Problemas do Capítulo 3: Análise de Probabilidades

21 Chapter 3 Problems 1 P6 different P6 differentPdifferent 1st 62nd 6 1s 62nd 6 56 P P t 2 16 56 56 13 could also have been solved by using reduced sample spacefor given that outcomes differ it is the same as asking for the probability that 6 is chosen when 2 of the numbers 1 2 3 4 5 6 are randomly chosen 2 P6 sum of 7 61 16 P 16 P6 sum of 8 6 2 536 P 15 P6 sum of 9 6 3 436 P 14 P6 sum of 10 6 4 336 P 13 P6 sum of 11 6 5 236 P 12 P6 sum of 12 1 3 PE has 3 N S has 8 has 3 has 8 has 8 P E N S P N S 13 39 5 21 52 26 8 18 3 10 26 13 13 39 52 8 18 26 339 4 Pat least one 6 sum of 12 1 Otherwise twice the probability given in Problem 2 5 6 5 9 8 15 14 13 12 6 In both cases the one black ball is equally likely to be in either of the 4 positions Hence the answer is 12 7 P1 g and 1 b at least one b 1 2 3 4 23 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 22 Chapter 3 8 12 9 PA w 2w 2 2 P A w w P w 2 P A w B w C w P A w B w C w P w 1 2 3 1 1 1 7 3 3 4 3 3 4 1 2 3 1 1 1 2 2 1 11 2 3 4 3 3 4 3 3 4 10 1150 11 a PBAs 1 3 3 1 1 52 21 52 51 2 17 52 s s P BA P A Which could have been seen by noting that given the ace of spades is chosen the other card is equally likely to be any of the remaining 51 cards of which 3 are aces b PBA 4 3 1 52 51 48 47 33 1 52 51 P B P A 12 a 987 504 b Let Fi denote the event that she failed the ith exam 1 2 2 1 2 3 92 1 504 496 c c c c c P F F P F F F F 3629 13 PE1 4 48 52 1 12 13 PE2E1 3 36 39 1 12 13 PE3E1E2 2 24 26 1 12 13 PE4E1E2E3 1 Hence p 4 48 52 3 36 39 2 24 26 1 12 13 1 12 13 1 12 13 14 5 7 7 9 35 12 14 16 18 768 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 23 15 Let E be the event that a randomly chosen pregnant women has an ectopic pregnancy and S the event that the chosen person is a smoker Then the problem states that PES 2PESc PS 32 Hence PSE PSEPE c C P E S P S P E S P S P E S P S 2 2 c P S P S P S 3266 4548 16 With S being survival and C being C section of a randomly chosen delivery we have that 98 PS PSC15 PSC2 85 9615 PSC2 85 Hence PSCc 9835 17 PD 36 PC 30 PCD 22 a PDC PD PCD 0792 b PDC PDCPC 07923 264 18 a PIndvoted voted Ind Ind voted ype type P P P t P 3546 3546 623 5824 331 b PLibvoted 6230 3546 623 5824 383 c PConvoted 5824 3546 623 5824 286 d Pvoted 3546 623 5824 4862 That is 4862 percent of the voters voted Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 24 Chapter 3 19 Choose a random member of the class Let A be the event that this person attends the party and let W be the event that this person is a woman a PWA P A W P W P A W P W P A M P M where M Wc 4838 4838 3762 443 Therefore 443 percent of the attendees were women b PA 4838 3762 4118 Therefore 4118 percent of the class attended 20 a PFC P FC P C 0205 40 b PCF PFCPF 0252 126 038 21 a Phusband under 25 212 36500 496 b Pwife overhusband over Pboth overPhusband over 54500 252500 314 214 c Pwife overhusband under 36248 145 22 a 6 5 4 5 6 6 6 9 b 1 1 3 6 c 5 1 5 9 6 54 23 Pww transferredPw tr PwR trPR tr 2 1 1 2 4 3 3 3 3 9 Pw transferred w 2 1 tr tr 3 3 4 9 P w w P w P w 12 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 25 24 a Pg gat least one g 1 4 3 4 13 b Since we have no information about the ball in the urn the answer is 12 26 Let M be the event that the person is male and let C be the event that he or she is color blind Also let p denote the proportion of the population that is male PMC 