Capítulo 2: Problemas e Questões

10 Chapter 2 Problems 1 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 xn1 n 1 xi 6 i 1 n 1 with the interpretation that the outcome is n x1 xn1 if the first 6 appears on roll n and xi appears on roll i i 1 n 1 The event 1 c n En is the event that 6 never appears 3 EF 1 2 1 4 1 6 2 1 4 1 6 1 E F occurs if the sum is odd or if at least one of the dice lands on 1 FG 1 4 4 1 EFc is the event that neither of the dice lands on 1 and the sum is odd EFG FG 4 A 100010000001 B 01 00001 00000001 A Bc 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 415 8 a 8 b 3 c 0 9 Choose a customer at random Let A denote the event that this customer carries an American Express card and V the event that he or she carries a VISA card PA V PA PV PAV 24 61 11 74 Therefore 74 percent of the establishments customers carry at least one of the two types of credit cards that it accepts Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 2 11 10 Let R and N denote the events respectively that the student wears a ring and wears a necklace a PR N 1 6 4 b 4 PR N PR PN PRN 2 3 PRN Thus PRN 1 11 Let A be the event that a randomly chosen person is a cigarette smoker and let B be the event that she or he is a cigar smoker a 1 PA B 1 07 28 05 7 Hence 70 percent smoke neither b PAcB PB PAB 07 05 02 Hence 2 percent smoke cigars but not cigarettes 12 a PS F G 28 26 16 12 4 6 2100 12 The desired probability is 1 12 12 b Use the Venn diagram below to obtain the answer 32100 14 10 10 S F 8 G 2 4 2 c since 50 students are not taking any of the courses the probability that neither one is taking a course is 50 100 2 2 49198 and so the probability that at least one is taking a course is 149198 13 a 20000 b 12000 c 11000 d 68000 e 10000 1000 7000 19000 I II 0 III 1000 3000 1000 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 12 Chapter 2 14 PM PW PG PMW PMG PWG PMWG 312 470 525 086 042 147 025 1057 15 a 13 52 4 5 5 b 4 12 4 4 4 52 13 2 3 1 1 1 5 c 13 4 4 44 52 2 2 2 1 5 d 4 12 4 4 52 13 3 2 1 1 5 e 4 48 52 13 4 1 5 16 a 5 6 5 4 3 2 6 b 5 6 5 5 4 3 2 6 c 5 6 5 3 2 4 2 2 6 d 5 6 5 4 3 21 e 5 5 6 5 3 6 f 5 5 6 5 4 6 g 5 6 6 17 8 2 1 64 63 58 i i 18 2 4 16 52 51 19 436 436 136 136 518 20 Let A be the event that you are dealt blackjack and let B be the event that the dealer is dealt blackjack Then PA B PA PB PAB 4 4 16 4 4 16 3 15 52 51 52 51 50 49 0983 where the preceding used that PA PB 2 4 16 52 51 Hence the probability that neither is dealt blackjack is 9017 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 2 13 21 a p1 420 p2 820 p3 520 p4 220 p5 120 b There are a total of 4 1 8 2 5 3 2 4 1 5 48 children Hence q1 448 q2 1648 q3 1548 q4 848 q5 548 22 The ordering will be unchanged if for some k 0 k n the first k coin tosses land heads and the last n k land tails Hence the desired probability is n 12n 23 The answer is 512 which can be seen as follows 1 Pfirst higher Psecond higher psame 2Psecond higher psame 2Psecond higher 16 Another way of solving is to list all the outcomes for which the second is higher There is 1 outcome 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 4 536 512 25 PEn 1 1 26 6 2 36 36 5 n n n P E 27 Imagine that all 10 balls are withdrawn PA 3 9 7 6 3 7 7 6 5 4 3 5 7 6 5 4 3 2 3 3 10 28 Psame 5 6 8 3 3 3 19 3 Pdifferent 5 6 8 19 1 1 1 3 If sampling is with replacement Psame 3 3 3 3 5 6 8 19 Pdifferent PRBG PBRG PRGB PGBR 3 6 5 6 8 19 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 14 Chapter 2 29 a 1 1 1 n n m m n m n m b Putting all terms over the common denominator n m2n m 1 shows