Soluto to Problem 1.31. (a) Let A be the evet that a 0 s trasmtted. Usg the total probablty theorem, the desred probablty s P(A)(1 ɛ ( 0)+ 1 P(A) ) (1 ɛ 1)=p(1 ɛ 0)+(1 p)(1 ɛ 1). (b) By depedece, the probablty that the strg 1011 s receved correctly s (1 ɛ 0)(1 ɛ 1) 3. (c) I order for a 0 to be decoded correctly, the receved strg must be 000, 001, 010, or 100. Gve that the strg trasmtted was 000, the probablty of recevg 000 s (1 ɛ 0) 3, ad the probablty of each of the strgs 001, 010, ad 100 s ɛ 0(1 ɛ 0) 2. Thus, the probablty of correct decodg s 3ɛ 0(1 ɛ 0) 2 +(1 ɛ 0) 3. (d) Whe the symbol s 0, the probabltes of correct decodg wth ad wthout the scheme of part (c) are 3ɛ 0(1 ɛ 0) 2 +(1 ɛ 0) 3 ad 1 ɛ 0, respectvely. Thus, the probablty s mproved wth the scheme of part (c) f or whch s equvalet to ɛ 0 < 1/2. (e) Usg Bayes rule, we have 3ɛ 0(1 ɛ 0) 2 +(1 ɛ 0) 3 > (1 ɛ 0), P(0 101) = (1 ɛ 0)(1 + 2ɛ 0) > 1, P(0)P(101 0) P(0)P(101 0) + P(1)P(101 1). The probabltes eeded the above formula are P(0) = p, P(1) = 1 p, P(101 0) = ɛ 2 0(1 ɛ 0), P(101 1) = ɛ 1(1 ɛ 1) 2. Soluto to Problem 1.32. The aswer to ths problem s ot uque ad depeds o the assumptos we make o the reproductve strategy of the kg s parets. Suppose that the kg s parets had decded to have exactly two chldre ad the stopped. There are four possble ad equally lkely outcomes, amely BB, GG, BG, ad GB (B stads for boy ad G stads for grl ). Gve that at least oe chld was a boy (the kg), the outcome GG s elmated ad we are left wth three equally lkely outcomes (BB, BG, ad GB). The probablty that the sblg s male (the codtoal probablty of BB) s 1/3. Suppose o the other had that the kg s parets had decded to have chldre utl they would have a male chld. I that case, the kg s the secod chld, ad the sblg s female, wth certaty. 11
Soluto to Problem 1.33. Flp the co twce. If the outcome s heads-tals, choose the opera. f the outcome s tals-heads, choose the moves. Otherwse, repeat the process, utl a decso ca be made. Let A k be the evet that a decso was made at the kth roud. Codtoal o the evet A k, the two choces are equally lkely, ad we have 1 P(opera) = P(opera A k )P(A k )= 2 P(A k)= 1 2. k=1 We have used here the property P(A k=0 k) = 1, whch s true as log as P(heads) > 0 ad P(tals) > 0. Soluto to Problem 1.34. The system may be vewed as a seres coecto of three subsystems, deoted 1, 2, ad 3 Fg. 1.19 the text. The probablty that the etre system s operatoal s p 1p 2p 3, where p s the probablty that subsystem s operatoal. Usg the formulas for the probablty of success of a seres or a parallel system gve Example 1.24, we have k=1 p 1 = p, p 3 =1 (1 p) 2, ad p 2 =1 (1 p) ( 1 p ( 1 (1 p) 3)). Soluto to Problem 1.35. Let A be the evet that exactly compoets are operatoal. The probablty that the system s operatoal s the probablty of the uo A, ad sce the A are dsjot, t s equal to P(A )= p(), where p() are the bomal probabltes. Thus, the probablty of a operatoal system s p (1 p). Soluto to Problem 1.36. (a) Let A deote the evet that the cty expereces a black-out. Sce the power plats fal depedet of each other, we have P(A) = p. (b) There wll be a black-out f ether all or ay 1 power plats fal. These two evets are dsjot, so we ca calculate the probablty P(A) of a black-out by addg ther probabltes: ( ) P(A) = p + (1 p ) p j. =1 =1 12 =1 j
