Experiments, Sample Spaces, and Events Definition of Probability Rules of Probability

77

Experiments, Sample Spaces, and EventsExperiments, Sample Spaces, and Events Definition of ProbabilityDefinition of Probability Rules of ProbabilityRules of Probability Use of Counting Techniques in ProbabilityUse of Counting Techniques in Probability Conditional Probability and Independent EventsConditional Probability and Independent Events Bayes’ TheoremBayes’ Theorem

Probability Probability

7.17.1Experiments, Sample Spaces, and EventsExperiments, Sample Spaces, and Events

HH (H, H, H)(H, H, H)

TT (H, H, T)(H, H, T)

H H (H, T, H)(H, T, H)

T T (H, T, T)(H, T, T)

H H (T, H, H)(T, H, H)

T T (T, H, T)(T, H, T)

HH (T, T, H)(T, T, H)

T T (T, T, T) (T, T, T)

FirstFirst SecondSecond ThirdThird SampleSampletosstoss tosstoss tosstoss pointspoints

HH

TT

HH

TT

HH

TT

TerminologyTerminology

ExperimentExperiment An experiment is an An experiment is an activityactivity with with

observable resultsobservable results.. The The resultsresults of an experiment are called of an experiment are called

outcomesoutcomes of the experiment. of the experiment.

ExamplesExamples

Tossing a coinTossing a coin and observing whether it falls and observing whether it falls headsheads or or tailstails

Rolling a dieRolling a die and observing which of the numbers and observing which of the numbers 11, , 22, , 33, , 44, , 55, or , or 66 shows up shows up

Testing a spark plugTesting a spark plug from a batch of from a batch of 100100 spark spark plugs and observing plugs and observing whether or notwhether or not it is it is defectivedefective


Sample Point, Sample Space, and EventSample Point, Sample Space, and Event Sample point: Sample point:

An An outcomeoutcome of an experiment of an experiment Sample space:Sample space:

The The setset consisting of all possible consisting of all possible sample sample pointspoints of an experiment of an experiment

Event:Event:A A subsetsubset of a of a sample spacesample space of an experiment of an experiment

ExampleExample

Describe the Describe the sample spacesample space associated with the associated with the experimentexperiment of of tossing a cointossing a coin and observing whether it falls and observing whether it falls headsheads or or tailstails..

What are the What are the eventsevents of this of this experimentexperiment??SolutionSolution The two The two outcomesoutcomes are are headsheads and and tailstails, and the required , and the required

sample spacesample space is given by is given by

S S = {H, T}= {H, T}

where where HH denotes the denotes the outcomeoutcome headsheads and and TT denotes the denotes the outcomeoutcome tailstails..

The The eventsevents of the experiment, the of the experiment, the subsetssubsets of of SS, are, are

Ø, {H}, {T}, Ø, {H}, {T}, SS

Example 1, page 354Example 1, page 354


Union of Two EventsUnion of Two Events The union of two events The union of two events EE and and F F is the is the

event event E E FF.. Thus, the event Thus, the event E E F F contains the contains the set of set of

outcomesoutcomes of of EE and/orand/or FF..


Intersection of Two EventsIntersection of Two Events The intersection of two events The intersection of two events EE and and F F is is

the event the event E E FF.. Thus, the event Thus, the event E E F F contains the contains the set of set of

outcomesoutcomes common to common to EE andand FF..


Complement of an EventComplement of an Event The complement of an event The complement of an event EE is the event is the event

EE cc..

Thus, the event Thus, the event EE cc is the is the setset containing containing allall

thethe outcomes outcomes in thein the sample space sample space SS that that areare notnot in in EE..

ExampleExample

Consider the Consider the experimentexperiment of of rolling a dierolling a die and observing the and observing the numbernumber that falls uppermost. that falls uppermost.

Let Let SS = {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}

denote the denote the sample spacesample space of the experiment and of the experiment and

E E = {2, 4, 6}= {2, 4, 6} and and F F = {1, 3}= {1, 3}

be be eventsevents of this experiment. of this experiment.

Compute Compute E E FF. Interpret your results.. Interpret your results.

SolutionSolution E E FF = {1, 2, 3, 4, 6} = {1, 2, 3, 4, 6} and is the and is the eventevent that the that the outcomeoutcome of of

the experiment is a the experiment is a 11, a , a 22, a , a 33, a , a 44, or a , or a 66. .


ExampleExample


Let Let SS = {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}


E E = {2, 4, 6}= {2, 4, 6} and and F F = {1, 3}= {1, 3}


Compute Compute E E FF. Interpret your results.. Interpret your results.

SolutionSolution

E E FF = = ØØ since there are since there are no elements in commonno elements in common between between the two sets the two sets EE and and FF. .


ExampleExample


Let Let SS = {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}


E E = {2, 4, 6}= {2, 4, 6} and and F F = {1, 3}= {1, 3}


Compute Compute FF cc. Interpret your results.. Interpret your results.

