Probability With Applications and R 1st Edition Dobrow Solutions Manual

Chapter1

1.1 (i)Anexperimentwhoseoutcomeisuncertain.(ii)Thesetofallpossibleoutcomes.(iii) Asetofoutcomes.(iv)Arandomvariableassignsnumericalvaluestotheoutcomes ofarandomexperiment.

1.2 (i)Rollfourdices.(ii)Ω= {1111, 1112,..., 6665, 6666} (iii)Event: {5555}.(iv)Let X denotethenumberoffivesinfourdicerolls.Then X istherandomvariable.The desiredprobabilityis P (X =4)

1.3 (i)Choosingtoppings.(ii)Let a, b, c denotepineapple,peppers,andpepperoni,respectively.ThenΩ= {Ø,a,b,c,ab,ac,bc,abc} (iii)Event: {ab,ac,bc}.(iv)Let X be thenumberoftoppings.Then X istherandomvariable.Thedesiredprobabilityis P (X =2)

1.4 (i)PlayingAngryBirdsuntilyouwin.(ii)Let W denotewinning,and L denotelosing. ThenΩ= {W,LW,LLW,...} (iii)Event: {X< 1000}, where X isthenumberof timesyouplaybeforeyouwin.(iv)Xistherandomvariable.Thedesiredprobability is P (X< 1000)

1.5 (i)Harvesting1000tomatoes.(ii)Ωisthesetofall1000-elementofsequencesconsisting of B s (bad)and G s (good).(iii)Event: {X ≤ 5}, where X isthenumberofbad tomatoes.(iv)Xistherandomvariable.Thedesiredprobabilityis P (X ≤ 5)

1.6 (a) {13, 22, 31};

(b) {36, 45, 54, 63};

(c) {13, 23, 33, 43, 53, 63};

(d) {11, 22, 33, 44, 55, 66};

(e) {31, 41, 51, 52, 61, 62}.

1.7 (a) {R =0};

(b) {R =1,B =2};

(c) {R + B =4};

(d) {R =2B}

1.8 Let B denoteaboyand G denoteagirl.ThenΩ= {G,BG,BBG,...BBBBBB} Therandomvariableisthenumberofboys.

1.14 (a)Thenumberofwaystoselectapresidentis10 ThenumberofwaystoselectTom tobethepresidentis1 Thusthedesiredprobabilityis1/10 (b)Thenumberofways toselectapresidentandatreasureris10 × 9=90. Thenumberofwaystoselect BrendatobethepresidentandLiztobethetreasureris1. Thedesiredprobabilityis 1/90.

1.15 Thenumberof6-elementsequenceswithfirsttwoelements H andlasttwoelements T is22 =4 Thenumberof6-elementsequencesof

6 =64 Thusthe desiredprobabilityis4/64=1/

1.19 Thereare k!orderingsofwhichoneisinincreasingorder.Thus,1/k!.

1.20 (a)0 2;(b)0 2;(c)0 6

1.21 (a)0 9;(b)0;(c)0 1;(d)0 9

1.22 Weknow

1.24

1.26

1.41 n<-10000 simlist<-vector(length=n) for(iin1:n):{ trial<-sample(0:1,4,replace=TRUE) success<-if(sum(trial)==1)1else0 simlist[i]<-success

1.42 simdivis<-function(){ num<-sample(1:5000,1)

if(num%%4==0||num%%7==0||num%%10==0)1else0 }

simlist<-replicate(1000,simdivis())

mean(simlist)

1.43 n<-10000 simlist<-vector(length=n) for(iin1:n):{

trial<-sample(1:6,2,replace=TRUE) success<-if(sum(trial)>=8)1else0 simlist[i]<-success }

mean(simlist)

1.44 n<-10000 simlist<-vector(length=n) for(iin1:n):{

trial<-sample(1:4,1,replace=TRUE) success<-if(trial>=2)1else0 simlist[i]<-success }

mean(simlist)

Chapter2

2.5 (a)0;(b)1;(c)

2.6 (1)Weknow

(2)Similarly,

2.10 Bythetreediagram,theprobabilitythatthefinalballiswhiteis2/

2.15 Applythe“birthdayproblem”with5,000“days”and100“peopleintheroom.”The desiredprobabilityis1

2.16 (a)Let H betheeventthattheselectedcardisaheart.Let M betheeventthatthe missingcardisheart.

