gain $8,000 Gain$8,000 Gain $13,000 Gain$13,000 Gain $3,000 Gain$3,000 Gain $8,000 Gain$8,000 Gain $2,000 Loss$2,000 Loss $12,000 Loss$12,000 Loss $22,000 Loss$22,000 Loss

.20.20

.08.08

.16.16

.26.26

.10.10

.12.12

.02.02

.06.06

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An An eventevent is a collection of sample points.is a collection of sample points. An An eventevent is a collection of sample points.is a collection of sample points.

The The probability of any eventprobability of any event is equal to the sum of is equal to the sum of the probabilities of the sample points in the event.the probabilities of the sample points in the event. The The probability of any eventprobability of any event is equal to the sum of is equal to the sum of the probabilities of the sample points in the event.the probabilities of the sample points in the event.

If we can identify all the sample points of anIf we can identify all the sample points of an experiment and assign a probability to each, weexperiment and assign a probability to each, we can compute the probability of an event.can compute the probability of an event.

If we can identify all the sample points of anIf we can identify all the sample points of an experiment and assign a probability to each, weexperiment and assign a probability to each, we can compute the probability of an event.can compute the probability of an event.

Events and Their ProbabilitiesEvents and Their Probabilities

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Event Event MM = Markley Oil Profitable = Markley Oil Profitable

MM = {(10, 8), (10, = {(10, 8), (10, 2), (5, 8), (5, 2), (5, 8), (5, 2)}2)}

PP((MM) = ) = PP(10, 8) + (10, 8) + PP(10, (10, 2) + 2) + PP(5, 8) + (5, 8) + PP(5, (5, 2)2)

= .20 + .08 + .16 + .26= .20 + .08 + .16 + .26

= .70= .70

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Event Event CC = Collins Mining Profitable = Collins Mining Profitable

CC = {(10, 8), (5, 8), (0, 8), ( = {(10, 8), (5, 8), (0, 8), (20, 8)}20, 8)}

PP((CC) = ) = PP(10, 8) + (10, 8) + PP(5, 8) + (5, 8) + PP(0, 8) + (0, 8) + PP((20, 8)20, 8)

= .20 + .16 + .10 + .02= .20 + .16 + .10 + .02

= .48= .48

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Some Basic Relationships of ProbabilitySome Basic Relationships of Probability

There are some There are some basic probability relationshipsbasic probability relationships that thatcan be used to compute the probability of an eventcan be used to compute the probability of an eventwithout knowledge of all the sample point probabilities.without knowledge of all the sample point probabilities.

Complement of an EventComplement of an Event Complement of an EventComplement of an Event

Intersection of Two EventsIntersection of Two Events Intersection of Two EventsIntersection of Two Events

Mutually Exclusive EventsMutually Exclusive Events Mutually Exclusive EventsMutually Exclusive Events

Union of Two EventsUnion of Two EventsUnion of Two EventsUnion of Two Events

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The complement of The complement of AA is denoted by is denoted by AAcc.. The complement of The complement of AA is denoted by is denoted by AAcc..

The The complementcomplement of event of event A A is defined to be the eventis defined to be the event consisting of all sample points that are not in consisting of all sample points that are not in A.A. The The complementcomplement of event of event A A is defined to be the eventis defined to be the event consisting of all sample points that are not in consisting of all sample points that are not in A.A.

Complement of an EventComplement of an Event

Event Event AA AAccSampleSpace SSampleSpace S

VennVennDiagraDiagra

mm

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The union of events The union of events AA and and BB is denoted by is denoted by AA BB The union of events The union of events AA and and BB is denoted by is denoted by AA BB

The The unionunion of events of events AA and and BB is the event containing is the event containing all sample points that are in all sample points that are in A A oror B B or both.or both. The The unionunion of events of events AA and and BB is the event containing is the event containing all sample points that are in all sample points that are in A A oror B B or both.or both.

