A First Course in Probability, 10th Edition Solution Manual

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SOLUTIONSMANUALAFIRSTCOURSEINPROBABILITYTENTHEDITIONSheldon RossUniversity of Southern California

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1Chapter 1Problems1.(a) By the generalized basic principle of counting there are26⋅26⋅10⋅10⋅10⋅10⋅10 = 67,600,000(b) 26⋅25⋅10⋅9⋅8⋅7⋅6 = 19,656,0002.64= 12963.An assignment is a sequencei1, …,i20whereijis the job to which personjis assigned. Sinceonly one person can be assigned to a job, it follows that the sequence is a permutation of thenumbers 1, …, 20 and so there are 20! different possible assignments.4.There are 4! possible arrangements. By assigning instruments to Jay, Jack, John and Jim, inthat order, we see by the generalized basic principle that there are 2⋅1⋅2⋅1 = 4 possibilities.5.There were 8⋅2⋅9 = 144 possible codes. There were 1⋅2⋅9 = 18 that started with a 4.6.Each kitten can be identified by a code number i, j, k, l where each of i, j, k, l is any of thenumbers from 1 to 7. The numberirepresents which wife is carrying the kitten,jthenrepresents which of that wife’s 7 sacks contain the kitten;krepresents which of the 7 cats insackjof wifeiis the mother of the kitten; andlrepresents the number of the kitten of catkinsackjof wifei. By the generalized principle there are thus 7⋅7⋅7⋅7 = 2401 kittens7.(a) 6! = 720(b) 2⋅3!⋅3! = 72(c) 4!3! = 144(d) 6⋅3⋅2⋅2⋅1⋅1 = 728.(a) 5! = 120(b)7!2!2!= 1260(c)11!4!4!2!= 34,650(d)7!2!2!= 12609.(12)!6!4!= 27,720

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2Chapter 110.(a) 8! = 40,320(b) 2⋅7! = 10,080(c) 5!4! = 2,880(d) 4!24= 38411.(a) 6!(b) 3!2!3!(c) 3!4!12.103−10⋅9⋅8 = 280 numbers have at least 2 equal values. 280−10 = 270 have exactly 2equal values.13.Withniequal to the number of lengthi,n1= 3,n2= 8,n3= 12,n4= 30,n5= 30, giving theanswer of 83.14.(a) 305(b) 30⋅29⋅28⋅27⋅2615.20216.52515.There are101255possible choices of the 5 men and 5 women. They can then be paired upin 5! ways, since if we arbitrarily order the men then the first man can be paired with any ofthe 5 women, the next with any of the remaining 4, and so on. Hence, there are10125!55possible results.18.(a)674222++= 42 possibilities.(b) There are 6⋅7 choices of a math and a science book, 6⋅4 choices of a math and aneconomics book, and 7⋅4 choices of a science and an economics book. Hence, there are94 possible choices.19.The first gift can go to any of the 10 children, the second to any of the remaining 9 children,and so on. Hence, there are 10⋅9⋅8⋅⋅⋅5⋅4 = 604,800 possibilities.

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Chapter 1320.564223= 60021.(a) There are8482433312+= 896 possible committees.There are8433that do not contain either of the 2 men, and there are824312thatcontain exactly 1 of them.(b) There are6626633123+= 1000 possible committees.(c) There are757575332332++= 910 possible committees. There are7533inwhich neither feuding party serves;7523in which the feuding women serves; and7532in which the feuding man serves.22.62666,51453 ++  23.7!3!4!= 35. Each path is a linear arrangement of 4r’s and 3u’s (rfor right andufor up). Forinstance the arrangementr,r,u,u,r,r,uspecifies the path whose first 2 steps are to the right,next 2 steps are up, next 2 are to the right, and final step is up.24.There are4!2!2!paths from A to the circled point; and3!2!1!paths from the circled point to B.Thus, by the basic principle, there are 18 different paths from A to B that go through thecircled point.25.3!2326.(a)()0221nnkknk==+(b)()01nnkknxxk==+

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4Chapter 128.5213,13,13,1330.1212!3, 4, 53!4!5!=31.Assuming teachers are distinct.(a) 48(b)488!2, 2, 2, 2(2)== 2520.32.(a) (10)!/3!4!2!(b)37!3 24!2!33.2⋅9!−228! since 2⋅9! is the number in which the French and English are next to each otherand 228! the number in which the French and English are next to each other and the U.S. andRussian are next to each other.34.(a) number of nonnegative integer solutions ofx1+x2+x3+x4= 8.Hence, answer is113= 165(b) here it is the number of positive solutions—hence answer is73= 3535.(a) number of nonnegative solutions ofx1+ … +x6= 8answer =135(b) (number of solutions ofx1+ … +x6= 5)×(number of solutions ofx1+ … +x6= 3) =10855