Here, (1 p ) pj s the probablty that 1 plats have faled ad plat s the j oe that has ot faled. Soluto to Problem 1.37. The probablty that k 1 voce users ad k 2 data users smultaeously eed to be coected s p 1(k 1)p 2(k 2), where p 1(k 1)adp 2(k 2) are the correspodg bomal probabltes, gve by p (k )= ( k ) p k (1 p) k, =1, 2. The probablty that more users wat to use the system tha the system ca accommodate s the sum of all products p 1(k 1)p 2(k 2)ask 1 ad k 2 rage over all possble values whose total bt rate requremet k 1r 1+k 2r 2 exceeds the capacty c of the system. Thus, the desred probablty s p 1(k 1)p 2(k 2). {(k 1,k 2 ) k 1 r 1 +k 2 r 2 >c, k 1 1,k 2 2 } Soluto to Problem 1.38. We have p T = P(at least 6 out of the 8 remag holes are wo by Tels), p W = P(at least 4 out of the 8 remag holes are wo by Wedy). Usg the bomal formulas, p T = 8 k=6 8 p k (1 p) 8 k, p W = k 8 k=4 8 (1 p) k p 8 k. k The amout of moey that Tels should get s 10 p T /(p T + p W ) dollars. Soluto to Problem 1.39. Let the evet A be the evet that the professor teaches her class, ad let B be the evet that the weather s bad. We have ad Therefore, P(A) =P(B) P(A) =P(B)P(A B)+P(B c )P(A B c ), P(A B) = P(A B c )= p b(1 p b ), p g(1 p g). p b(1 p b ) + ( 1 P(B) ) 13 p g(1 p g).
ot a kg s 47/51. We cotue smlarly utl the 12th card. The probablty that the 12th card s ot a kg, gve that oe of the precedg 11 was a kg, s 37/41. (There are 52 11 = 41 cards left, ad 48 11 = 37 of them are ot kgs.) Fally, the codtoal probablty that the 13th card s a kg s 4/40. The desred probablty s 48 47 37 4 52 51 41 40. Soluto to Problem 1.53. Suppose we label the classes A, B, adc. The probablty that Joe ad Jae wll both be class A s the umber of possble combatos for class A that volve both Joe ad Jae, dvded by the total umber of combatos for class A. Therefore, ths probablty s 88 28. 90 30 Sce there are three classes, the probablty that Joe ad Jae ed up the same class s 88 28 3. 90 30 A much smpler soluto s as follows. We place Joe oe class. Regardg Jae, there are 89 possble slots, ad oly 29 of them place her the same class as Joe. Thus, the aswer s 29/89, whch turs out to agree wth the aswer obtaed earler. Soluto to Problem 1.54. (a) Sce the cars are all dstct, there are 20! ways to le them up. (b) To fd the probablty that the cars wll be parked so that they alterate, we cout the umber of favorable outcomes, ad dvde by the total umber of possble outcomes foud part (a). We cout the followg maer. We frst arrage the US cars a ordered sequece (permutato). We ca do ths 10! ways, sce there are 10 dstct cars. Smlarly, arrage the foreg cars a ordered sequece, whch ca also be doe 10! ways. Fally, terleave the two sequeces. Ths ca be doe two dfferet ways, sce we ca let the frst car be ether US-made or foreg. Thus, we have a total of 2 10! 10! possbltes, ad the desred probablty s 2 10! 10!. 20! Note that we could have solved the secod part of the problem by eglectg the fact that the cars are dstct. Suppose the foreg cars are dstgushable, ad also that the US cars are dstgushable. Out of the 20 avalable spaces, we eed to choose 10 spaces whch to place the US cars, ad thus there are ( 20 10) possble outcomes. Out of these outcomes, there are oly two whch the cars alterate, depedg o 17
