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1Slide© 2005 Thomson/South-Western
Slides Prepared byJOHN S. LOUCKS
St. Edward’s University
2Slide© 2005 Thomson/South-Western
Chapter 4Introduction to Probability
� Experiments, Counting Rules, and Assigning Probabilities
� Events and Their Probability� Some Basic Relationships
of Probability� Conditional Probability� Bayes’ Theorem
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Probability as a Numerical Measureof the Likelihood of Occurrence
0 1.5
Increasing Likelihood of Occurrence
Probability:
The eventis very
unlikelyto occur.
The occurrenceof the event isjust as likely asit is unlikely.
The eventis almostcertain
to occur.
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An Experiment and Its Sample Space
An experiment is any process that generateswell-defined outcomes.
The sample space for an experiment is the set ofall experimental outcomes.
An experimental outcome is also called a samplepoint.
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Example: Bradley Investments
Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that thepossible outcomes of these investments three monthsfrom now are as follows.
Investment Gain or Lossin 3 Months (in $000)
Markley Oil Collins Mining1050
−20
8−2
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A Counting Rule for Multiple-Step Experiments
If an experiment consists of a sequence of k stepsin which there are n1 possible results for the first step,n2 possible results for the second step, and so on, then the total number of experimental outcomes isgiven by (n1)(n2) . . . (nk).
A helpful graphical representation of a multiple-stepexperiment is a tree diagram.
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Bradley Investments can be viewed as atwo-step experiment. It involves two stocks, eachwith a set of experimental outcomes.
Markley Oil: n1 = 4Collins Mining: n2 = 2Total Number of Experimental Outcomes: n1n2 = (4)(2) = 8
A Counting Rule for Multiple-Step Experiments
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Tree Diagram
Gain 5
Gain 8
Gain 8
Gain 10
Gain 8
Gain 8
Lose 20
Lose 2
Lose 2
Lose 2
Lose 2
Even
Markley Oil(Stage 1)
Collins Mining(Stage 2)
ExperimentalOutcomes
(10, 8) Gain $18,000
(10, -2) Gain $8,000
(5, 8) Gain $13,000
(5, -2) Gain $3,000
(0, 8) Gain $8,000
(0, -2) Lose $2,000
(-20, 8) Lose $12,000
(-20, -2) Lose $22,000
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A second useful counting rule enables us to count thenumber of experimental outcomes when n objects are tobe selected from a set of N objects.
Counting Rule for Combinations
CNn
Nn N nn
N =⎛⎝⎜
⎞⎠⎟
=−!
!( )!C
Nn
Nn N nn
N =⎛⎝⎜
⎞⎠⎟
=−!
!( )!
Number of Combinations of N Objects Taken n at a Time
where: N! = N(N − 1)(N − 2) . . . (2)(1)n! = n(n − 1)(n − 2) . . . (2)(1)0! = 1
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Number of Permutations of N Objects Taken n at a Time
where: N! = N(N − 1)(N − 2) . . . (2)(1)n! = n(n − 1)(n − 2) . . . (2)(1)0! = 1
P nNn
NN nn
N =⎛⎝⎜
⎞⎠⎟
=−
! !( )!
P nNn
NN nn
N =⎛⎝⎜
⎞⎠⎟
=−
! !( )!
Counting Rule for Permutations
A third useful counting rule enables us to count thenumber of experimental outcomes when n objects are tobe selected from a set of N objects, where the order ofselection is important.
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Assigning Probabilities
Classical Method
Relative Frequency Method
Subjective Method
Assigning probabilities based on the assumptionof equally likely outcomes
Assigning probabilities based on experimentationor historical data
Assigning probabilities based on judgment
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Classical Method
If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome.
Experiment: Rolling a dieSample Space: S = {1, 2, 3, 4, 5, 6}Probabilities: Each sample point has a
1/6 chance of occurring
Example
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Relative Frequency Method
Number ofPolishers Rented
Numberof Days
01234
46
18102
Lucas Tool Rental would like to assignprobabilities to the number of car polishersit rents each day. Office records show the followingfrequencies of daily rentals for the last 40 days.
