Contents
Preface1
SolutionstoChapter13
SolutionstoChapter229
SolutionstoChapter359
SolutionstoChapter491
SolutionstoChapter5113
SolutionstoChapter6129
SolutionstoChapter7155
SolutionstoChapter8167
SolutionstoChapter9197
SolutionstoChapter10205
SolutionstotheAppendix235 i
Preface
Thiscollectionofsolutionsisforreferencefortheinstructorswhouseourbook. Theauthorsfirmlybelievethatthebestwaytomasternewmaterialisviaproblem solving.Havingallthedetailedsolutionsreadilyavailablewouldunderminethis process.Hence,weaskthatinstructorsnotdistributethisdocumenttothestudents intheircourses.
Theauthorswelcomecommentsandcorrectionstothesolutions.Alistof correctionsandclarificationstothetextbookisupdatedregularlyatthewebsite https://www.math.wisc.edu/asv/
1.2. (a) SinceBobhastochooseexactlytwooptions,Ωconsistsofthe2-element subsetsoftheset {cereal, eggs, fruit}:
Ω= {{cereal, eggs}, {cereal, fruit}, {eggs, fruit}} TheitemsinBob’sbreakfastdonotcomeinanyparticularorder,hencethe outcomesaresetsinsteadoforderedpairs.
(b) Thetwooutcomesintheevent A are {cereal, eggs} and {cereal, fruit}.In symbols, A = {Bob’sbreakfastincludescereal} = {{cereal, eggs}, {cereal, fruit}}
1.3. (a) ThisisaCartesianproductwherethefirstfactorcoverstheoutcomeof thecoinflip({H,T } or {0, 1},dependingonhowyouwanttoencodeheads andtails)andthesecondfactorrepresentstheoutcomeofthedie.Hence Ω= {0, 1}×{1, 2,..., 6} = {(i,j): i =0or1and j ∈{1, 2,..., 6}}
(b) NowweneedalargerCartesianproductspacebecausetheoutcomehasto containthecoinflipanddierollofeachperson.Let ci betheoutcomeofthe coinflipofperson i,andlet di betheoutcomeofthedierollofperson i.Index i runsfrom1to10(oneindexvalueforeachperson).Each ci ∈{0, 1} andeach di ∈{1, 2,..., 6}.Herearevariouswaysofwritingdownthesamplespace:
{(ci,di)1≤i≤10 :each ci ∈{0, 1} andeach di ∈{1, 2,..., 6}}.
Thelastformulaillustratestheuseofindexingtoshortenthewritingofthe 20-tupleofalloutcomes.Thenumberofelementsis#Ω=210 610 =1210 = 61,917,364,224.
(c) Ifnobodyrolledafive,theneachdieoutcome di comesfromtheset {1, 2, 3, 4, 6} thathas5elements.Hencethenumberoftheseoutcomesis210 510 =1010 Togetthenumberofoutcomeswhereatleast1personrollsafive,subtract thenumberofoutcomeswherenoonerollsa5fromthetotal:1210 1010 = 51,917,364,224.
1.4. (a) Thisisanexampleofsamplingwithreplacement,whereordermatters. Thus,thesamplespaceis
Ω= {ω =(x1,x2,x3): xi ∈{statesintheU.S.}} Inotherwords,eachsamplepointisa3-tupleororderedtripleofU.S.states. Theproblemstatementcontainstheassumptionthateverydayeachstate isequallylikelytobechosen.Since#Ω=503 =125,000,eachsamplepoint ω hasequalprobability P {ω} =50 3 = 1 125,000 .Thisspecifiestheprobability measurecompletelybecausethentheprobabilityofanyevent A comesfrom theformula P (A)= #A 125,000 .