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Chapter 1536.(a)x1+x2+x3+x4= 20,x1≥2,x2≥2,x3≥3,x4≥4Lety1=x1−1,y2=x2−1,y3=x3−2,y4=x4−3y1+y2+y3+y4= 13,yi> 0Hence, there are123= 220 possible strategies.(b) there are152investments only in 1, 2, 3there are142investments only in 1, 2, 4there are132investments only in 1, 3, 4there are132investments only in 2, 3, 4152+142+1312223+= 552 possibilities37.(a)1410014=(b)101203=(c) There are132863=possible outcomes having 0 trout caught and122203=possibleoutcomes having 1 trout caught. Hence, using (a), there are1001286220495−−=possible outcomes in which at least 2 of the 10 are trout.

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6Chapter 1Theoretical Exercises2.1miin=3.n(n−1)⋅⋅⋅(n−r+ 1) =n!/(n−r)!4.Each arrangement is determined by the choice of therpositions where the black balls aresituated.5.There arenjdifferent 0−1 vectors whose sum isj, since any such vector can becharacterized by a selection ofjof thenindices whose values are then set equal to 1. Hencethere arenjknj=vectors that meet the criterion.6.nk7.111nnrr−−+−=(1)!(1)!!(1)!()!(1)!nnrnrnrr−−+−−−−=!!()!nnnrrrrnrnn−+=−8.There arenmr+gropus of sizer. As there arenmiri−groups of sizerthat consist ofimen andr−iwomen, we see that0rinmnmriri=+=−.9.02ninnnnini==−=20nini=10.Parts (a), (b), (c), and (d) are immediate. For part (e), we have the following:nkk=! !!()! !()!(1)!k nnnkknkk=−−−(1)1nnkk−+−=(1) !!(1)!(1)!()!(1)!nknnnkknkk−+=−+−−−11nnk−−=(1)!!()!(1)!()!(1)!n nnnkknkk−=−−−−

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Chapter 1711.The number of subsets of sizekthat haveias their highest numbered member is equal to11ik−−, the number of ways of choosingk−1 of the numbers 1, …,i−1. Summing overiyields the number of subsets of sizek.12.Number of possible selections of a committee of sizekand a chairperson isnkkand so1nknkk=represents the desired number. On the other hand, the chairperson can be anyone ofthenpersons and then each of the othern−1 can either be on or off the committee. Hence,n2n−1also represents the desired quantity.(i)2nkk(ii)n2n−1since there arenpossible choices for the combined chairperson and secretary andthen each of the othern−1 can either be on or off the committee.(iii)n(n−1)2n−2(c) From a set ofnwe want to choose a committee, its chairperson its secretary and itstreasurer (possibly the same). The result follows since(a) there aren2n−1selections in which the chair, secretary and treasurer are the sameperson.(b) there are 3n(n−1)2n−2selection in which the chair, secretary and treasurer jobs areheld by 2 people.(c) there aren(n−1)(n−2)2n−3selections in which the chair, secretary and treasurer areall different.(d) there are3nkkselections in which the committee is of sizek.13.(1−1)n=10( 1)nnini−=−14.(a)njnnijiiji−=−(b) From (a),njinjji==21nnijinniniji−=−=−(c)( 1)nnjjinjji−=−=( 1)1nnjjinniij−=−−−=0( 1)ninikknniik−− −=−−= 0