that we must prove that n2n m 1 m2n m 1 nn 1n m mm 1n m which is immediate upon multiplying through and simplifying 30 a 7 8 3 3 3 8 9 4 4 4 118 b 7 8 3 3 3 8 9 4 4 4 118 16 c 7 8 7 8 3 4 4 3 8 9 4 4 12 31 Pcomplete Psame 32 1 g b g g b g b g 33 5 15 2 2 70 20 323 4 34 32 52 13 13 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 2 15 35 a 12 16 18 3 2 2 46 7 b 34 12 34 7 1 6 1 46 46 7 7 c 12 16 18 7 7 7 46 7 d 3 3 3 3 3 3 12 34 16 30 12 16 18 3 4 3 4 3 3 1 46 46 46 7 7 7 P R B P R P B P R B 36 a 4 52 2 2 0045 b 4 52 13 2 2 117 0588 37 a 7 10 5 5 112 0833 b 7 3 10 4 1 5 112 12 38 12 3 2 2 n or nn 1 12 or n 4 39 5 4 3 12 5 5 5 25 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 16 Chapter 2 40 P1 4 1 44 64 P2 4 4 4 84 4 4 4 2 2 256 P3 4 4 3 36 4 4 3 1 2 64 P4 4 4 6 64 4 41 1 4 4 5 6 42 1 35 36 n 43 2 1 2 2 n n n n in a line 2 2 2 1 n n n n if in a circle n 2 44 a If A is first then A can be in any one of 3 places and Bs place is determined and the others can be arranged in any of 3 ways As a similar result is true when B is first we see that the probability in this case is 2 3 35 310 b 2 2 35 15 c 2 35 110 45 1n if discard 1 1k k n n if do not discard 46 If n in the room Pall different 12 11 13 12 12 12 n When n 5 this falls below 12 Its value when n 5 is 3819 47 121212 48 20 4 4 12 8 20 12 4 4 3 2 49 6 6 12 3 3 6 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 2 17 50 13 39 8 31 52 39 5 8 8 5 13 13 51 1 n m n n n N m 52 a 20 18 16 14 12 10 8 6 20 19 18 17 16 15 14 13 b 6 10 9 82 1 6 2 20 19 18 17 16 15 14 13 53 Let Ai be the event that couple i sit next to each other Then 2 3 4 4 1 2 7 2 6 2 5 2 4 4 6 4 8 8 8 8 i i P A and the desired probability is 1 minus the preceding 54 PS H D C PS PH PD PC PSH PSHDC 39 26 13 4 6 4 13 13 13 52 52 52 13 13 13 39 26 4 6 4 13 13 52 13 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 18 Chapter 2 55 a PS H D C PS PSHDC 3 4 2 2 2 48 2 46 2 44 4 6 4 2 2 2 9 2 7 2 5 52 52 52 52 13 13 13 13 50 48 46 44 4 6 4 11 9 7 5 52 13 b P1 2 13 48 13 44 13 40 13 9 2 5 3 1 52 52 52 13 13 13 56 Player B If Player A chooses spinner a then B can choose spinner c If A chooses b then B chooses a If A chooses c then B chooses b In each case B wins probability 59 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 2 19 Theoretical Exercises 5 Fi 1 1 i c i j E j E 6 a EFcGc b EFcG c E F G d EF EG FG e EFG f EcFcGc g EcFcGc EFcGc EcFGc EcFcG h EFGc i EFGc EFcG EcFG j S 7 a E b EF c EG F 8 The number of partitions that has n 1 and a fixed set of i of the elements 1 2 n as a subset is Tni Hence where T0 1 Hence as there are n i such subsets Tn1 1 0 0 1 1 1 n n n n i n i k i i k n n n T T T i i k 11 1 PE F PE PF PEF 12 PEFc EcF PEFc PEcF PE PEF PF PEF 13 E EF EFc Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 20 Chapter 2 15 M N k r k M N r 16 PE1 En PE1 En1 PEn 1 by Bonferonnis Ineq 1 1 n i P E n 2 PEn 1 by induction hypothesis 19 1 1 1 1 n m n r r k r n m n m k k 21 Let y1 y2 yk denote the successive runs of losses and x1 xk the successive runs of wins There will be 2k runs if the outcome is either of the form y1 x1 yk xk or x1y1 xk yk where all xi yi are positive with x1 xk n y1 yk m By Proposition 61 there are 1 1 2 1 1 n m k k number of outcomes and so P2k runs 1 1 2 1 1 n m m n k k n There will be 2k 1 runs if the outcome is either of the form x1 y1 xk yk xk1 or y1 x1 yk xk yk 1 where all are positive and ix n iy m By Proposition 61 there are 1 1 1 n m k k outcomes of the first type and 1 1 1 n m k k of the second Copyright 2010 Pearson Education Inc Publishing as Prentice Hall