05 05 00251 c c P C M P M p p p P C M P M P C M P M 27 Method b is correct as it will enable one to estimate the average number of workers per car Method a gives too much weight to cars carrying a lot of workers For instance suppose there are 10 cars 9 transporting a single worker and the other carrying 9 workers Then 9 of the 18 workers were in a car carrying 9 workers and so if you randomly choose a worker then with probability 12 the worker would have been in a car carrying 9 workers and with probability 12 the worker would have been in a car carrying 1 worker 28 Let A denote the event that the next card is the ace of spades and let B be the event that it is the two of clubs a PA Pnext card is an acePAnext card is an ace 3 1 3 32 4 128 b Let C be the event that the two of clubs appeared among the first 20 cards PB PBCPC PBCcPCc 19 1 29 29 0 48 32 48 1536 29 Let A be the event that none of the final 3 balls were ever used and let Bi denote the event that i of the first 3 balls chosen had previously been used Then PA PAB0PB0 PAB1PB1 PAB2PB2 PAB3PB3 3 0 6 6 9 3 3 15 15 3 3 i i i i 083 30 Let B and W be the events that the marble is black and white respectively and let B be the event that box i is chosen Then PB PBB1PB1 PBB2PB2 1212 2312 712 PB1W 1 1 1 21 2 512 P W B P B P W 35 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 26 Chapter 3 31 Let C be the event that the tumor is cancerous and let N be the event that the doctor does not call Then β PCN P NC P N c c P N C P C P N C P C P N C P C 1 1 2 α α α 2 1 α α α with strict inequality unless α 1 32 Let E be the event the child selected is the eldest and let Fj be the event that the family has j children Then PFjE j P EF P E j j j j j P F P E F P F P E F 1 1 251 2 3513 31 4 jp j 24 Thus PF1E 24 PF4E 18 33 Let E and R be the events that Joe is early tomorrow and that it will rain tomorrow a 77 93 76 c c P E P E R P R P E R P R b P E R P R P R E P E 4976 34 PGC c c P C G P G P C G P G P C G P G 5462 35 Let U be the event that the present is upstairs and let M be the event it was hidden by mom a 76 54 62 c c P U P U M P M P U M P M b down dadPdad 54 dad down 1019 1 62 38 c c P P M U P Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 27 36 PCwoman women women women women P C P C P A P A P B P B P C P C 7100 1 225 50 75 100 2 5 6 7 225 225 225 37 a Pfairh 1 1 1 2 2 1 1 1 3 2 2 2 b Pfairhh 1 1 1 4 2 1 1 1 5 4 2 2 c 1 38 Ptailsw 3 1 36 36 15 2 3 1 5 1 36 75 111 15 2 12 2 39 Paccno acc no acc acc no acc P P 3 7 46 28 46 10 10 3 7 185 6 8 10 10 40 a 7 8 9 12 1314 b 7 8 5 312 13 14 c 5 6 7 12 13 14 d 5 6 7 312 13 14 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 28 Chapter 3 41 Pace Paceinterchanged selected 1 27 Paceinterchanged not selected 26 27 1 3 26 129 127 51 27 51 27 42 PAfailure 025 10 025 033 052 29 43 P2 headedheads 1 1 4 4 3 1 1 1 1 3 4 2 3 9 3 1 3 2 3 4 45 P5thheads th th th heads 5 5 th i P P P h i P i 10 1 5 1 1 10 10 11 1 10 10 i i 46 Let M and F denote respectively the events that the policyholder is male and that the policyholder is female Conditioning on which is the case gives the following PA2A1 1 2 1 P A A P A 1 2 1 2 1 1 1 1 P A A M P A A F P A M P A F α α α α 2 2 1 1 m f m f p p p p α α α α Hence we need to show that 2 21 m f p p α α pmα pf1 α2 or equivalently that 2 2 2 2 1 1 m f p p a α α α 2α1 αpfpm Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 29 Factoring out α1 α gives