Problemas do Capítulo 4

47 Chapter 4 Problems 1 PX 4 4 2 6 14 91 2 PX 0 2 2 1 14 91 2 PX 2 4 2 2 1 8 14 91 2 PX 1 8 2 1 1 16 14 91 2 PX 1 4 8 1 1 32 14 91 2 PX 2 8 2 28 14 91 2 2 p1 136 p5 236 p9 136 p15 236 p24 236 p2 236 p6 436 p10 236 p16 136 p25 136 p3 236 p7 0 p11 0 p18 236 p30 236 p4 336 p8 236 p12 436 p20 236 p36 136 4 PX 1 12 PX 2 5 5 5 10 9 18 PX 3 5 4 5 5 10 9 8 36 PX 4 5 4 3 5 10 10 9 8 7 168 PX 5 5 4 3 2 5 5 10 9 8 7 6 252 PX 6 5 4 3 2 1 1 10 9 8 7 6 252 5 n 2i i 0 1 n 6 PX 3 18 PX 1 38 PX 1 38 PX 3 18 8 a p6 1 562 1136 p5 2 16 46 162 936 p4 2 16 36 162 736 p3 2 16 26 162 536 p2 2 16 16 162 336 p1 136 d p5 136 p4 236 p3 336 p2 436 p1 536 p0 636 pj pj j 0 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 48 Chapter 4 11 a Pdivisible by 3 333 1000 Pdivisible by 105 9 1000 Pdivisible by 7 142 1000 Pdivisible by 15 66 1000 In limiting cases probabilities converge to 13 17 115 110 b PμN 0 PN is not divisible by 2 ip i 1 i P N is not divisible by 2 ip 2 1 1 i i p 6π2 13 p0 Pno sale on first and no sale on second 74 28 p500 P1 sale and it is for standard P1 sale2 Psale no sale Pno sale sale2 34 762 27 p1000 P2 standard sales P1 sale for deluxe 3614 P1 sale2 045 27 315 p1500 P2 sales one deluxe and one standard 3612 09 p2000 P2 sales both deluxe 3614 045 14 PX 0 P1 loses to 2 12 PX 1 Pof 1 2 3 3 has largest then 1 then 2 1312 16 PX 2 Pof 1 2 3 4 4 has largest and 1 has next largest 1413 112 PX 3 Pof 1 2 3 4 5 5 has largest then 1 1514 120 PX 4 P1 has largest 15 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 49 15 PX 1 1166 PX 2 11 2 12 11 66 54 j j j PX 3 1 2 12 12 11 66 54 42 k j k j j k j j k PX 4 1 3 1 1 i P X 16 PY1 i 12 66 i PY2 i 12 12 66 54 j i j i j PY3 i 12 12 11 66 54 42 k j j i k i j k j k j All sums go from 1 to 11 except for prohibited values 20 a Px 0 Pwin first bet Plose win win 1838 203818382 5918 b No because if the gambler wins then he or she wins 1 However a loss would either be 1 or 3 c EX 11838 203818382 2038220381838 320383 108 21 a EX since whereas the bus driver selected is equally likely to be from any of the 4 buses the student selected is more likely to have come from a bus carrying a large number of students b PX i i148 i 40 33 25 50 EX 402 332 252 502148 3928 EY 40 33 25 504 37 22 Let N denote the number of games played a EN 2p2 1 p2 32p1 p 2 2p1 p The final equality could also have been obtained by using that N 2 where I is 0 if two games are played and 1 if three are played Differentiation yields that 2 4 d E N p dp and so the minimum occurs when 2 4p 0 or p 12 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 50 Chapter 4 b EN 3p3 1 p3 43p21 pp 3p1 p21 p 56p21 p2 6p4 12p3 3p2 3p 3 Differentiation yields d E N dp 24p3 36p2 6p 3 Its value at p 12 is easily seen to be 0 23 a Use all your money to buy 500 ounces of the commodity and then sell after one week The expected amount of money you will get is Emoney 1 1 500 2000 2 2 1250 b Do not immediately buy but use your money to buy after one week Then Eounces of commodity 1 1 1000 250 2 2 625 24 a 3 7 1 3 4 4 4 p p p b 3 11 1 2 2 4 4 p p p 7 11 3 4 2 1118 4 4 p p p maximum value 2372 c 3 1 4 q q d 3 21 4 q q minimax value 2372 attained