Union of Two EventsUnion of Two Events

SampleSpace SSampleSpace SEvent Event AA Event Event BB

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Union of Two EventsUnion of Two Events



MM CC = Markley Oil Profitable = Markley Oil Profitable oror Collins Mining Profitable Collins Mining Profitable

MM CC = {(10, 8), (10, = {(10, 8), (10, 2), (5, 8), (5, 2), (5, 8), (5, 2), (0, 8), (2), (0, 8), (20, 8)}20, 8)}

PP((MM C)C) = = PP(10, 8) + (10, 8) + PP(10, (10, 2) + 2) + PP(5, 8) + (5, 8) + PP(5, (5, 2)2)

+ + PP(0, 8) + (0, 8) + PP((20, 8)20, 8)

= .20 + .08 + .16 + .26 + .10 + .02= .20 + .08 + .16 + .26 + .10 + .02

= .82= .82

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The intersection of events The intersection of events AA and and BB is denoted by is denoted by AA The intersection of events The intersection of events AA and and BB is denoted by is denoted by AA

The The intersectionintersection of events of events AA and and BB is the set of all is the set of all sample points that are in bothsample points that are in both A A and and BB.. The The intersectionintersection of events of events AA and and BB is the set of all is the set of all sample points that are in bothsample points that are in both A A and and BB..


Intersection of Two EventsIntersection of Two Events

Intersection of A and BIntersection of A and B

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Intersection of Two EventsIntersection of Two Events



MM CC = Markley Oil Profitable = Markley Oil Profitable andand Collins Mining Profitable Collins Mining Profitable

MM CC = {(10, 8), (5, 8)} = {(10, 8), (5, 8)}

PP((MM C)C) = = PP(10, 8) + (10, 8) + PP(5, 8)(5, 8)

= .20 + .16= .20 + .16

= .36= .36

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The The addition lawaddition law provides a way to compute the provides a way to compute the probability of event probability of event A,A, or or B,B, or both or both AA and and B B occurring.occurring. The The addition lawaddition law provides a way to compute the provides a way to compute the probability of event probability of event A,A, or or B,B, or both or both AA and and B B occurring.occurring.

Addition LawAddition Law

The law is written as:The law is written as: The law is written as:The law is written as:

PP((AA BB) = ) = PP((AA) + ) + PP((BB) ) PP((AA BB

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Event Event MM = Markley Oil Profitable = Markley Oil ProfitableEvent Event CC = Collins Mining Profitable = Collins Mining Profitable

MM CC = Markley Oil Profitable = Markley Oil Profitable oror Collins Mining Profitable Collins Mining Profitable

We know: We know: PP((MM) = .70, ) = .70, PP((CC) = .48, ) = .48, PP((MM CC) = .36) = .36

Thus: Thus: PP((MM C) C) = = PP((MM) + P() + P(CC) ) PP((MM CC))

= .70 + .48 = .70 + .48 .36 .36

= .82= .82

Addition LawAddition Law

(This result is the same as that obtained earlier(This result is the same as that obtained earlierusing the definition of the probability of an event.)using the definition of the probability of an event.)

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Mutually Exclusive EventsMutually Exclusive Events

Two events are said to be Two events are said to be mutually exclusivemutually exclusive if the if the events have no sample points in common.events have no sample points in common. Two events are said to be Two events are said to be mutually exclusivemutually exclusive if the if the events have no sample points in common.events have no sample points in common.

Two events are mutually exclusive if, when one eventTwo events are mutually exclusive if, when one event occurs, the other cannot occur.occurs, the other cannot occur. Two events are mutually exclusive if, when one eventTwo events are mutually exclusive if, when one event occurs, the other cannot occur.occurs, the other cannot occur.


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Mutually Exclusive EventsMutually Exclusive Events

If events If events AA and and BB are mutually exclusive, are mutually exclusive, PP((AA BB = 0. = 0. If events If events AA and and BB are mutually exclusive, are mutually exclusive, PP((AA BB = 0. = 0.