SolutionSolution

FF cc = {2, 4, 5, 6} = {2, 4, 5, 6} is precisely the is precisely the eventevent that the event that the event FF does does

not occurnot occur. .



Mutually Exclusive EventsMutually Exclusive Events EE and and F F are mutually exclusive if are mutually exclusive if

E E F = F = ØØ

ExampleExample

An An experimentexperiment consists of consists of tossing a coin three timestossing a coin three times and and observing the resulting sequence of observing the resulting sequence of headsheads and and tailstails..✦ Describe the Describe the sample spacesample space SS of the experiment. of the experiment.

✦ Determine the Determine the eventevent EE that that exactly two headsexactly two heads appear. appear.

✦ Determine the Determine the eventevent FF that that at least one headat least one head appears. appears.


ExampleExampleSolutionSolution As the As the tree diagramtree diagram demonstrates, the demonstrates, the sample spacesample space is is

SS = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

HH (H, H, H)(H, H, H)

TT (H, H, T)(H, H, T)

H H (H, T, H)(H, T, H)

T T (H, T, T)(H, T, T)

H H (T, H, H)(T, H, H)

T T (T, H, T)(T, H, T)

HH (T, T, H)(T, T, H)

T T (T, T, T) (T, T, T)

FirstFirst SecondSecond ThirdThird SampleSampletosstoss tosstoss tosstoss pointspoints

HH

TT

HH

TT

HH

TT


ExampleExampleSolutionSolution Scanning the Scanning the sample spacesample space obtained obtained

SS = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

we see that the we see that the outcomesoutcomes in which in which exactly two headsexactly two heads appear are given by the appear are given by the eventevent

EE = {HHT, HTH, THH} = {HHT, HTH, THH}

We can also see that the We can also see that the outcomesoutcomes in which in which at least one at least one headhead appears are given by the appears are given by the eventevent

FF = {HHH, HHT, HTH, HTT, THH, THT, TTH} = {HHH, HHT, HTH, HTT, THH, THT, TTH}


Applied Example:Applied Example: Movie Attendance Movie Attendance

The manager of a local cinema The manager of a local cinema recordsrecords the the number of number of patronspatrons attending a first-run movie screening. attending a first-run movie screening.

The theatre has a The theatre has a seating capacityseating capacity of of 500500.. Determine an appropriate Determine an appropriate sample spacesample space for this experiment. for this experiment. Describe the Describe the eventevent EE that that fewer thanfewer than 5050 people attend the people attend the

screening.screening. Describe the Describe the eventevent FF that the theatre is that the theatre is more than half fullmore than half full

at the screening. at the screening.SolutionSolution The The number of patronsnumber of patrons at the screening could run from at the screening could run from

00 to to 500500. Therefore, a . Therefore, a sample spacesample space for this experiment is for this experiment is

SS = {0, 1, 2, 3, …, 500} = {0, 1, 2, 3, …, 500}

Applied Example 5, page 357Applied Example 5, page 357





at the screening. at the screening.SolutionSolution The The eventevent EE that that fewer thanfewer than 5050 people attend the screening people attend the screening

can be described ascan be described as

EE = {0, 1, 2, 3, …, 49} = {0, 1, 2, 3, …, 49}






at the screening. at the screening.SolutionSolution The The eventevent FF that the theatre is that the theatre is more than half fullmore than half full at the at the

screening can be described asscreening can be described as

FF = {251, 252, …, 500} = {251, 252, …, 500}


7.27.2Definition of ProbabilityDefinition of Probability

6

( ) [(1,1)] [(2,2)] [(6,6)]

1 1 1

36 36 361

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

1 1 1 1 1

36 36 36 36 365

36

P E P P P

P E P P P P P

6

( ) [(1,1)] [(2,2)] [(6,6)]

1 1 1

36 36 361

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

1 1 1 1 1

36 36 36 36 365

36

P E P P P

P E P P P P P

Probability of an Event in a Uniform Sample SpaceProbability of an Event in a Uniform Sample Space

If If SS = { = {ss11, , ss22, … , , … , ssnn}}

is the is the sample spacesample space for an for an experimentexperiment in in which the which the outcomesoutcomes are are equally likelyequally likely, , then we assign thethen we assign the probabilities probabilities

to each of the outcomes to each of the outcomes ss11,, ss22,, … , … , ssnn..

1 2

1( ) ( ) ( )nP s P s P s

n 1 2

1( ) ( ) ( )nP s P s P s

n

ExampleExample

A fair A fair diedie is rolled, and the number that falls uppermost is is rolled, and the number that falls uppermost is observed.observed.

Determine the Determine the probability distributionprobability distribution for the experiment. for the experiment.SolutionSolution The The sample spacesample space for the for the experimentexperiment is is

SS = {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}

and the and the simple eventssimple events are accordingly given by the sets are accordingly given by the sets

{1}{1}, , {2}{2}, , {3}{3}, , {4}{4}, , {5}{5}, and , and {6}{6}

Since the die is assumed to be Since the die is assumed to be fairfair, the , the six outcomessix outcomes are are equally likelyequally likely..