when q 1118 25 a PX 1 63 47 46 b EX 146 242 13 27 C Ap 1 10 10 A C A p 28 3 4 20 35 29 If check 1 then if desired 2 Expected Cost C1 1 pC2 pR1 1 pR2 if check 2 then 1 Expected Cost C2 pC1 pR1 1 pR2 so 1 2 best if C1 1 pC2 C2 pC1 or C1 2 1 p p C Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 51 30 EX 1 2 1 2 n n n a probably not b yes if you could play an arbitrarily large number of games 31 Escore p1 1 P2 1 p1 p2 d dp 21 pp 2p1 p 0 p p 32 If T is the number of tests needed for a group of 10 people then ET 910 111 910 11 10910 35 If X is the amount that you win then PX 110 49 1 PX 1 EX 1149 59 69 067 VarX 11249 59 692 1089 36 Using the representation N 2 I where I is 0 if the first two games are won by the same team and 1 otherwise we have that VarN VarI EI2 E2I Now EI2 EI PI 1 2p1 p and so VarN 2p1 p1 2p1 p 8p3 4p4 6p2 2p Differentiation yields Var d N dp 24p2 16p3 12p 2 and it is easy to verify that this is equal to 0 when p 12 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 52 Chapter 4 37 EX2 403 333 253 503148 16254 VarX EX2 EX2 822 EY2 402 332 252 5024 14535 VarrY 845 38 a E2 X2 Var2 X E2 X2 VarX 9 14 b Var4 3X 9 VarX 45 39 4 1 24 2 38 40 4 5 13 231 4 135 11243 41 10 10 7 10 1 2 i i 42 3 2 4 5 2 3 5 5 3 1 1 1 3 4 2 p p p p p p p p 6p3 15p2 12p 3 0 6p 12p 12 0 p 12 43 3 2 4 5 5 5 2 8 2 8 2 3 4 44 1 1 2 2 1 1 1 n n i n i i n i i k i k n n p p p p i i α α 45 with 3 Ppass 2 3 2 3 3 3 1 2 8 2 8 4 6 4 2 2 3 3 533 with 5 Ppass 5 5 5 5 3 3 5 5 1 2 8 2 4 6 3 3 i i i i i i i i 3038 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 53 46 Let C be the event that the jury is correct and let G be the event that the defendant is guilty Then 3 8 12 12 0 0 2 8 65 1 9 35 c c i i i i i i P C P C G P G P C G P G Let CV be the event that the defendant is convicted Then 3 8 12 12 0 0 3 12 12 12 0 9 2 8 65 1 1 9 35 2 8 65 1 9 35 c c i i i i i i i i i i i i P CV P CV G P G P CV G P G 47 a and b i 9 9 5 9 1 i i i p p i ii 8 8 5 8 1 i i i p p i iii 7 7 4 7 1 i i i p p i where p 7 in a and p 3 in b 48 The probability that a package will be returned is p 1 9910 1099901 Hence if someone buys 3 packages then the probability they will return exactly 1 is 3p1 p2 49 a 7 3 7 3 10 10 1 1 4 6 7 3 7 7 2 2 b 7 3 7 3 1 9 1 4 6 7 3 2 6 2 55 50 a PH T T6 heads PH T T and 6 headsP6 heads PH T TP6 headsH T TP6 heads 2 5 2 6 4 7 10 5 6 pq p q p q 110 b PT H T6 heads PT H T and 6 headsP6 heads PT H TP6 headsT H TP6 heads 2 5 2 6 4 7 10 5 6 q p p q p q 110 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 54 Chapter 4 51 a e2 b 1 e2 2e2 1 12e2 Since each letter has a small probability of being a typo the number of errors should approximately have a Poisson distribution 52 a 1 e35 35e35 1 45e35 b 45e35 Since each flight has a small probability of crashing it seems reasonable to suppose that the number of crashes is approximately Poisson distributed 53 a The probability that an arbitrary couple were both born on April 30 is assuming independence and an equal chance of having being born on any given date 13652 Hence the number of such couples is approximately Poisson with mean 800003652 6 Therefore the probability that at least one pair were both