The addition law for mutually exclusive events is:The addition law for mutually exclusive events is: The addition law for mutually exclusive events is:The addition law for mutually exclusive events is:

PP((AA BB) = ) = PP((AA) + ) + PP((BB))

there’s no need tothere’s no need toinclude “include “ PP((AA BB””

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The probability of an event given that another eventThe probability of an event given that another event has occurred is called a has occurred is called a conditional probabilityconditional probability.. The probability of an event given that another eventThe probability of an event given that another event has occurred is called a has occurred is called a conditional probabilityconditional probability..

A conditional probability is computed as follows :A conditional probability is computed as follows : A conditional probability is computed as follows :A conditional probability is computed as follows :

The conditional probability of The conditional probability of AA given given BB is denoted is denoted by by PP((AA||BB).). The conditional probability of The conditional probability of AA given given BB is denoted is denoted by by PP((AA||BB).).

Conditional ProbabilityConditional Probability

( )( | )

( )P A B

P A BP B

( )( | )

( )P A B

P A BP B

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We know:We know: P P((MM CC) = .36, ) = .36, PP((MM) = .70 ) = .70

Thus: Thus:

Conditional ProbabilityConditional Probability

( ) .36( | ) .5143

( ) .70P C M

P C MP M

( ) .36( | ) .5143

( ) .70P C M

P C MP M

= Collins Mining Profitable= Collins Mining Profitable givengiven Markley Oil Profitable Markley Oil Profitable

( | )P C M( | )P C M

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Multiplication LawMultiplication Law

The The multiplication lawmultiplication law provides a way to compute the provides a way to compute the probability of the intersection of two events.probability of the intersection of two events. The The multiplication lawmultiplication law provides a way to compute the provides a way to compute the probability of the intersection of two events.probability of the intersection of two events.


PP((AA BB) = ) = PP((BB))PP((AA||BB))

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We know:We know: P P((MM) = .70, ) = .70, PP((CC||MM) = .5143) = .5143

Multiplication LawMultiplication Law

MM CC = Markley Oil Profitable = Markley Oil Profitable andand Collins Mining Profitable Collins Mining Profitable

Thus: Thus: PP((MM C) C) = = PP((MM))PP((M|CM|C))= (.70)(.5143)= (.70)(.5143)

= .36= .36

(This result is the same as that obtained earlier(This result is the same as that obtained earlierusing the definition of the probability of an event.)using the definition of the probability of an event.)

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Independent EventsIndependent Events

If the probability of event If the probability of event AA is not changed by the is not changed by the existence of event existence of event BB, we would say that events , we would say that events AA and and BB are are independentindependent..

If the probability of event If the probability of event AA is not changed by the is not changed by the existence of event existence of event BB, we would say that events , we would say that events AA and and BB are are independentindependent..

Two events Two events AA and and BB are independent if: are independent if: Two events Two events AA and and BB are independent if: are independent if:

PP((AA||BB) = ) = PP((AA)) PP((BB||AA) = ) = PP((BB))oror

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The multiplication law also can be used as a test to seeThe multiplication law also can be used as a test to see if two events are independent.if two events are independent. The multiplication law also can be used as a test to seeThe multiplication law also can be used as a test to see if two events are independent.if two events are independent.


PP((AA BB) = ) = PP((AA))PP((BB))

Multiplication LawMultiplication Lawfor Independent Eventsfor Independent Events

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Multiplication LawMultiplication Lawfor Independent Eventsfor Independent Events


We know:We know: P P((MM CC) = .36, ) = .36, PP((MM) = .70, ) = .70, PP((CC) = .48) = .48 But: But: PP((M)P(C) M)P(C) = (.70)(.48) = .34, not .36= (.70)(.48) = .34, not .36

Are events Are events MM and and CC independent? independent?DoesDoesPP((MM CC) = ) = PP((M)P(C) M)P(C) ??

Hence:Hence: M M and and CC are are notnot independent. independent.

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Bayes’ TheoremBayes’ Theorem

NewNewInformationInformation

NewNewInformationInformation

ApplicationApplicationof Bayes’of Bayes’TheoremTheorem

ApplicationApplicationof Bayes’of Bayes’TheoremTheorem

PosteriorPosteriorProbabilitiesProbabilities


PriorPriorProbabilitiesProbabilities


Often we begin probability analysis with initial orOften we begin probability analysis with initial or prior probabilitiesprior probabilities..