We therefore assign a We therefore assign a probabilityprobability of of 1/61/6 to each of the to each of the simple eventssimple events. .

Example 1, page 364

ExampleExample

A fair A fair diedie is thrown, and the number that falls uppermost is is thrown, and the number that falls uppermost is observed.observed.

Determine the Determine the probability distributionprobability distribution for the experiment. for the experiment.SolutionSolution Thus, the Thus, the probability distributionprobability distribution of these simple events is: of these simple events is:

SimpleSimpleeventevent ProbabilityProbability

{1}{1} 1/61/6

{2}{2} 1/61/6

{3}{3} 1/61/6

{4}{4} 1/61/6

{5}{5} 1/61/6

{6}{6} 1/61/6Example 1, page 364

Finding the Probability of EventFinding the Probability of Event E E

1.1. Determine a Determine a sample spacesample space SS associated with associated with the the experimentexperiment..

2.2. Assign Assign probabilitiesprobabilities to the to the simple eventssimple events of of SS..3.3. If If

EE = { = {ss11, , ss22, … , , … , ssnn}}

where where {{ss11}}, , {{ss22}}, , {{ss33}}, … , , … , {{ssnn}} are are simple eventssimple events, ,

thenthen

PP((EE) = ) = PP((ss11) + ) + PP((ss22) + ) + PP((ss33) + ) + ······ + + PP((ssnn))

If If EE is the is the empty setempty set, , ØØ, then , then PP((EE) = 0) = 0..

Applied Example: Rolling DiceApplied Example: Rolling Dice

A A pairpair of fair of fair dicedice is rolled. is rolled. Calculate the Calculate the probabilityprobability that the two dice show the that the two dice show the

same numbersame number.. Calculate the Calculate the probabilityprobability that the that the sumsum of the numbers of the numbers

of the of the two dicetwo dice is is 66..

Applied Example 3, page 365


SolutionSolution The The sample spacesample space SS of the experiment has of the experiment has 3636 outcomesoutcomes

SS = {(1, 1), (1, 2), … , (6, 5), (6, 6)} = {(1, 1), (1, 2), … , (6, 5), (6, 6)}

Both dice are Both dice are fairfair, making each of the , making each of the 3636 outcomes outcomes equally likelyequally likely, so we assign the , so we assign the probabilityprobability of of 1/36 1/36 to each to each simple eventsimple event..

The The eventevent that the that the two dicetwo dice show the show the same numbersame number is is

EE = {(1, 1), (2, 2) , (3, 3), (4, 4), (5, 5), (6, 6)} = {(1, 1), (2, 2) , (3, 3), (4, 4), (5, 5), (6, 6)}

Therefore, the Therefore, the probabilityprobability that the that the two dicetwo dice show the show the same same numbernumber is given by is given by

( ) [(1,1)] [(2,2)] [(6,6)]

1 1 1 1

36 36 36 6

P E P P P

( ) [(1,1)] [(2,2)] [(6,6)]

1 1 1 1

36 36 36 6

P E P P P

Six termsSix terms



SolutionSolution The The eventevent that the that the sumsum of the numbers of the of the numbers of the two dicetwo dice is is 6 6

is given byis given by

EE66 = {(1, 5), (2, 4) , (3, 3), (4, 2), (5, 1)} = {(1, 5), (2, 4) , (3, 3), (4, 2), (5, 1)}

Therefore, the Therefore, the probabilityprobability that the that the sumsum of the numbers on of the numbers on the the two dicetwo dice is is 66 is given by is given by

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

1 1 1 1 1 5

36 36 36 36 36 36

P E P P P P P

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

1 1 1 1 1 5

36 36 36 36 36 36

P E P P P P P


Rules of ProbabilityRules of Probability

7.37.3

Property 1.Property 1. PP((EE) ) 0 0 for every for every EE..

Property 2.Property 2. PP(S) = 1(S) = 1. .

Property 3.Property 3. If If E E and and FF are mutually exclusive are mutually exclusive

((EE FF = = ØØ)), then, then

PP((EE FF) = ) = PP((EE) + ) + PP((FF))

Property 4.Property 4. If If E E and and FF are any two events of an are any two events of anexperiment, thenexperiment, then

PP((EE FF) = ) = PP((EE) + ) + PP((FF)) – – PP((E E FF))

Property 5.Property 5. If If E E is an event of an experiment andis an event of an experiment andEEcc denotes the complement of denotes the complement of EE, then, then

PP((EEcc) = 1 – ) = 1 – PP((EE))

Properties of the Probability FunctionProperties of the Probability Function

Property 1.Property 1. PP((EE) ) 0 0 for every for every EE..

Property 2.Property 2. PP(S) = 1(S) = 1. .