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8Chapter 115.(a) The number of vectors that havexk=jis equal to the number of vectorsx1≤x2≤…≤xk−1satisfying 1≤xi≤j. That is, the number of vectors is equal toHk−1(j), and the resultfollows.(b)H2(1) =H1(1) = 1H2(2) =H1(1) +H1(2) = 3H2(3) =H1(1) +H1(2) +H1(3) = 6H2(4) =H1(1) +H1(2) +H1(3) +H1(4) = 10H2(5) =H1(1) +H1(2) +H1(3) +H1(4) +H1(5) = 15H3(5) =H2(1) +H2(2) +H2(3) +H2(4) +H2(5) = 3516.(a) 1 < 2 < 3, 1 < 3 < 2, 2 < 1 < 3, 2 < 3 < 1, 3 < 1 < 2, 3 < 2 < 1,1 = 2 < 3, 1 = 3 < 2, 2 = 3 < 1, 1 < 2 = 3, 2 < 1 = 3, 3 < 1 = 2, 1 = 2 = 3(b) The number of outcomes in whichiplayers tie for last place is equal toni, the numberof ways to choose theseiplayers, multiplied by the number of outcomes of the remainingn−iplayers, which is clearly equal toN(n−i).(c)1(1)ninN ni=−=1()ninN nini=−−=10( )njnNjj−=where the final equality followed by lettingj=n−i.(d)N(3) = 1 + 3N(1) + 3N(2) = 1 + 3 + 9 = 13N(4) = 1 + 4N(1) + 6N(2) + 4N(3) = 7517.A choice ofrelements from a set ofnelements is equivalent to breaking these elements intotwo subsets, one of sizer(equal to the elements selected) and the other of sizen−r(equal tothe elements not selected).18.Suppose thatrlabeled subsets of respective sizesn1,n2, …,nrare to be made up fromelements 1, 2, …,nwheren=1riin=. As11,...,1,...irnnnn−−represents the number ofpossibilities when personnis put in subseti, the result follows.19.By induction:(x1+x2+ … +xr)n=1111210(...)ninirin xxxi−=++by the Binomial theorem

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Chapter 19=11110niin xi=2,...,...rii 22112...,...,iirrnixxii−21...riini++=−=1,...,...rii 111...,...,riirrnxxii12...riiin+++=where the second equality follows from the induction hypothesis and the last from theidentity1112,...,,...,rnninniiiii−=.20.The number of integer solutions ofx1+ … +xr=n,xi≥miis the same as the number of nonnegative solutions ofy1+ … +yr=n−1rim,yi≥0.Proposition 6.2 gives the result111rinmrr−+−−.21.There arerkchoices of thekof thex’s to equal 0. Given this choice the otherr−kof thex’s must be positive and sum ton.By Proposition 6.1, there are111nnrknrk−−=−−−+such solutions.Hence the result follows.22.11nrn+−−by Proposition 6.2.23.There are1jnj+−nonnegative integer solutions of1niixj==Hence, there are01kjjnj=+−such vectors.

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10Chapter 2Problems1.(a)S= {(r,r), (r,g), (r,b), (g,r), (g,g), (g,b), (b,r),b,g), (b,b)}(b)S= {(r,g), (r,b), (g,r), (g,b), (b,r), (b,g)}2.S= {(n,x1, …,xn−1),n≥1,xi≠6,i= 1, …,n−1}, with the interpretation that the outcome is(n,x1, …,xn−1) if the first 6 appears on rolln, andxiappears on roll,i,i= 1, …,n−1. Theevent1()cnnE∞=∪is the event that 6 never appears.3.EF= {(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)}.E∪Foccurs if the sum is odd or if at least one of the dice lands on 1.FG= {(1, 4), (4, 1)}.EFcis the event that neither of the dice lands on 1 and the sum is odd.EFG=FG.4.A= {1,0001,0000001, …}B= {01, 00001, 00000001, …}(A∪B)c= {00000 …, 001, 000001, …}5.(a) 25= 32(b)W= {(1, 1, 1, 1, 1), (1, 1, 1, 1, 0), (1, 1, 1, 0, 1), (1, 1, 0, 1, 1), (1, 1, 1, 0, 0), (1, 1, 0, 1, 0)(1, 1, 0, 0, 1), (1, 1, 0, 0, 0), (1, 0, 1, 1, 1), (0, 1, 1, 1, 1), (1, 0, 1, 1, 0), (0, 1, 1, 1, 0)(0, 0, 1, 1, 1) (0, 0, 1, 1, 0), (1, 0, 1, 0, 1)}(c) 8(d)AW= {(1, 1, 1, 0, 0), (1, 1, 0, 0, 0)}6.(a)S= {(1,g), (0,g), (1,f), (0,f), (1,s), (0,s)}(b)A= {(1,s), (0,s)}(c)B= {(0,g), (0,f), (0,s)}(d) {(1,s), (0,s), (1,g), (1,f)}7.(a) 615(b) 615−315(c) 4158.(a) .8(b) .3(c) 0