born on this date is approximately 1 e6 b The probability that an arbitrary couple were born on the same day of the year is 1365 Hence the number of such couples is approximately Poisson with mean 80000365 21918 Hence the probability of at least one such pair is 1 e21918 1 54 a e22 b 1 e22 22e22 1 32e22 55 3 42 1 1 2 2 e e 56 The number of people in a random collection of size n that have the same birthday as yourself is approximately Poisson distributed with mean n365 Hence the probability that at least one person has the same birthday as you is approximately 1 en365 Now ex 12 when x log2 Thus 1 en365 12 when n365 log2 That is there must be at least 365 log2 people 57 a 1 e3 3e3 e3 2 3 3 17 1 2 2 e b PX 3X 1 3 3 17 1 3 2 1 1 e P X P X e 59 a 1 e12 b 12 1 2 e c 12 12 12 1 3 1 1 2 2 e e e Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 55 60 Pbeneficial2 2 beneficial34 2 beneficial3 4 2 not beneficial1 4 P P P 2 3 2 2 3 5 3 3 2 4 3 3 5 1 2 4 2 4 e e e 61 1 e14 14e14 62 For i j say that trial pair i j is a success if the same outcome occurs on trials i and j Then i j is a success with probability 2 1 n k k p By the Poisson paradigm the number of trial pairs that result in successes will approximately have a Poisson distribution with mean 2 2 1 1 1 2 n n k k i j k k p n n p and so the probability that none of the trial pairs result in a success is approximately 2 1 exp 1 2 n k k n n p 63 a e25 b 1 e25 25e25 2 3 25 25 25 25 2 3 e e 64 a 1 7 4 0 4 i i e i p b 1 1 p12 12p1 p11 c 1 pi1p 65 a 1 e12 b PX 2X 1 12 12 12 1 1 2 1 e e e c 1 e12 d 1 exp 500 i1000 66 Assume n 1 a 2 2 n 1 b 2 2 n 2 c exp2n2n 1 e1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 56 Chapter 4 67 Assume n 1 a 2 n b Conditioning on whether the man of couple j sits next to the woman of couple i gives the result 2 1 1 2 2 2 3 1 1 1 1 1 n n n n n n n c e2 68 exp10e5 69 With Pj equal to the probability that 4 consecutive heads occur within j flips of a fair coin P1 P2 P 3 0 and P4 116 P5 12P4 116 332 P6 12P5 14P4 116 18 P7 12P6 14P5 18P4 116 532 P8 12P7 14P6 18P5 116P4 116 632 P9 12P8 14P7 18P6 116P5 116 111512 P10 12P9 14P8 18P7 116P6 116 2511024 2451 The Poisson approximation gives P10 1 exp632 116 1 e25 2212 70 eλt 1 eλtp 71 a 26 5 38 b 26 3 12 38 38 72 Pwins in i games 4 1 6 4 4 3 i i 73 Let N be the number of games played Then PN 4 2124 18 PN 5 4 2 4 1 1 21 2 14 PN 6 2 4 2 5 2 1 2 1 2 516 PN 7 516 EN 48 54 3016 3516 9316 58125 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 57 74 a 2 5 3 b 5 3 6 2 7 8 8 8 8 2 1 2 1 2 1 2 5 6 7 3 3 3 3 3 3 3 c 5 5 2 1 4 3 3 d 5 2 6 2 1 4 3 3 76 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 N N k N N k N N k N N k N N 77 2 2 2 1 2 N k N k N 2 1 2 1 2 1 2 1 2 1 N k N k N 79 a PX 0 94 10 100 10 b PX 2 1 94 94 6 94 6 10 9 1 8 2 100 10 80 Prejected1 defective 310 Prejected4 defective 1 6 10 3 3 56 P4 defectiverejected 5 3 6 10 5 3 3 7 6 10 10 10 75138 81 Prejected 1 94 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 58 Chapter 4 83 Let Xi be the number of accidents that occur on highway i Then 1 2 3 1 2 3 15 E X X X E X E X E X 84 Let Xi equal 1 if box i does not have any balls and let it equal 0 otherwise Then 