(b)Theselectedcardisequallylikelytobeoneofthefoursuits.Thus P (H)=1/4 2.17 (i)

TheprobabilitythatGummiBearsischosenis3/10+1/6=7/15.

(ii)Let G betheeventthatGummiBearsischosen.Let A betheeventthatthefirst bagischosenand B betheeventthatthesecondbagischosen.

2.18 (b)

2.20 Afteraddingawhitecountertherearethreeequallylikelystates:(i)Thebaginitially containsablackcounter B1.Awhitecounter W2 isputintothebagand W2 ispicked atthefirstdraw;(ii)Thebaginitiallycontainsawhitecounter W1.Awhitecounter W2 isputintothebagand W2 ispickedatthefirstdraw;and(iii)Thebaginitially containsawhitecounter W1.Awhitecounter W2 isputintothebagand W1 ispicked atthefirstdraw.Thustheprobabilitythattheseconddrawisawhitecounteris2/3.

2.22 Let A betheeventthat HH firstoccurs, B betheeventthat HT firstoccurs, H be theeventthatthefirstcoinflipisahead,and T betheeventthatthefirstcoinflip isatail.Then,

2.23 Let A betheeventthattheyouthisasmoker.Let

betheeventthatatleastone parentisasmoker.Wearegiven

2.24 Let B betheeventthatawomanhasbreastcancer,+betheeventthatamammogram givesapositiveresult,and betheeventthatamammogramgivesanegativeresult.

2.25 Let L betheeventthatapersonisaliar,+betheeventthatapolygraphtest concludeslying,and betheeventthatapolygraphtestconcludesnotlying.

(b)Want P (L|+) ≥ 0 8

2.26 Let B betheeventthatthecabisblue.Let b betheeventthatawitnessassertsthe cabisblue.

2.27 Let A betheeventthatthefairdieischosen,

betheeventthatthediewithall5 s ischosen,and C betheeventthatthediewiththree5

andthree4 s ischosen.

2.28 (a) n<-10000 simlist<-vector(length=n) for(iin1:n){

trial<-sample(1:365,23,replace=T) success<-if(2\%in\%table(trial))1else0 simlist[i]<-success } mean(simlist)

(b) k =47, (c)0 967, (d) k =28

2.29 n<-1000 simlist1<-vector(length=n) simlist2<-vector(length=n) simlist3<-vector(length=n) for(iin1:n){

trialA<-sample(c(3,3,5,5,7,7),1,replace=T) trialB<-sample(c(2,2,4,4,9,9),1,replace=T)

trialC<-sample(c(1,1,6,6,8,8),1,replace=T) success1<-if(trialA>trialB)1else0 success2<-if(trialB>trialC)1else0 success3<-if(trialC>trialA)1else0 simlist1[i]<-success1 simlist2[i]<-success2 simlist3[i]<-success3 }

mean(simlist1);mean(simlist2);mean(simlist3)

2.30 n<-1000 envelopes<-c("A","B","C","D")

Probability With Applications and R 1st Edition Dobrow Solutions Manual

noswitchwin<-vector(length=n)

switchwin<-vector(length=n) for(iin1:n){

win<-sample(envelopes,1)

pick<-sample(envelopes,1)

remove<-sample(envelopes[which(envelopes!=pick &envelopes!=win)],2)

switchyes<-envelopes[which(envelopes!=pick&envelopes!= remove[1]&envelopes!=remove[2])]

noswitch<-if(pick==win)1else0

switch<-if(switchyes==win)1else0 noswitchwin[i]<-noswitch switchwin[i]<-switch

}

mean(noswitchwin);mean(switchwin)

AfterIchooseanenvelope,theprobabilitythatitcontainsthebillis1/4 Theprobabilitythatthebillisinoneoftheotherthreeenvelopesis3/4 Aftertwoempty envelopesareremoved,theprobabilitythatmyenvelopecontainsthebillhasnot changed,sotheprobabilitythattheoneonthetablecontainsthebillis3/4 ThusI shouldswitch.