05 Probability

Chapter5

Probability: Review ofBasicConcepts

I. GraphicSummary

II. ProblemSolutions

III. ExerciseSet

IV. Self-Examination

Notes:

76 Introductionto BusinessStatisticsStudyGuide

I. Graphic Summary

Probability

SubjectiveApproachApproach

Numberof possibleoutcomesin which the

eventoccurs

Numberof possibleoutcomesin which the

eventoccursTotal numberof trials

Determinationofprobability is simply a"hunch"; an educatedguess

ClassicalApproach RelativeFrequency

P=Total numberof

possibleoutcomes

Chapter5 Probability: A Reviewof BasicConcepts 77

Representationsof SampleSpaces

Tabular

ContingencyTable

Visual

VennDiagram

A

MutuallyExclusive:

No eleat the

ment is insametime.

two categories Theareasfor eachcategorynot overlap,or, if they do, thecommonareacontainsno

Exhaustive: Everyof the

elementiscategories

in at leastonegiven,

elements.

[A B]

1cD

Every elementis in oneof theareasgiven. No elementis inthe shadedarea.

Intersection: All elementsthat are in bothcategoriesgiven.

All elementsin eithercategorygiven.

Union:

78 Introductionto BusinessStatisticsStudy Guide

SomeStraight TalkAbout Unions and Intersections

and the Addition Rulesfor Probability

Supposewe havetwo sets of elements,setA andsetB. Thenumberof elementsin Aor in B, takentogetherin theunionofA andB, cannotbe the simple sumof theelementsin A plus theelementsin B. Why? By addingall the elementsin B to all theelementsin A, someof the elementsare double-counted.Which ones?Theelementsin the intersectionof A andB are countedtwice, oncewhenwe count the membersofA andagainwhenwe add the membersof B. So, to get an accuratecountof thenumberof elementsin the unionofA andB, we haveto addB’s elementsto A’selementsandthensubtractthoseelementsthat arein the intersectionofA andB.

NumberofNumberof Numberof Numberof elementsin the

elementsin the elementsin + elementsin - intersectionofunionofAandB setA setB AandB

However,ifA andB are mutually exclusive,thereare no elementsin their intersection,and

Numberof Numberof Numberofelementsin the = elementsin + elementsin Zero

unionofAandB setA setB

So, if you rememberthat the intersectionof mutuallyexclusivesets is empty,you onlyneedto rememberthe first equationandsubstitutezero for thenumberin theintersectionof the two sets.

Now, rememberingthat a probabilityis the numberof elementsdivided by the totalnumberof trials, let’s convertthe first equationto probabilities.

PA or B = PA + PB - PA andB

And that rule is true whetherA andB aremutuallyexclusiveor not. If they aremutuallyexclusive,thenPA andB = 0, andPA or B = PA + PB.


Illustrations of Different Probabilities

Supposewe havethe following contingencytable,showingfour regionsof theUnitedStatesin which peoplelive andthreetypesof musicpeoplelistento.

Southeast

Southwest

Northeast

Northwest

Classical

50

105

80

65

Country Rock

40

85

70

55

85

160

125

80

Let’s total the numberof peopleeachof the typesof music.

in eachregionand the numberof peoplewho listen to

A Marginal Probability is the sumsitting in themarginor the marginat thebottomdivided by the totalnumberample,is 1,000.

So, themarginalprobability that a personin the sample:*listens to rock musicis 450/1000,or 0.450.*lives in the Northwestis 200/1000,or 0.200.*listens to countrymusic is 250/1000,or 0.250.*lives in the Southeastis 175/1000,or 0.175.

eitherthe marginat the rightof trials, which, in this ex

Classical Country Rock

Southeast 50 40 85 175

Southwest 105 85 160 350

Northeast 80 70 125 275

Northwest 65 55 80 200

300 250 450 1,000


More on Types of Probabilities

Continuingwith our example,let us definea joint probability.Ajoint probability isthe probability that two or moreeventswill all occur.



Southwest 105 85 160 350

Northeast 80 70 125 275


300 250 450 1,000

Sothe joint probabilitythat a personin the sample:*lives in the Southwestand listensto rock music is 160/1000,or 0.160.*lives in the Northeastandlistensto classicalis 80/1000,or 0.080.*listens to countrymusic andlives in the Southwestis 85/1000,or 0.085.

A Conditional Probability is theprobability thatan eventwill occurgiven that anothereventhasalreadyhappened.It is computedby taking thejoint probabilitythatbotheventsoccuranddividing by the marginalprobability that the given eventoccurs.