5 5 5 5 10 1 1 1 1 1 1 i i i i i i i i E X E X P X p Let Yi equal 1 if box i has exactly one ball and let it equal 0 otherwise Then 5 5 5 5 9 1 1 1 1 1 10 1 i i i i i i i i i E Y E Y P Y p p where we used that the number of balls that go into box i is binomial with parameters 10 and pi 85 Let Xi equal 1 if there is at least one type i coupon in the set of n coupons Then 1 1 1 1 1 1 1 1 1 k k k k k n n i i i i i i i i i i E X E X P X p k p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 59 Theoretical Exercises 1 Let Ei no type i in first n selections PT n 1 N i P i E 1 1 1 J n n n i i j k i I J i j k P P P p p 1N n i i P PT n PT n 1 PT n 2 Not true Suppose PX b ε 0 and bn b 1n Then lim n n b b P X b PX b PX b 3 When α 0 PαX β x x x P x F β β α α When α 0 PαX β x 0 1 lim 1 h x x P X F β β α α 4 1 i P N i 1 1 i k P N k 1 1 k i P N K 1 k kP N k E N 5 0 i i P N i 0 1 i k i i P N k 1 1 0 k k i P N k i 1 1 2 k P N k k k 2 1 1 2 k k k P N k kP N k Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 60 Chapter 4 6 EcX cp c11 p Hence 1 EcX if cp c11 p 1 or equivalently pc2 c 1 p 0 or pc 1 pc 1 0 Thus c 1 pp 7 EY EXσ μσ 1 σ E X μσ μσ μσ 0 VarY 1σ2 VarX σ2σ2 1 8 Let I equal 1 if X a and let it equal 0 if X b then X a b aI yielding the result VarX b a2VarI b a2p1 p 9 0 0 1 1 n n i n i i i n P X i p p i If x and y are positive numbers then letting p x x y gives 0 1 n n i n i n x y i x y x y or upon multiplying both sides by x yn that 0 n n i n i i n x y x y i Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 61 10 E1X 1 0 1 1 1 n i n i i n p p i n i i 0 1 1 n i n i i n p p n i i 1 0 1 1 1 1 1 n i n i i n p p i n p 1 1 1 1 1 1 1 n j n j j n p p j n p 0 1 0 1 1 1 1 0 1 n n p p n p 1 1 1 1 1 p n n p 11 For any given arrangement of k successes and n k failures Parrangementtotal of k successes arrangement 1 1 successes 1 k n k k n k P p p n n P k p p k k 12 Condition on the number of functioning components and then use the results of Example 4c of Chapter 1 Prob 0 1 1 n i n i i n i n p p i n i i where 1 i n i 0 if n i i 1 We are using the results of Exercise 11 13 Easiest to first take log and then determine the p that maximizes log PX k log PX k log n k k log p n k log 1 p log 1 k n k P x k p p p 0 p kn maximizes Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 62 Chapter 4 14 a 1 1 1 1 n n p p p α α b Condition on the number of children For k 0 Pk boys 1 children n n P k n p α 1 2n n n k n p k α P0 boys 1 1 1 2 1 n n n p p p α α 17 a If X is binomial n p then from exercise 15 PX is even 1 1 2pn2 1 1 2λnn2 when λ np 1 e2λ2 as n approaches infinity b PX is even eλ 2 2 n n n λ eλeλ eλ2 18 log PX k λ k log λ log k log 1 k P X k λ λ 0 λ k Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 63 19 EX n 0 n i i i e i λλ 1 n i i i e i λλ 1 1 1 n i i i e i λλ 1 1 0 1 n j j j e j λλ 1 0 1 n j j j e j λ λ λ 1 1 n λE X Hence X 3 λEX 12 2 0 1 i i i e i λ λ λ 2 0 0 0 2 i i i i i i i e i ie i e i λ λ λ λ λ λ λ 2 2 1 E X E X λ λVarX E2X 2EX 1 λλ λ2 2λ 1 λλ2 3λ 1 20 Let S denote the number of heads that occur when all n coins are tossed and note that S has a distribution that is approximately that of a Poisson random variable with mean λ Then