Then, from a sample, special report, or a productThen, from a sample, special report, or a product test we obtain some additional information.test we obtain some additional information. Given this information, we calculate revised orGiven this information, we calculate revised or posterior probabilitiesposterior probabilities..

Bayes’ theoremBayes’ theorem provides the means for revising the provides the means for revising the prior probabilities.prior probabilities.

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A proposed shopping center A proposed shopping center will provide strong competitionwill provide strong competitionfor downtown businesses likefor downtown businesses likeL. S. Clothiers. If the shoppingL. S. Clothiers. If the shoppingcenter is built, the owner of center is built, the owner of L. S. Clothiers feels it would be best toL. S. Clothiers feels it would be best torelocate to the center. relocate to the center.


Example: L. S. ClothiersExample: L. S. Clothiers

The shopping center cannot be built unless aThe shopping center cannot be built unless azoning change is approved by the town council. Thezoning change is approved by the town council. Theplanning board must first make a recommendation, forplanning board must first make a recommendation, foror against the zoning change, to the council.or against the zoning change, to the council.

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Prior ProbabilitiesPrior Probabilities

Let:Let:


AA11 = town council approves the zoning change = town council approves the zoning change

AA22 = town council disapproves the change = town council disapproves the change

P(P(AA11) = .7, P() = .7, P(AA22) = .3) = .3

Using subjective judgment:Using subjective judgment:

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New InformationNew Information

The planning board has recommended The planning board has recommended against against the zoning change. Let the zoning change. Let BB denote the denote the event of a negative recommendation by the event of a negative recommendation by the planning board.planning board.

Given that Given that BB has occurred, should L. S. has occurred, should L. S. Clothiers revise the probabilities that the town Clothiers revise the probabilities that the town council will approve or disapprove the zoning council will approve or disapprove the zoning change?change?


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Conditional ProbabilitiesConditional Probabilities

Past history with the planning board and Past history with the planning board and the town council indicates the following:the town council indicates the following:


PP((BB||AA11) = .2) = .2 PP((BB||AA22) = .9) = .9

PP((BBCC||AA11) = .8) = .8 PP((BBCC||AA22) = .1) = .1Hence:Hence:

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P(Bc|A1) = .8P(Bc|A1) = .8P(A1) = .7P(A1) = .7

P(A2) = .3P(A2) = .3

P(B|A2) = .9P(B|A2) = .9

P(Bc|A2) = .1P(Bc|A2) = .1

P(B|A1) = .2P(B|A1) = .2 P(A1 B) = .14P(A1 B) = .14

P(A2 B) = .27P(A2 B) = .27

P(A2 Bc) = .03P(A2 Bc) = .03

P(A1 Bc) = .56P(A1 Bc) = .56


Tree DiagramTree Diagram

Town CouncilTown Council Planning BoardPlanning Board ExperimentalExperimentalOutcomesOutcomes

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1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )i i

in n

P A P B AP A B

P A P B A P A P B A P A P B A

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | ) ... ( ) ( | )i i

in n

P A P B AP A B

P A P B A P A P B A P A P B A

To find the posterior probability that event To find the posterior probability that event AAii will will occur given that eventoccur given that event B B has occurred, we applyhas occurred, we apply Bayes’ theoremBayes’ theorem..

Bayes’ theorem is applicable when the events forBayes’ theorem is applicable when the events for which we want to compute posterior probabilitieswhich we want to compute posterior probabilities are mutually exclusive and their union is the entireare mutually exclusive and their union is the entire sample space.sample space.