So the conditionalprobabilitythat a personselectedat random:

*lives in the Southwestgiven you know he/shelistensto rockmusic is theprobability of living in the Southwestand listeningto rock dividedby the marginalprobability of listening to rock music,or

PSouthwest androck 0.160PSouthwestI rock

= Prock = 0.450= 0.3556.

*listens to countrygivenyou know he/shelives in the Southwestis the probabilityof listening to countryand living in the Southwestdividedby the marginalprobability of living in the Southwest,or

PcountryandSouthwest 0.085PcountryI Southwest

= PSouthwest = 0.350= 0.243.


Mutually Exclusive

versus

IndependentEvents

If two sets aremutually exclusive,thereis literally nothingin onethat is also in theother. If two eventsaremutuallyexclusive,oneeventcannotoccur if the othereventdoesoccur.Examplesinclude"on/off’, "yes/no", "win/lose". Mutually exclusiveeventsneednot appearin pairs.As long asno elementin onesetalsoappearsin another, the setsare mutuallyexclusive.If eventsA andB aremutuallyexclusive,then

PA andB = 0 andPB I A aswell asPA I B =0

If two eventsare independent,the fact that oneeventoccursdoesnot affect thechancesthat the othereventwill alsooccur. Unlike mutuallyexclusiveevents,independenteventsmay bothoccur.Or they may not. If the two setsare independent,theconditionalprobability that anelementof setA is also in setB is the sameasthe marginal probability that the elementis in setB to beginwith. Thus,the fact that the elementis in

*doesnotaffect 1setA *has nothingto do with r its chancesof alsobeing a memberof B.

t .is independentfrom J

Computationally,if two eventsareindependent,then theirjoint probabilityis theproductof their two marginalprobabilities.Why? Becausethe probabilitythat anelementis in A doesnot changeif the elementis also in B. So,

PA andB = PA x PB I A= PA x PB if the eventsA andB areindependent.

If the two eventsare notindependent,however,the fact that anelementis in setAdoesaffect theprobability that it is also in B. If two eventsare not independent,theconditionalprobability that anelementis in B givenyou know it is in A doesnotequalthe marginalprobabilitythat the elementis in B. So,

PA andB = PA x PB I A if the eventsA andB are dependent.


ContingencyTablesand Probability Trees

Let’s returnto the examplewe developedearlierin this chapter,anddevelopaprobability treethat displaysthe sameinformation asthe contingencytable.


300 250 450 1,000

Region Type ofMusic

______________________________

Notice that the regional

_________

probabilitiesare allmarginalprobabilities,andthe probabilitiesforeachtypeof music in

______________________________

eachof the four regions

_________

are all conditionalprobabilities.So, theprobability thatsomeonelistenstoclassical,for example,changesdependingonwhich regionhe/shelives in.


Southwest 105 85 160 350

Northeast 80 70 125 275


Southeast175/1000

50/17540/175

85/175

ClassicalCountryRock

Classical105/350- Southwest_

350/100085/350160/350

80/275

CountryRock

Classical

Northeast 70/275 Country

Rock275/1000 125/275

Northwest65/2005 5/200

ClassicalCountryRock200/1000 80/200


Using ContingencyTablesto Flip Probability Trees

Theprobability treewe developedon the prior pageshowsthe regionsfirst andthetypesof musicpreferredas contingentupon the regionin which someonelives. But,supposewe want to showthe treethe otherway around;that is, supposewe wantedtoshowthe typesof music first andthe regionalidentificationsascontingentupon thetypesof musicpreferred.

RockClassical Country


Southwest 105 85 160 350

Northeast 80 70 125 275


300 250

Region

450 1,000

SoutheastSouthwestNortheastNorthwest

SoutheastSouthwestNortheastNorthwest

SoutheastSouthwestNortheast

Notice now that themusic typesaremarginalprobabilities,andthe probabilitiesforeachregionare allconditionalprobabilities.So, theprobability thatsomeonelives in theSoutheast,for example,changesdependingonwhich type of musiche!shelistensto.

TypeofMusic

Classical

50/300

40/250

250/1000

Rock

85/450

450/1000Northwest


Tree-Flippingwith Bayes’Theorem

Wheneventsare sequentiallyarranged,aswith the two probability treeswe havejustdeveloped,information from a secondeventcanbe usedto revisethe probabilitythatthe first eventhasoccurred.Let’s revisit our first probability treefor an example.

We know from this displaythat PclassicalI Northwest= 65/200.And we know thatthe marginalprobability ofliving in the Northwestis200/1000.

