Introduction to Probability 1st Edition Anderson Solutions Manual

Page 1

IntroductiontoProbability DetailedSolutionstoExercises DavidF.Anderson TimoSepp¨al¨ainen BenedekValk´o
DavidF.Anderson,TimoSepp¨al¨ainenandBenedekValk´o2018 Introduction to Probability 1st Edition Anderson Solutions Manual Full Download: http://testbanktip.com/download/introduction-to-probability-1st-edition-anderson-solutions-manual/ Download all pages and all chapters at: TestBankTip.com
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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

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}}

Onesamplespaceis Ω= {1,..., 6}×{1,..., 6} = {(i,j): i,j ∈{1,..., 6}}, wherewevieworderasmattering.Notethat#Ω=62 =36.Sincealloutcomes areequallylikely,wetake P (ω)= 1 36 foreach ω ∈ Ω.Theevent A is A =            (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 3), (2, 4), (2, 5), (2, 6) (3, 4), (3, 5), (3, 6) (4, 5), (4, 6) (5, 6)            = {(i,j): i,j ∈{1, 2, 3, 4, 5, 6},i<j}, and P (A)= #A #Ω = 15 36 Onewaytocountthenumberofelementsin A withoutexplicitlywritingthemout istonotethatforafirstrollof i ∈{1, 2, 3, 4, 5},thereareonly6 i allowablerolls forthesecond.Hence, #A = 5 i=1 (6 i)=5+4+3+2+1=15
SolutionstoChapter1 1.1.
3

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

4 SolutionstoChapter1
Ω=(
1
andeach
=
{0, 1}×{1, 2,..., 6})10 = {(c1,d1,c2,d2,...,c10,d10):each ci ∈{0,
}
di ∈{1, 2,..., 6}}
1 · 1 · 49+1 · 49 · 1+49 · 1 · 1+1=3 ·
49+1=148

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

SolutionstoChapter1 5
Ω1 =
(x1,...,x5): xi ∈{1,..., 40},xi = xj if i = j},
{1, 2, 3,..., 40}.In thiscase#Ω1 =40 39 38 37 36and P1(ω)= 1 #Ω1 foreach ω ∈ Ω1.
Ω2 = {{x1,...,x5} : xi ∈{1, 2, 3,..., 40},xi = xj if i = j}, thesetof5-elementsubsetsoftheset {1, 2, 3,..., 40}.Inthiscase#Ω2 = 40 5 and P2(ω)= 1 #Ω2 foreach ω ∈ Ω2.
{
thesetofordered5-tuplesofdistinctelementsfromtheset
Ifwedonotletordermatter,thenwetake
P (exactlythreenumbersareeven)= 5 3 · 20 · 19 · 18 · 20 · 19 40 · 39 · 38 · 37 · 36 = 475 1443
P
20 3 · 20 2 40 5 = 475 1443
(exactlythreenumbersareeven)=
.
P (A)= 24 42 = 4 7
A
3 1 4 1 7 2 = 4 7
(
)=
.

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.

6 SolutionstoChapter1
= 3 2 4+3 4 2+4 3 2 7 6 5 = 72 210 = 12 35
i,j,k} : i,j,k ∈{1, 2,..., 7},i
Ω= {{
3 2 4 1 7 3
aj ,ai ∈{
i =
1, 2,..., 14}}

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} =

SolutionstoChapter1 7
P (C)= #C #Ω = 2880 24024 = 120 1001
Ω= {{a1,a2,a3,a4} : ai = aj ,ai ∈{1, 2,..., 14}} ThesizeofΩis
4 =1001.
14
1,a
5 2 4 1 3 1 =120.(Becausewecanchoosethetwo
a
2,a3
P (C)= #C #Ω =
120 1001 , thesameasinpart(a).
Ω:
5or ω
P (A)= P {ω ∈ [0,L]: ω ≤ L/5} + P {ω ∈ [0,L]: ω ≥ 4L/5} = L/5 L + L/5 L = 2 5
ω ∈
ω ≤ L/
5k 1 1 6k
5 6 k 1 1 6
=