the equivalent condition 2 2 2 m f m p p pf or pm pf2 0 which follows because pm pf Intuitively the inequality follows because given the information that the policyholder had a claim in year 1 makes it more likely that it was a type policyholder having a larger claim probability That is the policyholder is more likely to me male if pm pf or more likely to be female if the inequality is reversed than without this information thus raising the probability of a claim in the following year 47 Pall white 1 5 5 4 5 4 3 5 4 3 2 5 4 3 2 1 6 15 15 14 15 14 13 15 14 1312 15 14 1312 11 P3all white 1 5 4 3 6 15 14 13 Pall white 48 a Psilver in othersilver found in other found found P S S P S To compute these probabilities condition on the cabinet selected 1 2 found 12 found 12 P S A P S B 1 2 1 1 2 3 49 Let C be the event that the patient has cancer and let E be the event that the test indicates an elevated PSA level Then with p PC PCE c c P E C P C P E C P C P E C P C Similarly PCEc c c c c c P E C P C P E C P C P E C P C 732 732 8651 p p p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 30 Chapter 3 50 Choose a person at random Pthey have accident Pacc goodPg PaccavePave Paccbad Pb 052 155 303 175 PA is good no accident 952 825 PA is averageno accident 855 825 51 Let R be the event that she receives a job offer a PR PRstrongPstrong PRmoderatePmoderate PRweakPweak 87 42 11 65 b PstrongR strong strong P R P P R 87 56 65 65 Similarly PmoderateR 8 65 PweakR 1 65 c PstrongRc strong strong c c P R P PR 27 14 35 35 Similarly PmoderateRc 12 35 PweakRc 9 35 52 Let M T W Th F be the events that the mail is received on that day Also let A be the event that she is accepted and R that she is rejected a PM PMAPA PMRPR 156 054 11 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 31 b PTMc c P T P M 1 P T A P A P T R P R P M 26 14 16 89 89 c PAMcTcWc c c c c c c P M T W A P A P M T W 1 15 20 256 12 46 754 27 d PATh P Th A P A P Th 156 3 156 154 5 e PAno mail no mail no mail P A P A P 156 9 156 44 25 53 Let W and F be the events that component 1 works and that the system functions PWF 1 1 2 1 1 1 2 c n P WF P W P F P F 55 PBoy F 4 16 x PBoy 10 16 x PF 10 16 x so independence 4 10 10 16 x 4x 36 or x 9 A direct check now shows that 9 sophomore girls which the above shows is necessary is also sufficient for independence of sex and class 56 Pnew 1 new type 1 n i i i i i P i p p p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 32 Chapter 3 57 a 2p1 p b 2 3 1 2 p p c Pup on firstup 1 after 3 Pup first up 1 after 33p21 p p2p1 p3p21 p 23 58 a All we know when the procedure ends is that the two most flips were either H T or T H Thus Pheads PH TH T or T H 1 1 1 P H T p p P H T P T H p p p p 1 2 b No with this new procedure the result will be heads tails whenever the first flip is tails heads Hence it will be heads with probability 1 p 59 a 116 b 116 c The only way in which the pattern H H H H can occur first is for the first 4 flips to all be heads for once a tail appears it follows that a tail will precede the first run of 4 heads and so T H H H will appear first Hence the probability that T H H H occurs first is 1516 60 From the information of the problem we can conclude that both of Smiths parents have one blue and one brown eyed gene Note that at birth Smith was equally likely to receive either a blue gene or a brown gene from each parent Let X denote the number of blue genes that