Property 3.Property 3. If If E E and and FF are mutually exclusive are mutually exclusive ((EE FF = = ØØ)), then, then

PP((EE FF) = ) = PP((EE) + ) +

PP((FF))

The superintendent of a metropolitan school district has The superintendent of a metropolitan school district has estimated the estimated the probabilitiesprobabilities associated with the associated with the SAT verbal SAT verbal scoresscores of students from that district. of students from that district.

The results are shown in the The results are shown in the tabletable below. below. If a student is selected at If a student is selected at randomrandom, find the , find the probabilityprobability that that

his or her his or her SAT verbal scoreSAT verbal score will be will be

✦ More than More than 400400..✦ Less than or equal to Less than or equal to 500500..✦ Greater than Greater than 400400 but less but less

than or equal to than or equal to 600600..

Score, Score, xx ProbabilityProbability

xx > 700 > 700 .01.01

600 < 600 < xx 700 700 .07.07

500 < 500 < xx 600 600 .19.19

400 < 400 < xx 500 500 .23.23

300 < 300 < xx 400 400 .31.31

xx 300 300 .19.19

Applied Example:Applied Example: SAT Verbal Scores SAT Verbal Scores


Applied Example:Applied Example: SAT Verbal Scores SAT Verbal ScoresSolutionSolution Let Let AA, , BB, , CC, , DD, , EE, and , and FF denote, respectively, the denote, respectively, the mutually mutually

exclusiveexclusive eventsevents listed in the listed in the tabletable below. below. The The probabilityprobability that the student’s score will be that the student’s score will be more thanmore than

400400 is given by is given by

PP((DD C C B B AA)) = = PP((DD) + ) + PP((CC) + ) + PP((BB) + ) + PP((AA))

= .23 + .19 + .07 + .01 = .50= .23 + .19 + .07 + .01 = .50


xx > 700 > 700 .01.01

600 < 600 < xx 700 700 .07.07

500 < 500 < xx 600 600 .19.19

400 < 400 < xx 500 500 .23.23

300 < 300 < xx 400 400 .31.31

xx 300 300 .19.19

A A

BB

CC

DD

EE

FFApplied Example 1, page 372


exclusiveexclusive eventsevents listed in the listed in the tabletable below. below. The The probabilityprobability that the student’s score will be that the student’s score will be less than or less than or

equal toequal to 500500 is given by is given by

PP((DD E E FF)) = = PP((DD) + ) + PP((EE) + ) + PP((FF))

= .23 + .31 + .19 = .73= .23 + .31 + .19 = .73


xx > 700 > 700 .01.01

600 < 600 < xx 700 700 .07.07

500 < 500 < xx 600 600 .19.19

400 < 400 < xx 500 500 .23.23

300 < 300 < xx 400 400 .31.31

xx 300 300 .19.19

A A

BB

CC

DD

EE



exclusiveexclusive eventsevents listed in the listed in the tabletable below. below. The The probabilityprobability that the student’s score will be that the student’s score will be greater greater

thanthan 400400 but but less than or equal toless than or equal to 600600 is given by is given by

PP((CC DD)) = = PP((CC) + ) + PP((DD))

= .19 + .23 = .42= .19 + .23 = .42


xx > 700 > 700 .01.01

600 < 600 < xx 700 700 .07.07

500 < 500 < xx 600 600 .19.19

400 < 400 < xx 500 500 .23.23

300 < 300 < xx 400 400 .31.31

xx 300 300 .19.19

A A

BB

CC

DD

EE



Addition RuleAddition Rule

Property 4.Property 4. If If E E and and FF are any two events of an are any two events of an experiment, thenexperiment, then

PP((EE FF) = ) = PP((EE) + ) +

PP((FF)) – – PP((E E FF))

ExampleExample A card is drawn from a shuffled deck of A card is drawn from a shuffled deck of 5252 playing cardsplaying cards.. What is the What is the probabilityprobability that it is an that it is an aceace or a or a spadespade??SolutionSolution Let Let EE denote the denote the eventevent that the card drawn is an that the card drawn is an aceace, and , and

let let FF denote the denote the eventevent that the card drawn is a that the card drawn is a spadespade.. Then,Then,

Note that Note that EE and and FF are are notnot mutually exclusive mutually exclusive events: events:✦ E E FF is the is the eventevent that the card drawn is an that the card drawn is an ace of spadesace of spades..✦ Consequently,Consequently,

4 13( ) ( )

52 52P E P F and

4 13( ) ( )

52 52P E P F and

1( )

52P E F

1( )

52P E F

Example 2, page 373

ExampleExample A card is drawn from a shuffled deck of A card is drawn from a shuffled deck of 5252 playing cardsplaying cards.. What is the What is the probabilityprobability that it is an that it is an aceace or a or a spadespade??SolutionSolution The The eventevent that a card drawn is an that a card drawn is an ace ace or aor a spade spade is is EE FF, ,

with with probabilityprobability given by given by

( ) ( ) ( ) ( )

4 13 1

52 52 5216

524

13

P E F P E P F P E F

( ) ( ) ( ) ( )

4 13 1

52 52 5216

524

13

P E F P E P F P E F

Example 2, page 373

Applied Example:Applied Example: Quality Control Quality Control

The quality-control department of Vista Vision, The quality-control department of Vista Vision, manufacturer of the Pulsar plasma TV, has determined manufacturer of the Pulsar plasma TV, has determined from records obtained from the company’s service centers from records obtained from the company’s service centers that that 3%3% of the sets sold experience of the sets sold experience video problemsvideo problems, , 1%1% experience experience audio problemsaudio problems, and , and 0.1%0.1% experience both experience both video and audio problemsvideo and audio problems before the expiration of the before the expiration of the warranty.warranty.