Example: Lucas Tool Rental
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Each probability assignment is given bydividing the frequency (number of days) bythe total frequency (total number of days).
Relative Frequency Method
4/40
ProbabilityNumber of
Polishers RentedNumberof Days
01234
46
18102
40
.10
.15
.45
.25
.051.00
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Subjective Method
When economic conditions and a company’scircumstances change rapidly it might beinappropriate to assign probabilities based solely onhistorical data.We can use any data available as well as ourexperience and intuition, but ultimately a probabilityvalue should express our degree of belief that theexperimental outcome will occur.
The best probability estimates often are obtained bycombining the estimates from the classical or relativefrequency approach with the subjective estimate.
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Subjective Method
Applying the subjective method, an analyst made the following probability assignments.
Exper. Outcome Net Gain or Loss Probability(10, 8)(10, −2)(5, 8)(5, −2)(0, 8)(0, −2)(−20, 8)(−20, −2)
$18,000 Gain$8,000 Gain
$13,000 Gain$3,000 Gain$8,000 Gain$2,000 Loss
$12,000 Loss$22,000 Loss
.20
.08
.16
.26
.10
.12
.02
.06
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An event is a collection of sample points.
The probability of any event is equal to the sum ofthe probabilities of the sample points in the event.
If we can identify all the sample points of anexperiment and assign a probability to each, wecan compute the probability of an event.
Events and Their Probabilities
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Events and Their Probabilities
Event M = Markley Oil ProfitableM = {(10, 8), (10, −2), (5, 8), (5, −2)}
P(M) = P(10, 8) + P(10, −2) + P(5, 8) + P(5, −2)= .20 + .08 + .16 + .26= .70
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Events and Their Probabilities
Event C = Collins Mining ProfitableC = {(10, 8), (5, 8), (0, 8), (−20, 8)}
P(C) = P(10, 8) + P(5, 8) + P(0, 8) + P(−20, 8)= .20 + .16 + .10 + .02= .48
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Some Basic Relationships of Probability
There are some basic probability relationships thatcan be used to compute the probability of an eventwithout knowledge of all the sample point probabilities.
Complement of an Event
Intersection of Two Events
Mutually Exclusive Events
Union of Two Events
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The complement of A is denoted by Ac.
The complement of event A is defined to be the eventconsisting of all sample points that are not in A.
Complement of an Event
Event A AcSampleSpace S
VennDiagram
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The union of events A and B is denoted by A ∪ B.
The union of events A and B is the event containingall sample points that are in A or B or both.
Union of Two Events
SampleSpace SEvent A Event B
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Union of Two Events
Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable
M ∪ C = Markley Oil Profitable or Collins Mining Profitable
M ∪ C = {(10, 8), (10, −2), (5, 8), (5, −2), (0, 8), (−20, 8)}P(M ∪ C) = P(10, 8) + P(10, −2) + P(5, 8) + P(5, −2)
+ P(0, 8) + P(−20, 8)= .20 + .08 + .16 + .26 + .10 + .02= .82
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The intersection of events A and B is denoted by A ∩ Β.
The intersection of events A and B is the set of allsample points that are in both A and B.
SampleSpace SEvent A Event B
Intersection of Two Events
Intersection of A and B
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Intersection of Two Events
Event M = Markley Oil ProfitableEvent C = Collins Mining ProfitableM ∩ C = Markley Oil Profitable
and Collins Mining ProfitableM ∩ C = {(10, 8), (5, 8)}
P(M ∩ C) = P(10, 8) + P(5, 8)
= .20 + .16
= .36
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The addition law provides a way to compute theprobability of event A, or B, or both A and B occurring.
Addition Law
The law is written as:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
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Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable
M ∪ C = Markley Oil Profitable or Collins Mining Profitable
We know: P(M) = .70, P(C) = .48, P(M ∩ C) = .36Thus: P(M ∪ C) = P(M) + P(C) − P(M ∩ C)
= .70 + .48 − .36= .82
Addition Law
(This result is the same as that obtained earlierusing the definition of the probability of an event.)