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

8 SolutionstoChapter1
1= P (Ω)= P (∞)+ ∞ k=1 P (k) = P (∞)+ ∞ k=1 5 6 k 1 1 6 = P (∞)+ 1 6 ∞ j=0 5 6 j (reindex) = P (∞)+ 1 6 1 1 5/6 (geometricseries) = P (∞)+1
P (A)= areaof A areaoftheboard = π 22 202 = π 100
P (k)=( 5 6 )k 1 1 6 forpositiveintegers k
(a) P (needatmost3rolls)= P (1)+ P (2)+ P (3)= 1 6 1+ 5 6 +( 5 6 )2 = 91 216 (b) P (evennumberofrolls)= ∞ m=1 P (2m)= ∞ m=1 ( 5 6 )2m 1 1 6 = 1 5 ∞ m=1 ( 25 36 )m = 1 5 25 36 1 25 36 = 5 11 .
P (W cBc)=0 6,P (W )=0 25,P (B)=0 30

Introduction to Probability 1st Edition Anderson Solutions Manual

Weareaskedfor P (W ∪ B).BydeMorgan(oraVennDiagram)wehave

(b)

P (W ∩ B).Wehave

1.14. Fromtheinclusion-exclusionprincipleweget

1.15. (a) Theeventthatoneofthecolorsdoesnotappearis W

∪ R.Ifweuse theinclusion-exclusionprinciplethen

Wecomputeeachtermontheright-handside.Notethatthewecanlabelthe 4ballssothatwecandifferentiatebetweenthe2redballs.Thiswaythethree drawsleadtoequallylikelyoutcomes,eachwithprobability 1

(b) Thecomplementoftheeventis {allthreecolorsappear}.Letuscounthow manydifferentwayswecangetsuchanoutcome.Wehave2choicestodecide whichredballwillshowup,whilethereisonlyonepossibilityforthegreen andthewhite.Thenthereare3!=6differentwayswecanorderthethree colors.Thisgives2 · 6=12possibilities.Thus

P (allthreecolorsappear)= 12 43 = 3 16 fromwhich

P (oneofthecolorsdoesnotappear)=1 P (allthreecolorsappear)= 13 16

SolutionstoChapter1 9
P (W ∪ B)=1 P ((W ∪ B)c)=1 P (W cBc)=1 0 6=0 4
P (W ∩ B)= P (W )+ P (B) P (W ∪ B)=0.25+0.30 0.4=0.15.
Wewant
P (A ∪ B)= P (A)+ P (B) P (AB)=0 4+0 7 P (AB)=1 1 P (AB) Rearrangingthisweget P (AB)=1.1 P (A ∪ B). Since P (A ∪ B
P (AB)=1.1 P (A ∪ B) ≥ 1.1 1=0.1. Ontheotherhand, B ⊂ A ∪ B so P (A ∪ B) ≥ P (B)=0 7whichgives P (AB)=1 1 P (A ∪ B) ≤ 1 1 0 7=0 4 Puttingthesetogetherweget0 1 ≤ P (AB) ≤ 0 4
)isaprobability,itisatmost1,so
∪ G
P (W ∪ G ∪ R)= P (W )+ P (G)+ P (R) P (WG) P (GR) P (RW )+ P (WGR)
43
Wehave P (W )= P (eachpickisgreenorred)= 33 43 andsimilarly P (G)= 33 43 and P (R)= 23 43 .Also: P (WG)= P (eachpickisred)= 23 43 andsimilarly P (GR)= 1 43 and P (RW )= 1 43 .Finally, P (WGR
P (W ∪ G ∪ R)= 1 43 (33 +33 +23 23 1 1)= 13 16 .
.
)=0,sinceit isnotpossibletohavenoneofthecolorsinthesample. Puttingeverythingtogether:
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