Find the Find the probabilityprobability that a plasma TV purchased by a that a plasma TV purchased by a consumer will experience consumer will experience video or audio problemsvideo or audio problems before before the warranty expires.the warranty expires.


Applied Example:Applied Example: Quality Control Quality ControlSolutionSolution Let Let EE denote the denote the eventevent that a plasma TV purchased will that a plasma TV purchased will

experience experience video problemsvideo problems within the warranty period, and within the warranty period, and let let FF denote the denote the eventevent that a plasma TV purchased will that a plasma TV purchased will experience experience audio problemsaudio problems within the warranty period. within the warranty period.

Then, Then,

PP((EE) = .03 ) = .03 PP((FF) = .01 ) = .01 PP((E E FF) = .001) = .001 The The eventevent that a plasma TV purchased will experience that a plasma TV purchased will experience

video or audio problemsvideo or audio problems before the warranty expires is before the warranty expires is

EE FF, and the , and the probabilityprobability of this event is given by of this event is given by( ) ( ) ( ) ( )

.03 .01 .001

.039

P E F P E P F P E F

( ) ( ) ( ) ( )

.03 .01 .001

.039

P E F P E P F P E F



Rule of ComplementsRule of Complements

Property 5.Property 5. If If E E is an event of an experiment is an event of an experiment and and EEcc denotes the complement denotes the complement of of EE, then, then

PP((EEcc) = 1 – ) = 1 –

PP((EE))

Applied Example:Applied Example: Warranties Warranties

What is the What is the probabilityprobability that a Pulsar plasma TV (from the that a Pulsar plasma TV (from the last example) bought by a consumer will last example) bought by a consumer will notnot experience experience video or audio problemsvideo or audio problems before the warranty expires? before the warranty expires?

SolutionSolution Let Let EE denote the denote the eventevent that a plasma TV bought by a that a plasma TV bought by a

consumer will experience consumer will experience video or audio problemsvideo or audio problems before before the warranty expires.the warranty expires.

Then, the Then, the eventevent that the plasma TV that the plasma TV will will notnot experience experience either problemeither problem before the warranty expires is given by before the warranty expires is given by EEcc, , with with probabilityprobability

( ) 1 ( )

1 .039

.961

cP E P E

( ) 1 ( )

1 .039

.961

cP E P E


7.47.4Use of Counting Techniques in ProbabilityUse of Counting Techniques in Probability

6! 6 5 4( ) (6,3) 3!3! 3 2( )( ) 64 64 64

20 5.3125

64 16

n E CP E

n S

6! 6 5 4

( ) (6,3) 3!3! 3 2( )( ) 64 64 64

20 5.3125

64 16

n E CP E

n S

( ) 42 21( ) .65625

( ) 64 32

n FP F

n S

( ) 42 21( ) .65625

( ) 64 32

n FP F

n S

Computing the Probability of an Event Computing the Probability of an Event in a Uniform Sample Spacein a Uniform Sample Space

Let Let SS be a uniform sample space and let be a uniform sample space and let EE be be any event. Then,any event. Then,

Number of outcomes in


( )( )

( )

E n EP E

S n S



( )( )

( )

E n EP E

S n S

ExampleExample

An unbiased An unbiased coincoin is tossed is tossed six six times. times. Find the Find the probabilityprobability that the coin will land that the coin will land headsheads

✦ ExactlyExactly three three times.times.

✦ At mostAt most three three times.times.

✦ On the On the firstfirst and the and the lastlast toss. toss.

Example 1, page 381

ExampleExampleSolutionSolution Each outcome of the experiment may be represented as a Each outcome of the experiment may be represented as a

sequencesequence of of headsheads and and tailstails.. Using the Using the generalized multiplication principlegeneralized multiplication principle, we see that , we see that

the the number of outcomesnumber of outcomes of this experiment is of this experiment is 2266, or , or 6464.. Let Let EE denote the denote the eventevent that the coin lands that the coin lands headsheads exactlyexactly

threethree times. times. Since there are Since there are CC(6, 3)(6, 3) ways this can occur, we see that the ways this can occur, we see that the

required required probabilityprobability is is

6! 6 5 4( ) (6,3) 3!3! 3 2( )( ) 64 64 64

20 5.3125

64 16

n E CP E

n S

6! 6 5 4

( ) (6,3) 3!3! 3 2( )( ) 64 64 64

20 5.3125

64 16

n E CP E

n S

Example 1, page 381

ExampleExampleSolutionSolution Let Let FF denote the denote the eventevent that the coin lands that the coin lands headsheads at mostat most

threethree times. times. Then, Then, nn((FF)) is given by the is given by the sumsum of the number of of the number of waysways the the

coin lands coin lands headsheads zerozero times (no heads), exactly times (no heads), exactly onceonce, , exactly exactly twicetwice, and exactly , and exactly threethree times. That is, times. That is,