________

Now, supposewe observea

___________________________

personat randomandnoticehe/sheis listeningtoclassicalmusic.Givenweknow the personlistenstoclassicalmusic,what is theprobability thatpersonlivesin the Northwest?

Region

Southeast

Type

50/17540/175

ofMusic


Southwest105/35085/350


Northeast80/27570/275


Northwest65/20055/200


Whatwe wantto find,then, is thePNorthwestI classical.

Continuedon nextpage.


Tree-Flippingwith Bayes’Theorem, continued

PNorthwest andclassicalRememberthat the PNorthwestI classical=

Pclassical

So, our first stepis to computePclassical.To accomplishthat, we look for all thebranchesof the treewherepeoplelistento classicalmusic.

Southeast Southwest Northeast Northwest

PclassicalP175

=1 x[±000

1 Po+1

17J [±000x

1011I

35JP275

+1[±000

x80127J

P200+1

[±000x 651

I20j

50 105 80 65+ + +

1000 1000 1000 1000

300

1000

And, we know from the abovetreethatPNorthwestandclassical= 65/1000.

86 Introduction to BusinessStatisticsStudyGuide

Tree-Flipping:A Comparison of Methods

Now, comparethe resultwe just derivedwith Bayes’Theoremto the treewe flippedusingthe contingencytable.Notice thatboth resultsshowPNorthwestI classical65/300.

So, we are ableto accomplishthe sameresult two differentways.Being able to useBayes’ Theorem,however,is a realassetwhen you do nothave a completecontingencytable to work with. For the formal statementof the theorem,seeyourtextbook.For now, let’s reviewherethe generalform of the theorem.We will assumethat you aregiven enoughinformation to constructa completeprobabilitytreewithwhich to work.

1. First, focuson thequestionbeingasked.Write out the conditionalprobabilitythequestionposes,suchas "Find PB I A."

2. Next, write down the definitionof the conditionalprobability,suchas

PB I Athe joint probabilityof BandA - PBandA

- the marginalprobability of the condition - PA

3. Thenext stepis the cruxof your computation.To computethe marginalprobability of the condition,PA, identify all thebranchesof the treewherethatconditionholds.Usethe Multiplication Rule for Probability to computetheprobabilityassociatedwith eachof the branchesyou identified.And thenadduptheproductsyou computedfor the totalmarginalprobabilityof the condition,PA. This sumwill be the denominatorof youranswer.

4. Now, identify thebranchwherebothconditionsB andA exist.Again, usetheMultiplication Rule for Probability to computethe joint probability.This productwill be the numeratorof your final answer,PB andA.

5. Finally, form the ratio of the joint probabilityover the marginalprobabilitytodeterminethevalueof PB I A.

Rememberthatwhat Bayes’Theoremallows you to do is switch theconditions.UsingBayes’Theorem,you cancomputePB I A from a treethathasPA I B.

Chapter5 Probability: A Review of BasicConcepts 87

Orderings and Factorials

Thesamethreeitemscanbe arrangedsix different wayswhenthe orderin which theyappearis important. Thenumberof waysn thingscanbe arrangedis n! read "nfactorial", wheren! = n x n-i x n-2 x ... x 1. SupposeHometown,USA, justannouncedthe first, second,and third placewinnersof the Volunteerof theYearAward. The threecandidatescanbe uniquelyordered3! = 3 x 2 x 1 = 6 differentways. They might look like this:

1st 2nd 3rd

1st 2nd 3rd

1st 2nd 3rd

1st 2nd 3rd

1st 2nd 3rd

1st 2nd 3rd

Supposethat a newspaperreporteris trying to get to Hometownto coverthe banquetscheduledtomorrownight. Shecandriveoneof threecarsin the paper’scarpooi,andshecanselectfrom four different routesto get there. According to the Principleof Multiplication, thereare 3 x 4 or 12 different waysfor the reporterto get toHometown.


Permutations

Whenorderis important,as in theVolunteerof theYearAward, the arrangementsarecalledPermutations.

Supposewe wantedto selecttwo winners,a first and a second,out of four finalists.Thelocal newspapermight run oneof the following twelve pairof pictures.

TheFinalists

&a

ThePossibleWinners

Thus,thereare twelve uniquepairs of possiblewinnerswhenorderis important. Wecall this setthe numberof permutationsof 4 objectstaken2 at a time.Computationally,

n!n-r!

or 4!2!

4x3x2x 1= 12

2xi


Combinations

Whenorderis not important,the arrangementis calleda Combination.