Smith received a PSmith blue gene PX 1X 1 1 2 1 1 4 23 b Condition on whether Smith has a blueeyed gene Pchild blue Pblueblue gene23 Pblueno blue13 1223 13 c First compute PSmith bluechild brown child brown Smith blue23 23 P 12 Now condition on whether Smith has a blue gene given that first child has brown eyes Psecond child brown PbrownSmith blue12 PbrownSmith no blue12 14 12 34 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 33 61 Because the nonalbino child has an albino sibling we know that both its parents are carriers Hence the probability that the nonalbino child is not a carrier is PA AA a or a A or A A 1 3 Where the first gene member in each gene pair is from the mother and the second from the father Hence with probability 23 the nonalbino child is a carrier a Condition on whether the nonalbino child is a carrier With C denoting this event and Oi the event that the ith offspring is albino we have PO1 PO1CPC PO1CcPCc 1423 013 16 b 2 1 P O Oc 1 2 1 c c P O O P O 1 2 1 2 56 c c c c P O O C P C P O O C P C 3 41 423 013 3 56 20 62 a Pboth hitat least one hit both hit at least one hit P P p1p21 q1q2 b PBarb hitat least one hit p11 q1q2 Qi 1 pi and we have assumed that the outcomes of the shots are independent 63 Consider the final round of the duel Let qx 1 px a PA not hit PA not hitat least one is hit PA not hit B hitPat least one is hit qBpA1 qAqB b Pboth hit Pboth hitat least one is hit Pboth hitPat least one hit pApB1 qAqB c qAqBn 11 qAqB d Pn roundsA unhit Pn rounds A unhitPA unhit 1 1 n A B A B B A A B q q p q q p q q qAqBn 11 qAqB Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 34 Chapter 3 e Pn roundsboth hit Pn rounds both hitPboth hit 1 1 n A B A B B A A B q q p p p p q q qAqBn11 qAqB Note that c d and e all have the same answer 64 If use a will win with probability p If use strategy b then Pwin Pwinboth correctp2 Pwinexactly 1 correct2p1 p Pwinneither correct1 p2 p2 p1 p 0 p Thus both strategies give the same probability of winning 65 a Pcorrectagree Pcorrect agreePagree p2p2 1 p2 3652 913 when p 6 b 12 66 a I 1 P1P21 P3P4P5 P1P2 P3P4 P1P2P3P4P5 b Let E1 1 and 4 close E2 1 3 5 all close E3 2 5 close E4 2 3 4 close The desired probability is 67 PE1 E2 E3 E4 PE1 PE2 PE3 PE4 PE1E2 PE1E3 PE1E4 PE2E3 PE2E4 PE3E4 PE1E2E3 PE1E2E4 PE1E3E4 PE2E3E4 PE1E2E3E4 P1P4 P1P3P5 P2P5 P2P3P4 P1P3P4P5 P1P2P4P5 P1P2P3P4 P1P2P3P5 P2P3P4P5 2P1P2P3P4P5 3P1P2P3P4P5 a P1P21 P31 P4 P11 P2P31 P4 P11 P21 P3P4 P2P31 P11 P4 1 P1P21 P3P4 1 P11 P2P3P4 P1P2P31 P4 P1P21 P3P4 P11 P2P3P4 1 P1P2P3P4 P1P2P3P4 c 1 n i n i i k n p p i Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 35 68 Let Ci denote the event that relay i is closed and let F be the event that current flows from A to B PC1C2F 1 2 P C C F P F 1 2 1 2 5 1 2 3 4 1 2 3 4 P F C C P C C p p p p p p p p p 5 1 2 5 1 2 3 4 1 2 3 4 p p p p p p p p p p p p 69 1 a 1 3 1 3 1 9 2 4 2 4 2 128 2 a 1 1 1 1 1 1 2 2 2 2 2 32 b 1 3 1 3 1 9 2 4 2 4 2 128 b 1 1 1 1 1 1 2 2 2 2 2 32 c 18 128 c 1 16 d 110 128 d 15 16 70 a Pcarrier3 without 181 2 181 2 11 2 19 b 118 71 PBraves win PBB wins 3 of 3 18 PBB wins 2 of 3 38 PBB wins 1 of 3 38 PBB wins 0 of 3 18 1 3 1 1 3 3 3 1 38 8 8 4 2 4 8 4 2 64 where PBB wins i of 3 is obtained by conditioning on the outcome of the other series For instance PBB win 2 of 3 PBD or G win 3 of 3 B win 2 of 3 14 PBD or G win 2 of 3 B win 2 of 3 34 1 1 3 2 4 4 By symmetry PD win PG win and as the probabilities