(b) The3-tuple(Wisconsin,Minnesota,Florida)isaparticularoutcome,andhence asexplainedabove,
P ((Wisconsin,Minnesota,Florida))= 1 503
(c) ThenumberofwaystohaveWisconsincomeonMondayandTuesday,butnot Wednesdayis1 · 1 · 49,withsimilarexpressionsfortheothercombinations. Sincethereisonly1wayforWisconsintocomeeachofthethreedays,wesee thetotalnumberofpositiveoutcomesis
Thus
P (Wisconsin’sflaghungatleasttwoofthethreedays)
= 3 49+1 503 = 37 31250 =0 001184
1.5. (a) Therearetwonaturalsamplespaceswecanchoose,dependingupon whetherornotwewanttoletordermatter.
Ifwelettheorderofthenumbersmatter,thenwemaychoose
(b) Thecorrectcalculationforthisquestiondependsonwhichsamplespacewas choseninpart(a).
Whenordermatters,weimaginefillingthepositionsofthe5-tuplewith threeevenandtwooddnumbers.Thereare 5 3 waystochoosethepositions ofthethreeevennumbers.Theremainingtwopositionsareforthetwoodd numbers.Wefillthesepositionsinorder,separatelyfortheevenandodd numbers.Thereare20 19 18waystochoosetheevennumbersand20 19 waystochoosetheoddnumbers.Thisgives
Whenorderdoesnotmatter,wechoosesets.Thereare 20 3 waystochoose asetofthreeevennumbersbetween1and40,and 20 2 waystochooseasetof twooddnumbers.Therefore,theprobabilitycanbecomputedas
1.6. Wegivetwosolutions,firstwithanorderedsample,andthenwithoutorder.
(a) Labelthethreegreenballs1,2,and3,andlabeltheyellowballs4,5,6,and 7.Weimaginepickingtheballsinorder,andhencetake
Ω= {(i,j): i,j ∈{1, 2,..., 7},i = j}, thesetoforderedpairsofdistinctelementsfromtheset {1, 2,..., 7}.The eventoftwodifferentcoloredballsis, A = {(i,j):(i ∈{1, 2, 3} and j ∈{4,..., 7}) or (i ∈{4,..., 7} and j ∈{1, 2, 3})}
(b) Wehave#Ω=7 6=42and#A =3 4+4 3=24. Thus,
Alternatively,wecouldhavechosenasamplespaceinwhichorderdoesnot matter.Inthiscasethesizeofthesamplespaceis 7 2 .Thereare 3 1 waysto chooseoneofthegreenballsand 4 1 waystochooseoneyellowball.Hence, theprobabilityiscomputedas P
1.7. (a) Labeltheballs1through7,withthegreenballslabeled1,2and3,and theyellowballslabeled4,5,6and7.Let
Ω= {(i,j,k): i,j,k ∈{1, 2,..., 7},i = j,j = k,i = k}, whichcapturestheideathatordermattersforthisproblem.Notethat#Ω= 7 6 5.Thereareexactly 3 4 2=24
waystochoosefirstagreenball,thenayellowball,andthenagreenball.Thus thedesiredprobabilityis
P (green,yellow,green)= 24 7 6 5 = 4 35
(b) Wecanusethesamereasoningasinthepreviouspart,byaccountingforall thedifferentordersinwhichthecolorscancome:
P (2greensandoneyellow)= P (green,green,yellow)
P (green,yellow,green)+ P (yellow,green,green)
Alternatively,sincethisquestiondoesnotrequireorderingthesampleof balls,wecantake
= j,j = k,i = k}, thesetof3-elementsubsetsoftheset {1, 2,..., 7}.Now#Ω= 7 3 .Thereare 3 2 waystochoose2greenballsfromthe3greenballs,and 4 1 waystochoose oneyellowballfromthe4yellowballs.Sothedesiredprobabilityis
P (2greensandoneyellow)=
= 12 35
1.8. (a)
Labelthelettersfrom1to14sothatthefirst5are Es,thenext4are As, thenext3are Nsandthelast2are Bs.
OurΩconsistsof(ordered)sequencesoffourdistinctelements:
Ω= {(a1,a2,a3,a4): a
ThesizeofΩis14 13 12 11=24024.(Becausewecanchoose a1 14different ways,then a2 13differentwaysandsoon.)