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Mutually Exclusive Events
Two events are said to be mutually exclusive if theevents have no sample points in common.
Two events are mutually exclusive if, when one eventoccurs, the other cannot occur.
SampleSpace SEvent A Event B
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Mutually Exclusive Events
If events A and B are mutually exclusive, P(A ∩ B) = 0.
The addition law for mutually exclusive events is:
P(A ∪ B) = P(A) + P(B)
there’s no need toinclude “− P(A ∩ B)”
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The probability of an event given that another eventhas occurred is called a conditional probability.
A conditional probability is computed as follows :
The conditional probability of A given B is denotedby P(A|B).
Conditional Probability
( )( | )( )
P A BP A BP B
∩=
( )( | )( )
P A BP A BP B
∩=
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Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable
We know: P(M ∩ C) = .36, P(M) = .70
Thus:
Conditional Probability
( ) .36( | ) .5143( ) .70
P C MP C MP M
∩= = =
( ) .36( | ) .5143( ) .70
P C MP C MP M
∩= = =
= Collins Mining Profitablegiven Markley Oil Profitable
( | )P C M( | )P C M
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Multiplication Law
The multiplication law provides a way to compute theprobability of the intersection of two events.
The law is written as:
P(A ∩ B) = P(B)P(A|B)
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Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable
We know: P(M) = .70, P(C|M) = .5143
Multiplication Law
M ∩ C = Markley Oil Profitableand Collins Mining Profitable
Thus: P(M ∩ C) = P(M)P(M|C)= (.70)(.5143)= .36
(This result is the same as that obtained earlierusing the definition of the probability of an event.)
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Independent Events
If the probability of event A is not changed by theexistence of event B, we would say that events Aand B are independent.
Two events A and B are independent if:
P(A|B) = P(A) P(B|A) = P(B)or
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The multiplication law also can be used as a test to seeif two events are independent.
The law is written as:
P(A ∩ B) = P(A)P(B)
Multiplication Lawfor Independent Events
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Multiplication Lawfor Independent Events
Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable
We know: P(M ∩ C) = .36, P(M) = .70, P(C) = .48But: P(M)P(C) = (.70)(.48) = .34, not .36
Are events M and C independent?Does P(M ∩ C) = P(M)P(C) ?
Hence: M and C are not independent.
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Bayes’ Theorem
NewInformation
Applicationof Bayes’Theorem
PosteriorProbabilities
PriorProbabilities
Often we begin probability analysis with initial orprior probabilities.Then, from a sample, special report, or a producttest we obtain some additional information.Given this information, we calculate revised orposterior probabilities.
Bayes’ theorem provides the means for revising theprior probabilities.
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A proposed shopping center will provide strong competitionfor downtown businesses likeL. S. Clothiers. If the shoppingcenter is built, the owner of L. S. Clothiers feels it would be best torelocate to the center.
Bayes’ Theorem
Example: L. S. Clothiers
The shopping center cannot be built unless azoning change is approved by the town council. Theplanning board must first make a recommendation, foror against the zoning change, to the council.
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� Prior ProbabilitiesLet:
Bayes’ Theorem
A1 = town council approves the zoning changeA2 = town council disapproves the change
P(A1) = .7, P(A2) = .3
Using subjective judgment:
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� New InformationThe planning board has recommended against the
zoning change. Let B denote the event of a negative recommendation by the planning board.
Given that B has occurred, should L. S. Clothiers revise the probabilities that the town council will approve or disapprove the zoning change?
Bayes’ Theorem
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� Conditional ProbabilitiesPast history with the planning board and the
town council indicates the following:
Bayes’ Theorem
P(B|A1) = .2 P(B|A2) = .9
P(BC|A1) = .8 P(BC|A2) = .1Hence:
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P(Bc|A1) = .8P(A1) = .7
P(A2) = .3P(B|A2) = .9
P(Bc|A2) = .1
P(B|A1) = .2 P(A1 ∩ B) = .14
P(A2 ∩ B) = .27
P(A2 ∩ Bc) = .03
P(A1 ∩ Bc) = .56
Bayes’ Theorem
Tree Diagram
Town Council Planning Board ExperimentalOutcomes
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Bayes’ Theorem
1 1 2 2
( ) ( | )( | )( ) ( | ) ( ) ( | ) ... ( ) ( | )
i ii
n n
P A P B AP A BP A P B A P A P B A P A P B A
=+ + +1 1 2 2
( ) ( | )( | )( ) ( | ) ( ) ( | ) ... ( ) ( | )
i ii
n n
P A P B AP A BP A P B A P A P B A P A P B A
=+ + +
To find the posterior probability that event Ai willoccur given that event B has occurred, we applyBayes’ theorem.