must sum to 1 we have PD win PG win 13 64 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 36 Chapter 3 72 Let f denote for and a against a certain place of legislature The situations in which a given steering committees vote is decisive are as follows given member other members of SC other council members for both for 3 or 4 against for one for one against at least 2 for against one for one against at least 2 for against both for 3 of 4 against Pdecisive p34p1 p3 p2p1 p6p21 p2 4p31 p p4 1 p2p1 p6p21 p2 4p31 p p4 1 pp24p1 p3 73 a 116 b 132 c 1032 d 14 e 3132 74 Let PA be the probability that A wins when A rolls first and let PB be the probability that B wins when B rolls first Using that the sum of the dice is 9 with probability 19 we obtain upon conditioning on whether A rolls a 9 that PA 1 8 1 9 9 PB Similarly PB 5 311 36 36 PA Solving these equations gives that PA 919 and that PB 4576 75 a The probability that a family has 2 sons is 14 the probability that a family has exactly 1 son is 12 Therefore on average every four families will have one family with 2 sons and two families with 1 son Therefore three out of every four sons will be eldest sons Another argument is to choose a child at random Letting E be the event that the child is an eldest son letting S be the event that it is a son and letting A be the event that the childs family has at least one son PES P ES P S 2PE 3 1 2 4 4 c P E A P E A 1 3 1 2 0 2 4 4 34 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 37 b Using the preceding notation PES P ES P S 2PE 7 1 2 8 8 c P E A P E A 1 7 2 3 8 712 76 Condition on outcome of initial trial PE before F PE b FEPE PE b FFPF PE b Fneither E or F1 PE PF PE PE b F1 PE PF Hence PE b F P E P E P F 77 a This is equal to the conditional probability that the first trial results in outcome 1 F1 given that it results in either 1 or 2 giving the result 12 More formally with L3 being the event that outcome 3 is the last to occur PF1L3 3 1 1 3 1 213 1 2 13 P L F P F P L b With S1 being the event that the second trial results in outcome 1 we have PF1S1L3 3 1 1 1 1 3 1 219 16 13 P L F S P F S P L 78 a Because there will be 4 games if each player wins one of the first two games and then one of them wins the next two P4 games 2p1 pp2 1 p2 b Let A be the event that A wins Conditioning on the outcome of the first two games gives PA PAa ap2 PAa bp1 p PAb a1 pp PAb b1 p2 p2 PA2p1 p where the notation a b means for instance that A wins the first and B wins the second game The final equation used that PAa b PAb a PA Solving gives PA 2 1 2 1 p p p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 38 Chapter 3 79 Each roll that is either a 7 or an even number will be a 7 with probability p 7 16 7 even 16 1 2 P P P 14 Hence from Example 4i we see that the desired probability is 7 7 2 7 1 4 3 4 i i i i 1 347 734614 80 a PAi 12i if i n 12n1 if i n b 1 1 1 1 2 1 2 1 2 1 2 n i n i n n i n c Condition on whether they initially play each other This gives Pn 2 1 1 2 2 1 2 2 1 2 1 n n n n P where 1 2 2 is the probability they both win given they do not play each other d There will be 2n 1 losers and thus that number of games e Since the 2 players in game i are equally likely to be any of the 2 2 n pairs it follows that PBi 2 1 2 n f Since the events Bi are mutually exclusive P Bi 1 2 2 1 1 2 2 n n n P Bi 81 15 30 1 911 1 911 82 a PA 2 2 2 1 1 2 1 1 P P P P A or PA 2 1 2 2 2 2 