Supposethe HometownCity Council wantsto namea specialcommitteeto studymore effectiveways to utilize volunteersin the community. Froma panelof fourcandidates,the Council will selecttwo peopleto serveon thecommittee. Theorderinwhich the candidatesare selectedis not important. How manydifferent committeescanthe Council form from the four candidates?

C

TheCandidates

LNoticethat thereare six pairsof duplicatecommitteesin the twelve permutationsidentifiedon the prior page. While therearetwelveuniquepermutationsof 4 peopletaken2 at a time, thereare only six uniquecombinationsof 4 peopletaken2 at a timewhenthe orderin which they are selectedis not important.

6 uniquecommitteesof 2 peopleeachcanbe selectedfrom 4people

orr! n-r!

4!2! 4-2!

4x3x2xi2 x 1 x 2 x 1

Computationally,n‘4

90 Introduction to BusinessStatisticsStudy Guide

II. Problem Solutions

STUDY GUIDE PROBLEM:

Fconti

atalitiesfrom poisoningby gasesandvaporsduring oneyeararedescribedin thefollowingngencytableof frequencies,whereM = male,F female.

<5 5-14 15-24 25-44 45-64 65M 14 20 137 311 188 117F 11 9 66 65 57 84

For a randomlyselectedvictim of poisoningby gasesandvapors,determine:a PMb PFc PFand5-14d PForS-14e PM and 25-44f PM or 25-44g PM I25-44h PF I25-44i PF or 5-14 or 45-64j P[F] and [5-14 or 45-64]

THIS PROBLEM DEALS WITH THE DIFFERENT TYPES OF PROBABILITIES.

ANSWERS NOTES

a Pmale Marginal or simpleprobability

Takethenumberofmalesanddivide by thetotal numberofpeople.

ANSWER:= 1079


STUDY GUIDE PROBLEM, continued

b Pfemale Marginal or simpleprobability

Take thenumberoffemalesanddivide by thetotal numberofpeople.*

ANSWER: =

*Anotherway to work this is to recognizethatbeingmale andbeingfemalearecomplementaryevents. SoPfemale = 1 - Pmale

= 1 - 787/1079= 292/1079

c Pfemaleand5-14 Jointprobability

Intersectionoftwo events

Takethe numberofpeoplewhoare bothfemaleand aged5-14, and divide by the totalnumberofpeople.

ANSWER:= 1079

d Pfemaleor 5-14 Union oftwo events

Notmutuallyexclusiveevents

292 29 9 UseAdditionRulefor Probability,PA or B- 1079 + 1079 - 1079 = PA + PB - PA andB

ANSWER: =



e Pmaleand 25-44

ANSWER:= 79

Jointprobability

Intersectionoftwo events

Takethenumberofpeoplewho are bothmaleand aged25-44, anddivide by the totalnumberofpeople.

1 Pmaleor 25-44

787 376 311= 1079 + 1079 - 1079

852ANSWER: = 1079

Union oftwo events

Notmutuallyexclusiveevents

UseAdditionRulefor Probability, PA or B= PA + PB - PA andB

g PmaleI 25-44 ConditionalProbability

Take thenumberofpeoplewho are bothmaleand25-44,anddivideby thenumberofpeoplewhoare 25-44.

ANSWER: =

h PfemaleI25-44 ConditionalProbability

Take thenumberofpeoplewho are bothfemaleand 25-44,and divide by thenumberofpeoplewhoare 25-44.

ANSWER: =



i Pfemaleor 5-14or 45-64 Union ofthreeevents

Visualize theVenndiagramrelating the threeeventsofinterest:

292 29 245 9 571079 + 1079 + 1079 - 1079 - 1079-0+0

ANSWER: =5001079

j P[female]and [5-14 or45-64] Visualizethe Venndiagramrelating the threeeventsofinterest:

Use thegeneralrule ofaddition,PA or B or C = PA + PB + PC -

PA and B - PA and C - PB and C +PA andB and C



- 9 57 Pfemaleand [5-14 or 45-64] =

- 1079 + 1079 0 Pfemaleand5-14+ Pfemaleand45-64- P5-14and45-64

ANSWER:= 1079

Chapter5 Probability: A Review of Basic Concepts 95

PROBLEM 5.69:

Datafrom theFederalBureauofInvestigationshow that 1 ofevery 155 motorvehicleswasstolenduring 1997. Applying this statisticto 5 motorvehiclesrandomlyselectedfrom thenation’sautopopulation:

a Whatis theprobability thatnoneof the 5 motorvehicleswill be stolen?b Whatis theprobability thatall 5 motorvehicleswill be stolen?c How manypossibilitiesexist in which 2 of the 5 motorvehiclesare stolen?