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Chapter 2119.Choose a customer at random. Let A denote the event that this customer carries an AmericanExpress card and V the event that he or she carries a VISA card.P(A∪V) =P(A) +P(V)−P(AV) = .24 + .61−.11 = .74.Therefore, 74 percent of the establishment’s customers carry at least one of the two types ofcredit cards that it accepts.10.LetRandNdenote the events, respectively, that the student wears a ring and wears anecklace.(a)P(R∪N) = 1−.6 = .4(b) .4 =P(R∪N) =P(R) +P(N)−P(RN) = .2 + .3−P(RN)Thus,P(RN) = .111.LetAbe the event that a randomly chosen person is a cigarette smoker and letBbe the eventthat she or he is a cigar smoker.(a) 1−P(A∪B) = 1−(.07 + .28−.05) = .7. Hence, 70 percent smoke neither.(b)P(AcB) =P(B)−P(AB) = .07−.05 = .02. Hence, 2 percent smoke cigars but notcigarettes.12.(a)P(S∪F∪G) = (28 + 26 + 16−12−4−6 + 2)/100 = 1/2The desired probability is 1−1/2 = 1/2.(b) Use the Venn diagram below to obtain the answer 32/100.141010SF8G242(c) Since 50 students are not taking any of the courses, the probability that neither one istaking a course is5010022= 49/198 and so the probability that at least one is takinga course is 149/198.

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12Chapter 213.(a)20,000(b)12,000(c)11,000(d)68,000(e)10,00014.P(M) +P(W) +P(G)−P(MW)−P(MG)−P(WG) +P(MWG) = .312 + .470 + .525−.086−.042−.147 + .025 = 1.05715.(a)1352455(b)4124445213 231115(c)1344445222215   (d)412445213 32115(e)4485213 41516.(a)56 5 4 3 26⋅⋅⋅⋅(b)5565 4 326⋅⋅(c)565342226(d)56 5 4 321⋅⋅(e)556 5 36⋅(f)556 5 46⋅(g)5661000700019000III0III100030001000

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Chapter 21317.15107881.1102259161      =   18.2 4 1652 51⋅⋅⋅19.4/36 + 4/36 +1/36 + 1/36 = 5/1820.Let A be the event that you are dealt blackjack and let B be the event that the dealer is dealtblackjack. Then,P(A∪B) =P(A) +P(B)−P(AB)= 4 4 164 4 16 3 1552 5152 51 50 49⋅⋅⋅⋅⋅⋅+⋅⋅⋅⋅= .0983where the preceding used thatP(A) =P(B) = 2×4 1652 51⋅⋅. Hence, the probability that neitheris dealt blackjack is .9017.21.(a)p1= 4/20,p2= 8/20,p3= 5/20,p4= 2/20,p5= 1/20(b) There are a total of 4⋅1 + 8⋅2 + 5⋅3 + 2⋅4 + 1⋅5 = 48 children. Hence,q1= 4/48,q2= 16/48,q3= 15/48,q4= 8/48,q5= 5/4822.The ordering will be unchanged if for somek, 0≤k≤n, the firstkcoin tosses land heads andthe lastn−kland tails. Hence, the desired probability is (n+ 1/2n23.The answer is 5/12, which can be seen as follows:1 =P{first higher} +P{second higher} +p{same}= 2P{second higher} +p{same}= 2P{second higher} + 1/6Another way of solving is to list all the outcomes for which the second is higher. There is 1outcome when the second die lands on two, 2 when it lands on three, 3 when it lands on four,4 when it lands on five, and 5 when it lands on six. Hence, the probability is(1 + 2 + 3 + 4 + 5)/36 = 5/12.25.P(En) =112662,()36365nnnP E−∞==

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A First Course in Probability, 10th Edition Solution Manual - Page 16 preview image

14Chapter 227.Imagine that all 10 balls are withdrawnP(A) =3 9!7 6 3 7!7 6 5 4 3 5!7 6 5 4 3 2 3 3!10!⋅+⋅⋅⋅+⋅⋅⋅⋅⋅+⋅⋅⋅⋅⋅⋅⋅28.P{same} =568333193++P{different} =568191113If sampling is with replacementP{same} =3333568(19)++P{different} =P(RBG) +P{BRG) +P(RGB) + … +P(GBR)=36 5 6 8(19)⋅⋅⋅29.(a)(1)(1)()(1)n nm mnmnm−+−++−(b) Putting all terms over the common denominator (n+m)2(n+m−1) shows that we mustprove thatn2(n+m−1) +m2(n+m−1)≥n(n−1)(n+m) +m(m−1)(n+m)which is immediate upon multiplying through and simplifying.30.(a)78 3!3389 4!44= 1/18(b)78 3!3389 4!44−1/18 = 1/6(c)787834438944+= 1/2

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