because X is distributed as the conditional distribution of S given that S 0 PX 1 PS 1S 0 1 0 1 P S e P S e λ λ λ 21 i 1365 ii 1365 iii 1 The events though independent in pairs are not independent 22 i Say that trial i is a success if the ith pair selected have the same number When n is large trials 1 k are roughly independent ii Since Ptrial i is a success 12n 1 it follows that when n is large Mk is approximately Poisson distributed with mean k2n 1 Hence PMk 0 expk2n 1 iii and iv PT αn PMαn 0 expαn2n 1 eα2 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 64 Chapter 4 23 a PEi 1 2 365 0 365 1365 364365 j j j j b exp365PE1 24 a There will be a string of k consecutive heads within the first n trials either if there is one within the first n 1 trials or if the first such string occurs at trial n the latter case is equivalent to the conditions of 2 b Because cases 1 and 2 are mutually exclusive Pn Pn1 1 Pnk11 Ppk 25 Pm counted events n n P m n e n λλ 1 m n m n n m n p p e n m λλ 1 1 m n m p p n m p p e e m n m λ λ λ λ m p p e m λ λ Intuitively the Poisson λ random variable arises as the approximate number of successes in n large independent trials each having a small success probability α and λ nα Now if each successful trial is counted with probability p than the number counted is Binomial with parameters n large and αp small which is approximately Poisson with parameter αpn λp 27 PX n kX n P X n k P X n 1 1 1 n k n p p p p1 pk1 If the first n trials are fall failures then it is as if we are beginning anew at that time 28 The events X n and Y r are both equivalent to the event that there are fewer than r successes in the first n trials hence they are the same event 29 1 P X k P X k 1 1 Np N np k n k Np N Np k n k 1 1 Np k n k k N Np n k Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 65 30 PY j 1 1 j N n n n j N EY 1 1 N j n j N n n N j n j n N n n 1 1 1 1 1 N i n i n N n n 1 1 N n N n n 1 1 n N n 31 Let Y denote the largest of the remaining m chips By exercise 28 PY j 1 1 j m n m m m j n m Now X n m Y and so PX i PY m n i 1 1 m n i m n m m i n 32 PX k 2 0 1 k i k n i n n k 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 66 Chapter 4 34 EX 1 0 2 2 1 2 1 n n n n k n k k n EX 2 2 2 0 2 1 2 1 2 1 n n n n k n k k n n VarX EX 2 EX2 2 2 2 2 2 12 2 1 n n n n n n 2 2 2 2 4 2 n n n n EY 1 2 n EY 2 1 2 2 2 1 1 x 3 n n i dx n i n n VarY 2 2 2 1 3 2 12 n n n 35 a PX i 1 2 1 2 3 1 1 i i i b PX lim i P X i lim1 1 1 1 i i c EX i iP X i 1 i i P X i P X i 1 1 1 i i i i 1 i i 1 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 4 67 36 a This follows because X Y zk k i j A X xi Y yj and the events X xi Y yj i j Ak are mutually exclusive b k k k k k k k i j k i j A k i j k i j A i j i j k i j A E X Y z P X Y z z P X x Y y z P X x Y y x y P X x Y y c This follows from b because every pair i j is in exactly one of the sets Ak d This follows because i j i j X x X x Y y e From the preceding k i j i j k i j A i j i j i j i i j j i j i j i j i i j j i j i j j i i i j j i j E X Y x y P X x Y y x y P X x Y y x P X x Y y y P X x Y y x P X x y y y P X x Y y x P X x y P Y y E X E Y Copyright 2010 Pearson Education Inc Publishing as Prentice Hall