THIS PROBLEMDEALS WITH THE MULTIPLICATION RULES AND COMBINATIONS.

ANSWERS NOTES

a Pnoneof 5 carsstolen Jointprobability

Eachofthefive eventsis independentoftheothers.

= PA x PB x PC x PD x PE UsetheMultiplication Rulefor independentevents,wheretheprobability ofa singleeventis 154/155.

r1n iANSWER: = I-- I = 0.9682

Li55]

b Pall 5 carsstolen Jointprobability

Eachof thefive eventsis independentof theothers.

= PA x PB x PC x PD x PE Use theMultiplication Rulefor independentevents,wheretheprobability ofa singleeventis 1/155.

r 1 i Notice thatpartsa andb areANSWER:

= [] = 0.03 18 complementsofoneanother. Sooneansweris 1 minusthe otheranswer.


PROBLEM5.69,continued

c How manywayscan2 of 5 motor Order in which themotorvehiclesarestolenvehiclesbe stolen? is not important.

Combinationofindependentevents2

HANDHELD CALCULATOR:5! Mostcalculatorswill automaticallycalculate

= 2 x combinations.Checkyourcalculatorkeypador lookin yourowner’smanualforinstructions.

MICROSOFT EXCEL:In a cell on an Excelspreadsheet,type:

= combin5,2

Youcan also activateExcel’scombinationfunction underthefunctionkeyon the topmenubar. SelecteitherALLor MATH &TRIGcategoriesanddoubleclick onCOMBINATIONto bring up thedialog box.

ANSWER: = 10


PROBLEM 5.74:

SpecialNote: The solutionsare developedwith andwithout theprobability tree. Notice howmucheasiertheproblemis to work onceitsprobability tree is developed.Spendsometimerelating thesolutionsusingprobability rules with thoseusing the tree.

In examiningborrowercharacteristicsversusloan delinquency,a bankhascollectedthefollowing information: 115%of theborrowerswho havebeenemployedattheir presentjob forless than3 yearsarebehindin theirpayments;2 5% oftheborrowerswho havebeenemployedattheir presentjob for at least3 yearsarebehindin their payments;3 80% oftheborrowershavebeenemployedat their presentjob for at least3 years. Giventhis information:

a Whatis theprobability that a randomlyselectedloanaccountwill be for apersonin thesamejob for at least3 yearswho is behindin making payments?

b Whatis theprobability that a randomlyselectedloanaccountwill be for a personin thesamejob for less than3 yearsorwho is behindin makingpayments?

c If a loan accountis behind,what is theprobability that the loan is for apersonwhohasbeenin thesamejob for lessthan3 years?

THIS PROBLEM DEALS WITH CONDITIONAL PROBABILITIES AND TREE-FLIPPING.

ANSWERS NOTES

a Patleast3 yearsandbehind Jointprobability

Intersectionofevents

PbehindI atleast3 yrs = 0.05 List the informationgivenin theproblem.Patleast3 yrs = 0.80

PbehindIat least3 yrs Write out thedefinitionoftheconditional- Pbehind and at least 3yrs probability.- Pat least3 yrs

x Answeris thenumeratorof thefractionPbehindI atleast3 yrs

= Patleast3 yrs definingthe conditionalprobability.

x Solvefor thenumerator.0.05 = j- so 0.05 0.80 = x

ANSWER: = 0.05 0.80 = 0.04

OR

Constructthe appropriateprobability treefrom the informationgiven.



0.05

0.g5

Behind 0.03

Not behind 0.17

Behind 0.04

Not behind 0.76

Notethattheanswerto part a is theprobabilityof beingat the third branchofthetreeshownbelow.

= 0.80 0.05

ANSWER: = 0.04

b Plessthan3 yrs or behind

= Plessthan3 yrs + Pbehind-

Plessthan3 yrs andbehind

PbehindI lessthan3 yrs- Pbehind and less than 3yrs- Plessthan 3 yrs

PbehindI lessthan3 yrs

x- 0.15

- 0.20

Use theMultiplication Ruleto computeprobability for the third branch.

Union ofevents.Use theAdditionRule.

First, we needtofindPlessthan 3 yrs andbehind

Write out definition ofconditionalprobability.

Thejoint probability Plessthan 3 yrs andbehindis thenumeratorof thefractiondefining theconditionalprobability.