Thus, the required Thus, the required probabilityprobability is is

( ) (6,0) (6,1) (6,2) (6,3)

6! 6! 6! 6! 6 5 6 5 41 6

0!6! 1!5! 2!4! 3!3! 2 3 242

n F C C C C

( ) (6,0) (6,1) (6,2) (6,3)

6! 6! 6! 6! 6 5 6 5 41 6

0!6! 1!5! 2!4! 3!3! 2 3 242

n F C C C C

( ) 42 21( ) .6563

( ) 64 32

n FP F

n S

( ) 42 21( ) .6563

( ) 64 32

n FP F

n S

Example 1, page 381

ExampleExampleSolutionSolution Let Let HH denote the denote the eventevent that the coin lands that the coin lands headsheads on the on the

firstfirst and the and the lastlast toss. toss. Then, Then,

Therefore, the required Therefore, the required probabilityprobability is is

4( ) 1 2 2 2 2 1 2n H 4( ) 1 2 2 2 2 1 2n H

4

6 2

( ) 2 1 1( )

( ) 2 2 4

n HP H

n S

4

6 2

( ) 2 1 1( )

( ) 2 2 4

n HP H

n S

Example 1, page 381

7.57.5Conditional Probability and Independent EventsConditional Probability and Independent Events

( )( )

( )

P A BP B A

P A

( )( )

( )

P A BP B A

P A

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

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

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

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

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

Conditional Probability of an EventConditional Probability of an Event

If If AA and and BB are events in an experiment and are events in an experiment and

PP((AA) ) 0 0, then the conditional probability that , then the conditional probability that the event the event BB will occur given that the event will occur given that the event AA has already occurred ishas already occurred is

( )( )

( )

P A BP B A

P A

( )( )

( )

P A BP B A

P A

ExampleExample

A A pairpair of fair of fair dicedice is rolled. What is the is rolled. What is the probabilityprobability that that the the sumsum of the numbers falling uppermost is of the numbers falling uppermost is 77 if it is if it is known that known that exactlyexactly oneone of the numbers is a of the numbers is a 55??

SolutionSolution Let Let AA denote the denote the eventevent that that exactlyexactly oneone of the numbers is a of the numbers is a

55 and let and let BB denote the denote the eventevent that the that the sumsum of the numbers of the numbers falling uppermost is falling uppermost is 77. Thus, . Thus,

so so

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

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

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

A

B

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

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

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

A

B

{(5,2),(2,5)}A B {(5,2),(2,5)}A B

Example 2, page 390

ExampleExample

A A pairpair of fair of fair dicedice is rolled. What is the is rolled. What is the probabilityprobability that that the the sumsum of the numbers falling uppermost is of the numbers falling uppermost is 77 if it is if it is known that known that exactlyexactly oneone of the numbers is a of the numbers is a 55??

SolutionSolution Since the dice are Since the dice are fairfair, each outcome of the experiment is , each outcome of the experiment is

equally likelyequally likely; therefore,; therefore,

Thus, the Thus, the probabilityprobability that the that the sumsum of the numbers falling of the numbers falling uppermost is uppermost is 77 given that given that exactlyexactly oneone of the numbers is a of the numbers is a 55 is given by is given by

2 11( ) ( )

36 36P A B P A and

2 11( ) ( )

36 36P A B P A and

2( ) 236( )

11( ) 1136

P A BP B A

P A

2( ) 236( )

11( ) 1136

P A BP B A

P A

Example 2, page 390

Applied Example:Applied Example: Color Blindness Color Blindness

In a test conducted by the U.S. Army, it was found that of In a test conducted by the U.S. Army, it was found that of 10001000 new recruits ( new recruits (600600 men and men and 400400 women), women), 5050 of the of the menmen and and 44 of the of the womenwomen were were red-green color-blindred-green color-blind..

Given that a recruit selected at Given that a recruit selected at randomrandom from this group is from this group is red-green color-blindred-green color-blind, what is the , what is the probabilityprobability that the that the recruit is a recruit is a malemale??

SolutionSolution Let Let CC denote the event that a randomly selected subject is denote the event that a randomly selected subject is

red-green color-blindred-green color-blind, and let , and let MM denote the event that the denote the event that the subject is a subject is a malemale recruit. recruit.