1 2 1 2 P P P P P c similar to a with 3 iP replacing 2 iP Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 39 b and d Let ij ij P P denote the probability that A wins when A needs i more and B needs j more and AB is to flip Then Pij P1Pi1j 1 1 ij P P ijP 2 1 2 1 i j ij P P P P These equations can be recursively solved starting with P01 1 P10 0 83 a Condition on the coin flip Pthrow n is red 1 4 1 2 1 2 6 2 6 2 b Prrr 3 3 2 2 1 2 1 1 3 2 3 2 3 5 1 2 1 1 2 3 2 3 P rrr P rr c PArr 2 2 2 2 1 3 2 2 1 1 1 3 2 3 2 P rr A P A P rr 45 84 b PA wins 4 8 7 6 4 8 7 6 5 4 3 4 8 7 6 5 4 3 12 12 1110 9 12 1110 9 8 7 6 12 1110 9 8 7 PB wins 8 4 8 7 6 5 4 8 7 6 5 4 3 2 4 12 11 12 1110 9 8 12 1110 9 8 7 6 5 PC wins 8 7 4 8 7 6 5 4 4 8 7 6 5 4 3 2 1 12 1110 12 1110 9 8 7 12 1110 9 8 7 6 5 85 Part a remains the same The possibilities for part b become more numerous 86 Using the hint PA B 0 0 2 2 2 2 4 n n i n n i n i i n n i i 34n where the final equality uses 0 21 n i n i i n i 2 1n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 40 Chapter 3 b PAB φ PA Bc 34n by part a since Bc is also equally likely to be any of the subsets 87 Pithall heads 0 n k n j i k j k 88 Nothey are conditionally independent given the coin selected 89 a PJ3 votes guiltyJ1 and J2 vote guilty PJ1 J2 J3 all vote guiltyPJ1 and J2 vote guilty 3 3 2 2 7 3 7 2 97 10 10 7 3 142 7 2 10 10 b PJ3 guiltyone of J1 J2 votes guilty 7 3 7273 2228 15 10 10 7 3 26 273 228 10 10 c PJ3 guilty J1 J2 vote innocent 2 2 2 2 7 3 73 28 33 10 10 7 3 102 3 8 10 10 Ei are conditionally independent given the guilt or innocence of the defendant 90 Let Ni denote the event that none of the trials result in outcome i i 1 2 Then PN1 N2 PN1 PN2 PN1N2 1 p1n 1 p2n 1 p1 p2n Hence the probability that both outcomes occur at least once is 1 1 p1n 1 p2n p0n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 41 Theoretical Exercises 1 PABA P AB P AB P A P A B PABA B 2 If A B PAB P A P B PABc 0 PBA 1 PBAc c c P BA P A 3 Let F be the event that a first born is chosen Also let Si be the event that the family chosen in method a is of size i PaF 1 i i i i i n P F S P S i m PbF i i m in Thus we must show that 2 i i i i in n i m or equivalently i j i j i j i j in n j n n or i j i j i j i j i n n n n j Considering the coefficients of the term ninj shows that it is sufficient to establish that 2 i j j i or equivalently i2 j2 2ij which follows since i j2 0 Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 42 Chapter 3 4 Let Ni denote the event that the ball is not found in a search of box i and let Bj denote the event that it is in box j PBjNi i j j c c i i i i i i P N B P B P N B P B P N B P B 1 1 j i i i P P P α if j i 1 1 1 i i i i i P P P α α if j i 5 None are true 6 1 1 1 1 1 1 n n n c i i i P E P E P E 7 a They will all be white if the last ball withdrawn from the urn when all balls are withdrawn is white As it is equally likely to by any of the n m balls the result follows b PRBG last g g b P RBG G r b g r b g r b Hence the answer is bg b g r b r b g r b g r g 8 a PA PACPC PAC cPC c PBCPC PBC c PC c PB b For the events given in the hint PAC 1616 13 336 336 P C A P A Because 16 PA is a weighted average of PAC and PACc it follows from the result PAC PA that PAC c PA Similarly 13 PBC PB PBC c However PABC 0 PABC c 9 PA PB PC 12 PAB PAC PBC 14 But PABC 14 10 PAij 1365 For i j k PAijAjk 3653653 13652 Also for i j k r PAijAkr 13652 11 1 1 pn 12 