1 1 © 2003 Thomson /South-Western Slide Slides Prepared by JOHN S. LOUCKS St. Edward’s University.
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Slides by
JOHNLOUCKSSt. Edward’sUniversity
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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’ Bayes’ TheoremTheorem
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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 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 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.
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 a two-Bradley Investments can be viewed as a two-stepstep
experiment. It involves two stocks, each with a set experiment. It involves two stocks, each with a set ofof
experimental outcomes.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 MethodClassical MethodClassical Method
Relative Frequency MethodRelative Frequency MethodRelative Frequency MethodRelative Frequency Method
Subjective MethodSubjective MethodSubjective 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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Relative Frequency MethodRelative Frequency Method
Number ofNumber ofPolishers RentedPolishers Rented
NumberNumberof Daysof Days
0011223344
44 6618181010 22
Lucas Tool Rental would like to assign Lucas Tool Rental would like to assign probabilitiesprobabilitiesto the number of car polishers it rents each day. to the number of car polishers it rents each day. OfficeOfficerecords show the following frequencies of daily records show the following frequencies of daily rentals for the last 40 days.rentals for the last 40 days.
Example: Lucas Tool Rental
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Each probability assignment is given by dividing theEach probability assignment is given by dividing thefrequency (number of days) by the total frequency (totalfrequency (number of days) by the total frequency (totalnumber of days).number of days).
Relative Frequency MethodRelative Frequency Method
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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Subjective MethodSubjective Method
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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Subjective MethodSubjective Method
Applying the subjective method, an analyst madeApplying the subjective method, an analyst madethe following probability assignments.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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Events and Their ProbabilitiesEvents and Their Probabilities
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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Events and Their ProbabilitiesEvents and Their Probabilities
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
Event Event MM = Markley Oil Profitable = Markley Oil Profitable
Event Event CC = Collins Mining Profitable = Collins Mining Profitable
MM CC = Markley Oil Profitable = Markley Oil Profitable oror Collins Mining Profitable (or both) Collins Mining Profitable (or both)
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..
SampleSpace SSampleSpace SEvent Event AA Event Event 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
Event Event MM = Markley Oil Profitable = Markley Oil Profitable
Event Event CC = Collins Mining Profitable = Collins Mining Profitable
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 BBPP((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 (or both) Collins Mining Profitable (or both)
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.
SampleSpace SSampleSpace SEvent Event AA Event Event BB
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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))PP((AA BB) = ) = PP((AA) + ) + PP((BB))
There is no need toThere is no need toinclude “include “ PP((AA BB””There is no need toThere is 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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Event Event MM = Markley Oil Profitable = Markley Oil Profitable
Event Event CC = Collins Mining Profitable = Collins Mining Profitable
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.
The law is written as:The law is written as: The law is written as:The law is written as:
PP((AA BB) = ) = PP((BB))PP((AA||BB))PP((AA BB) = ) = 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
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((AA||BB) = ) = PP((AA)) PP((BB||AA) = ) = PP((BB))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.
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) = ) = PP((AA))PP((BB))
Multiplication LawMultiplication Lawfor Independent Eventsfor Independent Events
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Multiplication LawMultiplication Lawfor Independent Eventsfor Independent Events
Event Event MM = Markley Oil Profitable = Markley Oil ProfitableEvent Event CC = Collins Mining Profitable = Collins Mining Profitable
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
PosteriorPosteriorProbabilitiesProbabilities
PriorPriorProbabilitiesProbabilities
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 will provide strongA proposed shopping center will provide strongCompetition for downtown businesses like L. S.Competition for downtown businesses like L. S.Clothiers. If the shopping center is built, the owner of Clothiers. If the shopping center is built, the owner of L. S. Clothiers feels it would be best to relocate to theL. S. Clothiers feels it would be best to relocate to thecenter. center.
Bayes’ TheoremBayes’ Theorem
Example: L. S. Clothiers
The shopping center cannot be built unless a The shopping center cannot be built unless a zoningzoningchange is approved by the town council. The change is approved by the town council. The planningplanningboard must first make a recommendation, for or board must first make a recommendation, for or againstagainstthe zoning change, to the council.the zoning change, to the council.
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Prior ProbabilitiesPrior Probabilities
Let:Let:
Bayes’ TheoremBayes’ Theorem
AA11 = town council approves the zoning change = town council approves the zoning change
AA22 = town council disapproves the change = town council disapproves the change
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) = .3P(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 againstagainst 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?
Bayes’ TheoremBayes’ Theorem
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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:
Bayes’ TheoremBayes’ Theorem
PP((BB||AA11) = .2) = .2PP((BB||AA11) = .2) = .2 PP((BB||AA22) = .9) = .9PP((BB||AA22) = .9) = .9
PP((BBCC||AA11) = .8) = .8PP((BBCC||AA11) = .8) = .8 PP((BBCC||AA22) = .1) = .1PP((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
Bayes’ TheoremBayes’ Theorem
Tree DiagramTree Diagram
Town CouncilTown Council Planning BoardPlanning Board ExperimentalExperimentalOutcomesOutcomes
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Bayes’ TheoremBayes’ Theorem
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
Bayes’ TheoremBayes’ Theorem
= .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.
Bayes’ TheoremBayes’ Theorem
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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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Tabular ApproachTabular Approach
(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)
EventsEvents
AAii
PriorPriorProbabilitiesProbabilities
PP((AAii))
ConditionalConditionalProbabilitiesProbabilities
PP((BB||AAii))
AA11
AA22
.7.7
.3.3
1.01.0
.2.2
.9.9
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Tabular ApproachTabular Approach
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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Tabular ApproachTabular Approach
(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)
EventsEvents
AAii
PriorPriorProbabilitiesProbabilities
PP((AAii))
ConditionalConditionalProbabilitiesProbabilities
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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Tabular ApproachTabular Approach
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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Tabular ApproachTabular Approach
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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Tabular ApproachTabular Approach
(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)
EventsEvents
AAii
PriorPriorProbabilitiesProbabilities
PP((AAii))
ConditionalConditionalProbabilitiesProbabilities
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))
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.
Tabular ApproachTabular Approach
)(
)()|(
BP
BAPBAP i
i
)(
)()|(
BP
BAPBAP i
i
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(1)(1) (2)(2) (3)(3) (4)(4) (5)(5)
EventsEvents
AAii
PriorPriorProbabilitiesProbabilities
PP((AAii))
ConditionalConditionalProbabilitiesProbabilities
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))
PP((BB) = .41) = .41
Tabular ApproachTabular Approach
.14/.4.14/.411
.14/.4.14/.411
PosteriorPosteriorProbabilitiesProbabilities
PP((AAii ||BB))
..34153415
.6585.6585
1.00001.0000
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End of Chapter 4End of Chapter 4