Solvefor thenumerator.

Less then 3 jrs0.20

0.15

0.85

t least 3 yrs0.80

x- Plessthan3 yrs

so 0.150.20=x

0. 15 0.20 = 0.03



Pbehind= Next,we needtofindPbehind.Plessthan3 yrs x PbehindI lessthan 3 yrs +Patleast3 yrs x PbehindIatleast3 yrs

= 0.200.15 + 0.80 0.05= 0.03 + 0.04Pbehind= 0.07

= Plessthan 3 yrs + Pbehind- Finally, weput it all together,usingthePlessthan3 yrs andbehind AdditionRulewe alreadycited.

= 0.20 + 0.07 - 0.03

ANSWER: = 0.24

OR

Partb, alternatesolution Constructtheappropriateprobabilitytreefrom the informationgiven. Seethe treeshownin part a solution.

Note that theansweris theprobabilityoflessthan 3 yrsplus theprobability ofthosewho aremorethan threeyrs andbehind.

= 0.20 + 0.04 To thePlessthan 3 yrs, then, we addtheprobability ofbeing on the thirdbranchof the tree.

ANSWER: = 0.24


PROBLEM 5.74,continued

c Plessthan3 yrs Ibehind

Plessthan3 yrs Ibehind- Pless than 3 yrs andbehind- Pbehind

Plessthan3 yrs andbehind= 0.03

Pbehind= 0.07

Plessthan3 yrs Ibehind=

ANSWER:

Conditionalprobability

Write out thedefinition oftheconditionalprobability.

Repeatcalculationfor Plessthan 3 yrs andbehindcompletedin Part b above.

Repeatcalculationfor Pbehindcompletedin Part b above.

Substitutevaluesinto thedefinitionfor theconditionalprobability.


PROBLEM 5.77:

Thechairpersonof theaccountingdepartmenthasthreesummercoursesavailable:Accounting201, Accounting202, andAccounting305. Twelvefaculty areavailablefor assignmentto thesecoursesandno faculty membercanbe assignedto more than oneof them. In how many wayscanthechairpersonassignfaculty to thesecourses?

THIS PROBLEMDEALS WITH PROBABILITIES AND PERMUTATION.

ANSWER NOTES

How many wayscanthreecoursesbe filled Order is important.by thepool of twelvefaculty?

12! Permutation,12 thingstaken3 at a time=l2xllxlO

HANDHELD CALCULATOR:Most calculatorswill automaticallycalculatepermutations.Checkyourcalculatorkeypador lookin yourowner’smanualforinstructions.

MICROSOFT EXCEL:In a cell on an Excelspreadsheet,type:

= permut12,3

Youcan also activateExcel’scombinationfunction underthefunctionkeyon the topmenubar. SelecttheSTATISTICALcategoryand doubleclick on PERMUTATIONtobring up thedialog box.

ANSWER: = 1,320ways


III. ExerciseSet

Exercise Set 5

1. Supposethat fatalities from poisoning by gasesand vapors rose in anotheryear see the extraproblem. And supposeyou found the following probability tree in the next year’s literaturerelating

A. Fromthe probability treebelow, reconstructthecontingencytableof frequencies.

B. For a randomly selectedvictim of poisoning by gasesand vapors,determinethe followingprobabilities:

a Pmale

c Pfemaleand5-14

e Pmaleand 25-44

g PmaleI 25-44

b Pfemale

d Pfemaleor 5-14

f Pmaleor 25-44

h PfemaleI 25-44

sexofthe victim with agebrackets.

Male811/1153

t2BlUl 1

Female342/1153

17/flu Less than 55-1415-2425-4445-64Greater than 65

Less than 55-1415-2425-4445-64Greater than 65

B/342

65/342

i Pfemaleor 5-14or 45-64 j P[femalejand [5-14 or 45-64]


2. Supposethat datafrom the localcounty sheriff’s office showedthat 2 ofevery 541 automobilesin thecountywerevandalizedlastyear.

a Is this a populationor a samplemeasure?

b Applying this to 7 cars randomlyselectedfrom the county’s automobilepopulation, what is theprobabilitythat all ofthe 7 carswill be vandalized?

c Applying this to 7 cars randomlyselectedfrom the county’s automobilepopulation, what is theprobability thatnoneof the 7 carswill be vandalized?

d How many possibilitiesexist in which 3 ofthe 7 automobileswill be vandalized?