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Posterior ProbabilitiesPosterior Probabilities

Given the planning board’s Given the planning board’s recommendation not to approve the zoning recommendation not to approve the zoning change, we revise the prior probabilities as change, we revise the prior probabilities as follows:follows:

1 11

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )P A P B A

P A BP A P B A P A P B A

1 11

1 1 2 2

( ) ( | )( | )

( ) ( | ) ( ) ( | )P A P B A

P A BP A P B A P A P B A

(. )(. )(. )(. ) (. )(. )

7 27 2 3 9

(. )(. )(. )(. ) (. )(. )

7 27 2 3 9


= .34= .34

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ConclusionConclusion

The planning board’s recommendation is The planning board’s recommendation is good news for L. S. Clothiers. The posterior good news for L. S. Clothiers. The posterior probability of the town council approving the probability of the town council approving the zoning change is .34 compared to a prior zoning change is .34 compared to a prior probability of .70.probability of .70.


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Tabular ApproachTabular Approach

Step 1Step 1

Prepare the following three columns:Prepare the following three columns:

Column 1Column 1 The mutually exclusive events for which The mutually exclusive events for which posterior probabilities are desired.posterior probabilities are desired.

Column 2Column 2 The prior probabilities for the events. The prior probabilities for the events.

Column 3Column 3 The conditional probabilities of the new The conditional probabilities of the new information information givengiven each event. each event.

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(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)

EventsEvents

AAii


PP((AAii))

ConditionalConditionalProbabilitiesProbabilities

PP((BB||AAii))

AA11

AA22

.7.7

.3.3

1.01.0

.2.2

.9.9

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Step 2Step 2

Column 4Column 4

Compute the joint probabilities for each Compute the joint probabilities for each event and the new information event and the new information BB by using the by using the multiplication law.multiplication law.

Multiply the prior probabilities in column 2 Multiply the prior probabilities in column 2 by the corresponding conditional probabilities by the corresponding conditional probabilities in column 3. That is, in column 3. That is, PP((AAi i BB) = ) = PP((AAii) ) PP((BB||AAii). ).

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(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)

EventsEvents

AAii


PP((AAii))


PP((BB||AAii))

AA11

AA22

.7.7

.3.3

1.01.0

.2.2

.9.9

.14.14

.27.27

JointJointProbabilitiesProbabilities

PP((AAi i BB))

.7 x .2.7 x .2.7 x .2.7 x .2

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Step 2 (continued)Step 2 (continued)

We see that there is a .14 probability of the townWe see that there is a .14 probability of the town council approving the zoning change and a negativecouncil approving the zoning change and a negative recommendation by the planning board. recommendation by the planning board. There is a .27 probability of the town councilThere is a .27 probability of the town council disapproving the zoning change and a negativedisapproving the zoning change and a negative recommendation by the planning board.recommendation by the planning board.

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Step 3Step 3

Column 4Column 4 Sum the joint probabilities. The sum is theSum the joint probabilities. The sum is theprobability of the new information, probability of the new information, PP((BB). The sum). The sum.14 + .27 shows an overall probability of .41 of a.14 + .27 shows an overall probability of .41 of anegative recommendation by the planning board.negative recommendation by the planning board.

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(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)

EventsEvents

AAii


PP((AAii))


PP((BB||AAii))

AA11

AA22

.7.7

.3.3

1.01.0

.2.2

.9.9

.14.14

.27.27


PP((AAi i BB))

PP((BB) = .41) = .41

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Step 4Step 4

Column 5Column 5

Compute the posterior probabilities using Compute the posterior probabilities using the basic relationship of conditional probability.the basic relationship of conditional probability.

The joint probabilities The joint probabilities PP((AAi i BB) are in ) are in column 4 and the probability column 4 and the probability PP((BB) is the sum of ) is the sum of column 4.column 4.


)(

)()|(

BP

BAPBAP i

i

)(

)()|(

BP

BAPBAP i

i

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(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)

EventsEvents

AAii


PP((AAii))


PP((BB||AAii))

AA11

AA22

.7.7

.3.3

1.01.0

.2.2

.9.9

.14.14

.27.27


PP((AAi i BB))

PP((BB) = .41) = .41


.14/.4.14/.411

.14/.4.14/.411


PP((AAii ||BB))

..34153415

.6585.6585

1.00001.0000

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End of Chapter 4End of Chapter 4

Documents

Slides Prepared by JOHN S. LOUCKS St. Edward’s University