from a l-letter alphabet s equal to l 1 l!. w 1 Soluto to Problem 1.58. (a) The sample space cossts of all ways of drawg 7 elemets out of a 52-elemet set, so t cotas 52 7 possble outcomes. Let us cout those outcomes that volve exactly 3 aces. We are free to select ay 3 out of the 4 aces, ad ay 4 out of the 48 remag cards, for a total of ( 4 48 ) 3)( 4 choces. Thus, 4 48 P(7 cards clude exactly 3 aces) = 3 4. 52 7 (b) Proceedg smlar to part (a), we obta P(7 cards clude exactly 2 kgs) = 4 48 2 5. 52 7 (c) If A ad B stad for the evets parts (a) ad (b), respectvely, we are lookg for P(A B) =P(A) +P(B) P(A B). The evet A B (havgexactly3aces ad exactly 2 kgs) ca occur by choosg 3 out of the 4 avalable aces, 2 out of the 4 avalable ) kgs, ad 2 more cards out of the remag 44. Thus, ths evet cossts of dstct outcomes. Hece, ( 4 )( 4 )( 44 3 2 2 P(7 cards clude 3 aces ad/or 2 kgs) = 4 48 3 4 + 4 48 4 4 44 2 5 3 2 2. 52 7 Soluto to Problem 1.59. Clearly f > m,or > k,orm > 100 k, the probablty must be zero. If m, k, adm 100 k, the we ca fd the probablty that the test drve foud of the 100 cars defectve by coutg the total umber of sze m subsets, ad the the umber of sze m subsets that cota lemos. Clearly, there are 100 m dfferet subsets of sze m. To cout the umber of sze m subsets wth lemos, we frst choose lemos from the k avalable lemos, ad the choose m good cars from the 100 k avalable good cars. Thus, the umber of ways to choose a subset of sze m from 100 cars, ad get lemos, s k 100 k, m 19
ad the desred probablty s k 100 k m. 100 m Soluto to Problem 1.60. The sze of the sample space s the umber of dfferet ways that 52 objects ca be dvded 4 groups of 13, ad s gve by the multomal formula 52! 13! 13! 13! 13!. There are 4! dfferet ways of dstrbutg the 4 aces to the 4 players, ad there are 48! 12! 12! 12! 12! dfferet ways of dvdg the remag 48 cards to 4 groups of 12. Thus, the desred probablty s 48! 4! 12! 12! 12! 12!. 52! 13! 13! 13! 13! A alteratve soluto ca be obtaed by cosderg a dfferet, but probablstcally equvalet method of dealg the cards. Each player has 13 slots, each oe of whch s to receve oe card. Istead of shufflg the deck, we place the 4 aces at the top, ad start dealg the cards oe at a tme, wth each free slot beg equally lkely to receve the ext card. For the evet of terest to occur, the frst ace ca go aywhere; the secod ca go to ay oe of the 39 slots (out of the 51 avalable) that correspod to players that do ot yet have a ace; the thrd ca go to ay oe of the 26 slots (out of the 50 avalable) that correspod to the two players that do ot yet have a ace; ad fally, the fourth, ca go to ay oe of the 13 slots (out of the 49 avalable) that correspod to the oly player who does ot yet have a ace. Thus, the desred probablty s 39 26 13 51 50 49. By smplfyg our prevous aswer, t ca be checked that t s the same as the oe obtaed here, thus corroboratg the tutve fact that the two dfferet ways of dealg the cards are probablstcally equvalet. 20