Theevent C consistsofsequences(a1,a2,a3,a4)consistingoftwonumbers between1and5,onebetween6and9andonebetween10and12.Wecan countthesebyconstructingsuchasequencestep-by-step:wefirstchoosethe positionsofthetwo Es:wecandothat 4 2 =6ways.Thenwechooseafirst E outofthe5choicesandplaceittothefirstchosenposition.Thenwechoose thesecond E outoftheremaining4andplaceittothesecond(remaining) chosenposition.Thenwechoosethe A outofthe4choices,anditsposition (thereare2possibilitiesleft),Finallywechoosetheletter N outofthe3choices andplaceitintheremainingposition(weonlyhaveonepossibilityhere).In eachstepthenumberofchoicesdidnotdependonthepreviouschoicessowe canjustmultiplythenumberstogethertoget6 5 4 4 2 3 1=2880.
Theprobabilityof C is
(b) Asbefore,welabelthelettersfrom1to14sothatthefirst5are Es,thenext 4are As,thenext3are Nsandthelast2are Bs.OurΩisthesetofunordered samplesofsize4,orinotherwords:allsubsetsof {1, 2,..., 14} ofsize4:
Theevent C isthat {
,a4} hastwonumbersbetween1and5,one between6and9andonebetween10and12.Thenumberofwayswecan choosesuchasetis
Esout of5possibilities,thesingle A outof4possibilitiesandthesingle N outof3 possibilities.)
Thisgives
1.9. Wemodelthepointatwhichthestickisbrokenasbeingchosenuniformly atrandomalongthelengthofthestick,whichwetaketobe L (insomearbitrary units).Thus,Ω=[0,L].Theeventwecareaboutis A = {
4L/5}.Hence,sincethetwoeventsaremutuallyexclusive,
1.10. (a) Sincetheoutcomeoftheexperimentisthenumberoftimeswerollthe die(asinExample1.16),wetake Ω= {∞, 1, 2, 3,... }.
Element k inΩmeansthatittook k rollstoseethefirstfour.Element ∞ meansthatfourneverappeared.
Nextwededucetheprobabilitymeasure P onΩ.SinceΩisadiscrete samplespace(countablyinfinite), P isdeterminedbygivingtheprobabilities ofalltheindividualsamplepoints.
Foraninteger k ≥ 1,wehave
P (k)= P {needed k rolls} = P {nofoursinthefirst k 1rolls,thena4}.
Eachrollhas6outcomessothetotalnumberofoutcomesfrom k rollsis6k Eachrollcanfailtobeafourin5ways.Hencebytakingtheratioofthe numberoffavorableoutcomesoverthetotalnumberofoutcomes,
P (k)= P {nofoursinthefirst k 1rolls,thena4} =
Tocompletethespecificationofthemeasure P ,wefindthevalue P (∞).Since theoutcomesaremutuallyexclusive,
Thus, P (∞)=0.
(b) Wealreadydeducedabovethat P (thenumberfourneverappears)= P (∞)=0 Hereisanalternativesolution.
P (thenumberfourneverappears) ≤ P (nofoursinthefirst n rolls)= 5 6
n .
Since( 5 6 )n → 0as n →∞ andtheinequalityholdsforany n,theprobability ontheleftmustbezero.
1.11. ThesamplespaceΩthatrepresentsthedartboarditselfisasquareofside length20inches.Wecanassumethatthecenteroftheboardisattheorigin.The event A,thatthedarthitswithin2inchesofthecenter,isthenthesubsetofΩ describedby A = {x : |x|≤ 2}.Probabilityisnowproportionaltoarea,andso
1.12. Thesamplespaceandprobabilitymeasureforthisexperimentweredescribed inthesolutiontoExercise1.10:
.
1.13. (a) Imagineselectingonestudentuniformlyatrandomfromtheschool. Thus,Ωisthesetofstudentsandeachoutcomeisequallylikely.Let W bethesubsetofΩconsistingofthosestudentswhowearawatch.Let B be thesubsetofstudentswhowearabracelet.Wearetoldthat