or n log2 log1 p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 43 12 1 1 1 i i j j a a is the probability that the first head appears on the ith flip and 1 1 i i a is the probability that all flips land on tails 13 Condition on the initial flip If it lands on heads then A will win with probability Pn1m whereas if it lands tails then B will win with probability Pmn and so A will win with probability 1 Pmn 14 Let N go to infinity in Example 4j 15 Pr successes before m failures Prth success occurs before trial m r 1 1 1 1 m r r n r n r n p p r 16 If the first trial is a success then the remaining n 1 must result in an odd number of successes whereas if it is a failure then the remaining n 1 must result in an even number of successes 17 P1 13 P2 1345 2315 25 P3 134567 234517 131517 37 P4 49 b Pn 2 1 n n c Condition on the result of trial n to obtain Pn 1 Pn1 1 1 2 2 1 2 1 n n P n n d Must show that 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 n n n n n n n n n or equivalently that 1 1 2 2 1 2 1 2 1 2 1 2 1 n n n n n n n n n But the right hand side is equal to 2 1 2 12 1 2 1 n n n n n n n Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 44 Chapter 3 18 Condition on when the first tail occurs 19 Pni 1 1 1 1 1 P n i n i p p P 20 αn1 αnp 1 αn1 p1 Pn αnp 1 αnp1 21 b Pn1 PA receives first 2 votes 1 1 1 1 n n n n n n Pn 2 PA receives first 2 and at least 1 of the next 2 1 2 1 2 1 2 1 1 2 n n n n n n n n c Pnm n m n m n m d Pnm PA always ahead PA alwaysA receives last vote n n m PA alwaysB receives last vote m n m n n m Pn1m n m 1 m P n m e The conjecture of c is true when n m 1 n 1 m 0 Assume it when n m k Now suppose that n m k 1 By d and the induction hypothesis we have that Pnm 1 1 1 1 n n m m n m n m n m n m n m n m n m which completes the proof 22 Pn Pn1p 1 Pn11 p 2p 1Pn1 1 p 1 1 1 2 1 2 1 2 2 n p p 1 p by the induction hypothesis 2 1 1 2 1 1 2 2 n p p p 1 1 2 1 2 2 n p Copyright 2010 Pearson Education Inc Publishing as Prentice Hall Chapter 3 45 23 P11 12 Assume that Pab 12 when k a b and now suppose a b k 1 Now Pab Plast is whitefirst a are white 1 a b a Plast is whitefirst b are black 1 b a b Plast is whiteneither first a are white nor first b are black 1 1 1 a b b a a b 1 1 2 a b a b a b a b a b a b 1 2 where the induction hypothesis was used to obtain the final conditional probability above 24 The probability that a given contestant does not beat all the members of some given subset of k other contestants is by independence 1 12k Therefore PBi the probability that none of the other n k contestants beats all the members of a given subset of k contestants is 1 12knk Hence Booles inequality we have that P Bi 1 1 2 k n k n k Hence if 1 1 2 k n k n k 1 then there is a positive probability that none of the n k events Bi occur which means that there is a positive probability that for every set of k contestants there is a contestant who beats each member of this set 25 PEF PEFPF PEFGPGF P EFG P FG P EFG P FG P F P F PEFGcPGcF P EFGc P F The result now follows since PEF PEFG PEFGc Copyright 2010 Pearson Education Inc Publishing as Prentice Hall 46 Chapter 3 27 E1 E2 En are conditionally independent given F if for all subsets i1 ir of 1 2 n 1 1 r j r i i i F j P E E F P E 28 Not true Let F E1 29 Pnext m headsfirst n heads Pfirst n m are headsPfirst n heads 1 1 0 0 1 1 n m n n p dp p dp n m Copyright 2010 Pearson Education Inc Publishing as Prentice Hall