clear all, close all; load sequeces varble load sequece1.mat; sequece1 = test_sequece; p_1 = sum ( 1 == cov ( 1 /1 * [1 ], sequece1 ) ) / legth(sequece1); p_2 = sum ( 1 == cov ( 1 / 2 * [1 1 ], sequece1 ) ) /legth(sequece1); p_3 = sum ( 1 == cov ( 1 / 3 * [1 1 1 ], sequece1 ) ) / legth(sequece1); p_4 = sum ( 1 == cov ( 1 / 4 * [1 1 1 1 ], sequece1 ) ) / legth(sequece1); p_5 = sum ( 1 == cov ( 1 / 5 * [1 1 1 1 1 ], sequece1 ) ) / legth(sequece1); Probablty_sequece1=[p_1;p_2;p_3;p_4;p_5]; load sequece2.mat; sequece2 = test_sequece; p_1 = sum ( 1 == cov ( 1 /1 * [1 ], sequece2 ) ) / legth(sequece2); p_2 = sum ( 1 == cov ( 1 / 2 * [1 1 ], sequece2 ) ) /legth(sequece2); p_3 = sum ( 1 == cov ( 1 / 3 * [1 1 1 ], sequece2 ) ) / legth(sequece2); p_4 = sum ( 1 == cov ( 1 / 4 * [1 1 1 1 ], sequece2 ) ) / legth(sequece2); p_5 = sum ( 1 == cov ( 1 / 5 * [1 1 1 1 1 ], sequece2 ) ) / legth(sequece2); Probablty_sequece2=[p_1;p_2;p_3;p_4;p_5]; load sequece3.mat; sequece3 = test_sequece; p_1 = sum ( 1 == cov ( 1 /1 * [1 ], sequece3 ) ) / legth(sequece3); p_2 = sum ( 1 == cov ( 1 / 2 * [1 1 ], sequece3 ) ) /legth(sequece3); p_3 = sum ( 1 == cov ( 1 / 3 * [1 1 1 ], sequece3 ) ) / legth(sequece3); p_4 = sum ( 1 == cov ( 1 / 4 * [1 1 1 1 ], sequece3 ) ) / legth(sequece3); p_5 = sum ( 1 == cov ( 1 / 5 * [1 1 1 1 1 ], sequece3 ) ) / legth(sequece3); Probablty_sequece3=[p_1;p_2;p_3;p_4;p_5]; load sequece4.mat; sequece4 = test_sequece; p_1 = sum ( 1 == cov ( 1 /1 * [1 ], sequece4 ) ) / legth(sequece4); p_2 = sum ( 1 == cov ( 1 / 2 * [1 1 ], sequece4 ) ) /legth(sequece4); p_3 = sum ( 1 == cov ( 1 / 3 * [1 1 1 ], sequece4 ) ) / legth(sequece4); p_4 = sum ( 1 == cov ( 1 / 4 * [1 1 1 1 ], sequece4 ) ) / legth(sequece4); p_5 = sum ( 1 == cov ( 1 / 5 * [1 1 1 1 1 ], sequece4 ) ) / legth(sequece4); Probablty_sequece4=[p_1;p_2;p_3;p_4;p_5]; load sequece5.mat; sequece5 = test_sequece; p_1 = sum ( 1 == cov ( 1 /1 * [1 ], sequece5 ) ) / legth(sequece5); p_2 = sum ( 1 == cov ( 1 / 2 * [1 1 ], sequece5 ) ) /legth(sequece5); p_3 = sum ( 1 == cov ( 1 / 3 * [1 1 1 ], sequece5 ) ) / legth(sequece5); p_4 = sum ( 1 == cov ( 1 / 4 * [1 1 1 1 ], sequece5 ) ) / legth(sequece5); p_5 = sum ( 1 == cov ( 1 / 5 * [1 1 1 1 1 ], sequece5 ) ) / legth(sequece5);
Probablty_sequece5=[p_1;p_2;p_3;p_4;p_5]; load sequece6.mat; sequece6 = test_sequece; p_1 = sum ( 1 == cov ( 1 /1 * [1 ], sequece6 ) ) / legth(sequece6); p_2 = sum ( 1 == cov ( 1 / 2 * [1 1 ], sequece6 ) ) /legth(sequece6); p_3 = sum ( 1 == cov ( 1 / 3 * [1 1 1 ], sequece6 ) ) / legth(sequece6); p_4 = sum ( 1 == cov ( 1 / 4 * [1 1 1 1 ], sequece6 ) ) / legth(sequece6); p_5 = sum ( 1 == cov ( 1 / 5 * [1 1 1 1 1 ], sequece6 ) ) / legth(sequece6); Probablty_sequece6=[p_1;p_2;p_3;p_4;p_5]; Result=[Probablty_sequece1 Probablty_sequece2 Probablty_sequece3 Probablty_sequ