SolutionSolution Since Since 5454 out of out of 10001000 subjects are subjects are color-blindcolor-blind, we see that , we see that

Furthermore, since Furthermore, since 5050 of the subjects are of the subjects are colorblind colorblind andand malemale, we see that, we see that

50( ) .05

1000P M C I

50( ) .05

1000P M C I

54( ) .054

1000P C

54( ) .054

1000P C





SolutionSolution Therefore, the Therefore, the probabilityprobability that a subject is that a subject is malemale given thatgiven that

the subject is the subject is red-green color-blindred-green color-blind is given by is given by

( ) .05( ) .926

( ) .054

P M CP M C

P C

I( ) .05( ) .926

( ) .054

P M CP M C

P C

I


Product RuleProduct Rule

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

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

ExampleExample

Two Two cardscards are drawn without replacement from a well- are drawn without replacement from a well-shuffled deck of shuffled deck of 5252 playing cards. playing cards.

What is the What is the probabilityprobability that the that the first cardfirst card drawn is an drawn is an aceace and the and the second cardsecond card drawn is a drawn is a face cardface card??

SolutionSolution Let Let AA denote the denote the eventevent that the that the firstfirst card card drawn is an drawn is an aceace, ,

and let and let FF denote the denote the eventevent that the that the second cardsecond card drawn is a drawn is a face cardface card. .

Then,Then,4

( )52

P A 4

( )52

P A

Example 5, page 392

ExampleExample



SolutionSolution After drawing the After drawing the first cardfirst card, there are , there are 5151 cards left in the cards left in the

deck, of which deck, of which 1212 are are face cardsface cards.. Therefore, the Therefore, the probabilityprobability of drawing a of drawing a face cardface card given given

thatthat the the first cardfirst card drawn was an drawn was an aceace is given by is given by

12( )

51P F A

12( )

51P F A

Example 5, page 392

ExampleExample



SolutionSolution By the By the product ruleproduct rule, the , the probabilityprobability that the that the first cardfirst card

drawn is an drawn is an aceace and the and the second cardsecond card drawn is a drawn is a face cardface card is given byis given by

( ) ( ) ( )

4 12

52 514

0.018221

P A F P A P F A

( ) ( ) ( )

4 12

52 514

0.018221

P A F P A P F A

Example 5, page 392

Independent EventsIndependent Events

If If AA and and BB are independent events, then are independent events, then

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

and ( ) ( ) ( ) ( )P A B P A P B A P B

and

Test for the Independence of Two EventsTest for the Independence of Two Events

Two events Two events AA and and BB are independent if and are independent if and only ifonly if

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

ExampleExample

Consider the experiment consisting of Consider the experiment consisting of tossing tossing a fair coina fair coin twicetwice and observing the outcomes. and observing the outcomes.

Show that obtaining Show that obtaining headsheads on the on the first tossfirst toss and and tailstails on the on the second tosssecond toss are are independent eventsindependent events..

SolutionSolution Let Let AA denote the denote the eventevent that the outcome of the that the outcome of the first tossfirst toss is is

a a headhead, and let , and let BB denote the denote the eventevent that the outcome of the that the outcome of the second tosssecond toss is a is a tailtail..

The The sample spacesample space of the experiment is of the experiment is

soso

{(HH), (HT), (TH), (TT)}

{(HH), (HT)}

{(HT), (TT)}

S

A

B

{(HH), (HT), (TH), (TT)}

{(HH), (HT)}

{(HT), (TT)}

S

A

B

{(HT)}A B {(HT)}A B Example 8, page 396

ExampleExample

Consider the experiment consisting of Consider the experiment consisting of tossing tossing a fair coina fair coin twicetwice and observing the outcomes. and observing the outcomes.

Show that obtaining Show that obtaining headsheads in the in the first tossfirst toss and and tailstails in the in the second tosssecond toss are are independent eventsindependent events..

SolutionSolution Next, we computeNext, we compute

and observe that the and observe that the test for independent eventstest for independent events is is satisfiedsatisfied::

1 1 1( ) ( ) ( )

4 2 2P A B P A P B

1 1 1( ) ( ) ( )

4 2 2P A B P A P B

( ) ( ) ( )

1 1 1

4 2 2

P A B P A P B

( ) ( ) ( )

1 1 1

4 2 2

P A B P A P B

Example 8, page 396

Applied Example:Applied Example: Medical Survey Medical Survey

A survey conducted by an independent agency for the A survey conducted by an independent agency for the National Lung Society found that, of National Lung Society found that, of 20002000 women, women, 680680 were were heavy smokersheavy smokers and and 5050 had had emphysemaemphysema..

Of those who had Of those who had emphysemaemphysema, , 4242 were were heavy smokersheavy smokers.. Using the data in this survey, determine whether the Using the data in this survey, determine whether the

eventsevents being a heavy smokerbeing a heavy smoker and and having emphysemahaving emphysema are are independentindependent eventsevents..