3. In examiningcustomertelephonehabitsversusincome levels,a regional telephonestudy reportedthefollowing information: 1 25% of the heavy long-distanceusers had annualhouseholdincome inexcessof $50,000, 2 90% of the medium-to-light long-distancecustomershad annualhouseholdincome below $50,000,and 3 70% of the long-distanceusersfall into the medium-to-lightusergroup. Given this information:

a What is the probability that a randomly selectedtelephoneaccount will be for a heavy long-distanceuserwith an annualhouseholdincomein excessof $50,000?

b Whatis the probability that a randomlyselectedtelephoneaccountwill be for a medium-to-lightlong-distanceuserwith an annualhouseholdincomein excessof $50,000?

c If a telephoneaccountto a householdwith annualincomein excessof $50,000is selected,what istheprobability that the accountis for a personwho is a heavylong-distanceuser?

4. Five new drivers report to the regional office of the National ParcelDelivery NPD, where sevendeliveryroutesremainuncovered.Eachdriver canonly be assignedto oneroute. In how manywayscanthenewdriversbe assignedto the availabledelivery routes?


Answers:ExerciseSet5

l.A.

AGE

SEXFemale

<5 5-14 15-24 25-44 45-64 65 Totals:

811

342

a 811/1153 b 342/1153 c 8/1153

d 367/1153 e 327/1153 0 896/1153

g 327/412

j 69/ 1153

h 85/412

r 2b I-

L 541

i 563/1153

rcL 541

0.075c 0.145

Male 17 25 128 327 196 118

13 8 74 85 61 101

1.B.

Totals 30 33 202 412 257 211 1153

2. a samplemeasure

7 7!d 3 = 3!4! = 35

3. a 0.075 b 0.07

4. = 42


IV. Self-Examination: Chapter 5

You haveappliedfor a job andaretold by the interviewerthat theprobabilityofyourbeinghired is0.60. Your oddsof beinghiredare:a. 0.6to4,or3to50b. 6tolO,or3to5c. 6to4,or3to2d. 40to60,or2to3e. 100 to 60, or5to3

2. Your instructorannouncedthat he gives a pop quiz in 20% ofthe classmeetings.What aretheoddsthat you will havea pop quiz in a givenclassmeeting?a. 0.2 to 80, or 1 to 400b. 2tol0,orlto5c. 2to8,orlto4d. 80to20,or4tole. 100 to 20, or 5 to 1

USETHE FOLLOWING FOR THE NEXT TEN QUESTIONS#3 - #12:The tablebelow givesthe probabilitiesof religiousandpolitical affiliations in a majorU.S. city.

A B C DProtestant Catholic Jewish Other

E Democrat 0.30 0.10 0.05 0.02F Republican 0.25 0.10 0.02 0.02G Independent 0.05 0.05 0.02 0.02

3. Whatis theprobability that a personselectedat randomis a Catholic?a. 0.10b. 0.25c. 0.47d. 0.75e. 0.82

4. Whatare theoddsthata personselectedat randomis a Catholic?a. lOto9O,orlto9b. 25to75,orlto3c. 47to53d. 75to25,or3tole. 100 to 25, or 4 to 1

5. Whatis theprobability that a personselectedat randomis a Catholicanda Republican?a. 0.10b. 0.25c. 0.39d. 0.40e. 0.43


6. Whatis the probabilitythat a personselectedat randomis a Catholicandat the sametime eitheraDemocrator a Republican?a. 0.11b. 0.20c. 0.25d. 0.86e. 0.90

7. Whatis the probability that a randomlyselectedpersonis a Democratwhosereligion is neitherCatholicnorJewish?a. 0.15b. 0.20c. 0.28d. 0.32e. 0.40

8. Given a personrandomlyselectedis knownto be Protestant,what is theprobability thepersonis aDemocrat?a. 0.28b. 0.47C. 0.50d. 0.64e. 0.70

9. Given a personrandomlyselectedis knownto be Democrat,what is theprobability thepersonis aProtestant?a. 0.28b. 0.47C. 0.50d. 0.64e. 0.70

10. Given a personrandomlyselectedis knownto be not a Protestant,whatis the probability thepersonis a Democrat?a. 0.17b. 0.19c. 0.27d. 0.425e. 0.452

11. Find theprobabilityPFIB.a. 0.10b. 0.26c. 0.40d. 0.64e. 0.70


12. Find theprobabilityPBIF.a. 0.10b. 0.26c. 0.40d. 0.64e. 0.70

USE THE FOLLOWING FOR THE NEXT SIX QUESTIONS#13 - #18:A personnelmanageris reviewingthe numberof overtimehoursworkedby employeesin herplant lastmonth. Shecompiledthe following data:

No. of Overtime No. of ProbabilityHours Worked Employees Px

0 15 0.1251-2 21 0.1753-4 24 0.2005-6 42 0.3507-8 12 0.100

More than 8 6 0.050

13. Which variablemeasuresthe frequency?a. Numberof overtimehoursworkedb. Numberof employeesc. Probabilityd. Probabilitytimesnumberof overtimehoursworkede. Probabilitytimesnumberofemployees

14. Whatis theprobability that a randomlyselectedemployeeworkedno overtimehourslastmonth?a. 0.125b. 0.30C. 0.50d. 0.875e. 0.925

15. Whatis theprobability that a randomlyselectedemployeeworkedat most4 overtimehourslastmonth?a. 0.125b. 0.30C. 0.50d. 0.875e. 0.925


16. Whatis theprobability that a randomlyselectedemployeeworkedless than7 overtimehourslastmonth?a. 0.10b. 0.85c. 0.90d. 0.95e. 0.975

17. Whatis theprobability that a randomlyselectedemployeeworkedlessthan7 butmorethan4overtimehourslastmonth?a. 0.35b. 0.65c. 0.85d. 0.95e. 0.975

18. Whatis the probabilitythat a randomlyselectedemployeeworkedat least3 but lessthan 7overtimehours?a. 0.20b. 0.35C. 0.55d. 0.65e. 0.95

USE THE FOLLOWING FOR THE NEXT FOUR QUESTIONS#19 - #22:A sidewalkyogurt vendorsells threeflavors: chocolate,vanilla,andfruit. Forty percentof salesarechocolate,while 35%arevanilla with the rest fruit flavored. Salesareby theconeor thecup. Thepercentagesof conessalesfor chocolate,vanilla andfruit, respectively,are80%, 60% , and40%. For arandomlyselectedsale,let

A1 = eventchocolatechosenA2 = eventvanillachosenA3 = eventfruit chosenB = eventyogurt on a cone

19. Find theprobabilityPB.a. 0.20b. 0.37c. 0.63d. 0.80e. 0.95

20. Find thePA2IB.a. 0.1419b. 0.210c. 0.320d. 0.3333...e. 0.420


21. Find theprobabilityPB ‘.a. 0.20b. 0.37c. 0.63d. 0.80e. 0.95

22. Find theposteriorprobability that, given the yogurtwas sold in a cup, it was fruit yogurt.a. 0.1419b. 0.210c. 0.320d. 0.3333...e. 0.420

23. Six studentsfrom a statisticsclasshaveformed a study group. Eachmay or maynot attendastudysession.Assumingthat thememberswill be making independentdecisionson whetherornot to attend,how manydifferentpossibilitiesexist for the compositionof the studysession?a. 6b. 64c. 120d. 720e. 980

24. At a trackmeet,six runnersarecompetingin the 100-yarddash. First, secondandthird placetrophieswill be awarded. In how manywayscanthetrophiesbe awarded?a. 6b. 64c. 120d. 720e. 980

25. An investmentcounselorwould like to meetwith 12 of her clientson Wednesday,but sheonlyhastime for 8 appointments.How manydifferentwayscan theclientsbe consideredfor inclusioninto herlimited schedulefor the day if the orderin which theclients arescheduledis notimportant?a. 96b. 495c. 95,040d. 79,833,600e. 695,782,043,300


Answers to Self-Examination: Chapter 5

1. C Topic: odds

2. C Topic: odds

3. B Topic: marginalprobability

4. B Topic: odds

5. A Topic: joint probability,unions andintersections

6. B Topic: unionsandintersections

7. D Topic: unionsandintersections

8. C Topic: conditionalprobability,

9. D Topic: conditionalprobability,

10. D Topic: conditionalprobability,

11. C Topic: conditionalprobability,

12. B Topic: conditionalprobability,

13. B Topic: frequencydistributions

14. A Topic: simpleprobability

15. C Topic: unions andintersections

16. B Topic: unions andintersections

17. A Topic: unions andintersections

18. C Topic: unionsandintersections

19. C Topic: Bayes

20. D Topic: Bayes

21. B Topic: complements

22. A Topic: Bayes

23. B Topic: multiplicationrule

24. C Topic: permutation

25. B Topic: combination

unions andinterseëtions

unions andintersections

unionsandintersections



05 Probability

Documents

number of elements

sets of elements

bs elements

number ofnumber of number

andnumber of number

number ofelements

number of peopleeach

intersection ofa