Bayes’ theorem is applicable when the events forwhich we want to compute posterior probabilitiesare mutually exclusive and their union is the entiresample space.
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� Posterior ProbabilitiesGiven the planning board’s recommendation not
to approve the zoning change, we revise the prior probabilities as follows:
1 11
1 1 2 2
( ) ( | )( | )( ) ( | ) ( ) ( | )
P A P B AP A BP A P B A P A P B A
=+
1 11
1 1 2 2
( ) ( | )( | )( ) ( | ) ( ) ( | )
P A P B AP A BP A P B A P A P B A
=+
=+
(. )(. )(. )(. ) (. )(. )
7 27 2 3 9
=+
(. )(. )(. )(. ) (. )(. )
7 27 2 3 9
Bayes’ Theorem
= .34
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� ConclusionThe planning board’s recommendation is good
news for L. S. Clothiers. The posterior probability of the town council approving the zoning change is .34 compared to a prior probability of .70.
Bayes’ Theorem
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Tabular Approach
� Step 1Prepare the following three columns:Column 1 − The mutually exclusive events for whichposterior probabilities are desired.Column 2 − The prior probabilities for the events.Column 3 − The conditional probabilities of the newinformation given each event.
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Tabular Approach
(1) (2) (3) (4) (5)
EventsAi
PriorProbabilities
P(Ai)
ConditionalProbabilities
P(B|Ai)
A1
A2
.7
.31.0
.2
.9
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Tabular Approach
� Step 2Column 4
Compute the joint probabilities for each event and the new information B by using the multiplication law.
Multiply the prior probabilities in column 2 by the corresponding conditional probabilities in column 3. That is, P(Ai ∩B) = P(Ai) P(B|Ai).
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Tabular Approach
(1) (2) (3) (4) (5)
EventsAi
PriorProbabilities
P(Ai)
ConditionalProbabilities
P(B|Ai)
A1
A2
.7
.31.0
.2
.9.14.27
JointProbabilities
P(Ai ∩B)
.7 x .2
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Tabular Approach
� Step 2 (continued)We see that there is a .14 probability of the town
council approving the zoning change and a negativerecommendation by the planning board.
There is a .27 probability of the town councildisapproving the zoning change and a negativerecommendation by the planning board.
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Tabular Approach
� Step 3Column 4
Sum the joint probabilities. The sum is theprobability of the new information, P(B). The sum.14 + .27 shows an overall probability of .41 of anegative recommendation by the planning board.
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Tabular Approach
(1) (2) (3) (4) (5)
EventsAi
PriorProbabilities
P(Ai)
ConditionalProbabilities
P(B|Ai)
A1
A2
.7
.31.0
.2
.9.14.27
JointProbabilities
P(Ai ∩B)
P(B) = .41
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� Step 4Column 5
Compute the posterior probabilities using the basic relationship of conditional probability.
The joint probabilities P(Ai ∩B) are in column 4 and the probability P(B) is the sum of column 4.
Tabular Approach
)()()|(
BPBAPBAP i
i∩
=)(
)()|(BP
BAPBAP ii
∩=
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(1) (2) (3) (4) (5)
EventsAi
PriorProbabilities
P(Ai)
ConditionalProbabilities
P(B|Ai)
A1
A2
.7
.31.0
.2
.9.14.27
JointProbabilities
P(Ai ∩B)
P(B) = .41
Tabular Approach
.14/.41
PosteriorProbabilities
P(Ai |B)
.3415
.65851.0000
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End of Chapter 4