SolutionSolution Let Let AA denote the denote the eventevent that a woman is a that a woman is a heavy smokerheavy smoker, ,

and let and let BB denote the denote the eventevent that a woman has that a woman has emphysemaemphysema.. Then, the Then, the probabilitiesprobabilities that a woman is a that a woman is a heavy smokerheavy smoker, ,

has has emphysemaemphysema, or , or bothboth are given by, respectively, are given by, respectively,

680 50 42( ) =.34 ( ) 0.025 ( ) .021

2000 2000 2000P A P B P A B

680 50 42( ) =.34 ( ) 0.025 ( ) .021

2000 2000 2000P A P B P A B


Applied Example:Applied Example: Medical Survey Medical Survey

A survey conducted by an independent agency for the A survey conducted by an independent agency for the National Lung Society found that, of National Lung Society found that, of 20002000 women, women, 680680 were were heavy smokersheavy smokers and and 5050 had had emphysemaemphysema..

Of those who had Of those who had emphysemaemphysema, , 4242 were were heavy smokersheavy smokers.. Using the data in this survey, determine whether the Using the data in this survey, determine whether the

events events being a heavy smokerbeing a heavy smoker and and having emphysemahaving emphysema are are independentindependent events. events.

SolutionSolution Next, we see that the Next, we see that the test for independent eventstest for independent events is is notnot

satisfiedsatisfied::

soso

and conclude that and conclude that AA and and B B are are notnot independent eventsindependent events..

( ) ( ) (.34)(.025) .0085 .021 ( )P A P B P A B ( ) ( ) (.34)(.025) .0085 .021 ( )P A P B P A B

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


7.67.6Bayes’ TheoremBayes’ Theorem

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )

(.2)(.02)

(.5)(.01) (.3)(.02) (.2)(.02)

.27

P C P D CP C D

P A P D A P B P D B P C P D C

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )

(.2)(.02)

(.5)(.01) (.3)(.02) (.2)(.02)

.27

P C P D CP C D


Bayes’ TheoremBayes’ Theorem

Let Let AA11, , AA22, … , , … , AAnn be a partition of a be a partition of a sample spacesample space SS, ,

and let and let EE be an be an eventevent of the experiment such that of the experiment such that

PP((EE) ) 0 0 and and PP((AAii) ) 0 0 for for 1 1 ≤ ≤ ii ≤ ≤ nn Then the Then the a posteriori probabilitya posteriori probability

is given byis given by

( ) (1 )iP A E i n

( ) (1 )iP A E i n

1 1 2 2

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )i i

in n

P A P E AP A E

P A P E A P A P E A P A P E A

1 1 2 2

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )i i

in n

P A P E AP A E

P A P E A P A P E A P A P E A

Applied Example:Applied Example: Quality Control Quality Control

The panels for the Pulsar widescreen LCD HDTVs are The panels for the Pulsar widescreen LCD HDTVs are manufactured in manufactured in three locationsthree locations and then shipped to the and then shipped to the main plant of Vista Vision for final assembly.main plant of Vista Vision for final assembly.

Plants APlants A, , BB, and , and CC supply supply 50%50%, , 30%30%, and , and 20%20%, , respectively, of the panels used by Vista Vision. respectively, of the panels used by Vista Vision.

The quality-control department of the company has The quality-control department of the company has determined that determined that 1%1% of the panels produced by of the panels produced by plant Aplant A are are defective, whereas defective, whereas 2%2% of the panels produced by of the panels produced by plants Bplants B and and CC are are defectivedefective..

If a Pulsar widescreen TV is selected at If a Pulsar widescreen TV is selected at randomrandom and the and the panel is found to be panel is found to be defectivedefective, what is the , what is the probabilityprobability that that the panel was manufactured in the panel was manufactured in plant Cplant C??


Applied Example:Applied Example: Quality Control Quality ControlSolutionSolution Let Let AA, , BB, and , and C C denote the denote the eventevent that the set chosen has a that the set chosen has a

panel manufactured in panel manufactured in plant Aplant A, , plant Bplant B, or , or plant Cplant C, , respectively.respectively.

Also, let Also, let DD denote the denote the eventevent that a set has a that a set has a defective paneldefective panel.. We can draw a We can draw a tree diagramtree diagram with this information: with this information:

DD

DDcc

DD

DDcc

DD

DDcc

AA

BB

CC

P(A) = .5

P(B) = .3

P(C) = .2

P(D | A) = .01

P(D | B) = .02

P(D | C) = .02


Applied Example:Applied Example: Quality Control Quality ControlSolutionSolution Next, using Next, using Bayes’ theoremBayes’ theorem, we find that the required , we find that the required

a posteriori probabilitya posteriori probability is given by is given by

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )

(.2)(.02)

(.5)(.01) (.3)(.02) (.2)(.02)

.27

P C P D CP C D


( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( )

(.2)(.02)

(.5)(.01) (.3)(.02) (.2)(.02)

.27

P C P D CP C D



End of End of Chapter Chapter

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Experiments, Sample Spaces, and Events Definition of Probability Rules of Probability

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