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Homework 4 Solutions 1. Let p1 be the probability that Andre wins a set against his father, and let p 2 be the probability that Andre wins a set against the club champion. Clearly, because the club champion is a better player than Andre’s father, we have p1 p 2 . For sequence (a) father-champion-father, the probability of Andre winning at least 2 tennis sets in a row is just the probability of him winning the first 2 sets plus the probability of him losing the 1st set and winning the 2nd and 3rd sets: P( Andre wins sets 1 2) P( Andre loses set 1 and wins sets 2 3) p1 p 2 (1 p1 ) p 2 p1 (2 p1 ) p1 p 2 For sequence (b) champion-father-champion, the probability of Andre winning at least 2 tennis sets in a row is instead p 2 p1 (1 p 2 ) p1 p 2 (2 p 2 ) p1 p 2 . Comparing the probabilities for (a) and (b), note that because p1 p 2 , it follows that 2 p1 2 p2 ; this means that Andre would have a higher probability of winning at least 2 sets in a row if he chose to play the champion-father-champion sequence. 2. Ross 3.80 p. 122 a) Let Ai be the event that A plays in exactly i contests, where i can be any integer value between 1 nd n, inclusive. In order for A to play in exactly i contests, A must lose the i-th contests for values of i less than n ( i n 1 ). Why does this not hold when i n ? Because once A reaches the n-th contest, there are none after, so A will play in exactly n contests regardless of whether he wins or loses the n-th contest. In other words, the probability that A plays in exactly n contests is just the probability that A wins all contests up to and including contest n 1 . These statements give us P ( Ai ) as (1 / 2) i i n 1 P( Ai ) n 1 . (1 / 2) in b) Let E denote the event that contestants A and B ever play each other. We can obtain this probability by conditioning on the events Ai . In other words, we know that n P( E ) P( E | Ai ) P( Ai ) . i 1 Note that Ai means A faced i different contestants, and each of the 2 n 1 other contestants have an equal chance of ever playing with A; hence, P( E | Ai ) i /(2 n 1) . Using what we determined for P ( Ai ) from (a), we have n 1 i n P( E ) n (1 / 2) n i (1 / 2) n 1 i 1 2 1 2 1 1 n 1 n i (1 / 2) n n 1 n i (1 / 2) 2 1 i 1 2 1 1 1 1 n(1 / 2) (n 1)(1 / 2) 1 n 1 n n n 1 n n(1 / 2) 2 1 2 (1 1 / 2) 2 2 1 1 n n 1 2(1 n(1 / 2) (n 1)(1 / 2) ) n(1 / 2) n n 1 2 1 1 n 2 n(1 / 2) 2n(1 / 2) 2(1 / 2) 2 1 n 1 n n 2 2 1 n [1 (1 / 2) ] n n 1 n 2 1 2 2 c) Let Pn P (E ) be the probability that A and B meet in a contest with 2 n players. Now, let F be the event that A and B play in the first contest; then P ( E ) Pn P( E | F ) P( F ) P( E | F C ) P ( F C ) . But since we know that the probability that A and B meet in the first contest is just 1 /(2 n 1) , we get that 1 C 2 2 n Pn (1) n 2 1 . P( E | F ) n 2 1 Now, let G be the event that both A and B advance to the second round of this contest (i.e., the event that both win their first match). Then we have that P( E | F C ) P( E | F C G ) P(G | F C ) P( E | F C G C ) P(G C | F C ) 2 2 1 1 P( E | F C ) Pn 1 0 Pn 1 2 2 In other words, the probability that they meet given that they both advance to the second round and don’t meet in the first round is just the probability that out of 2 n 1 possible contestants, we see A and B meet ( 2 n 1 because the number of contestants left after 1 round has been completed is half the original number). Hence, we have that 2 n 2 1 2 1 P( E ) Pn n n Pn 1 . 2 1 2 1 2 To verify that this holds, let us plug in P ( E ) Pn 1 /(2 n 1 ) (and hence Pn 1 1 /(2 n 2 ) ) to verify that this identity holds: 2 n 2 1 2 n 2 1 1 2 2 1 1 n Pn 1 n n 2 n 1 2 1 2 2 1 2 1 2 2 n 2 2 n 2 1 2n 2n 2 n n 1 1 n 2 1 2 2 n 2n 1 2 1 2 n 1 2 1 2(2 n 1) 2 n n 1 1 2 1 2 2 n 1 2 n 2 n 1 Pn n 2 (Checks out) d) Note that every game played yields 1 loser. Also, with 2 n contestants and only 1 winner, there must be 2 n 1 losers; hence, we require 2 n 1 games in order to obtain 2 n 1 losers. e) Let B i denote the event that A and B play each other in game i, where i 1, ,2 n 1 . 2n Note that there are possible pairings that can occur. Also, since players are equally 2 2n skilled, the 2 players in game i are equally likely to be any of the pairs. Hence, it 2 2 n 2 follows that P( Bi ) 1 n n 2 2 (2 1) . f) Since the events B i are mutually exclusive, then the event E is just the union of the events B i ; hence, 2 2 1 P( E ) P( Bi ) (2 n 1) n n 2 (2 1) 2 n 2 n1 . 3. Let a k denote the probability of having no two consecutive heads in a sequence of k tosses of a fair coin. Then let bk (i) be the probability of having at least 2 heads in a row if there are k tosses remaining, i consecutive heads have already appeared, and there has not already been a run of at least 2 heads prior to the current run of i heads in a row. A recursive formula using bk (i ) ’s will give the complement to a k . Adapting the FT shooting recursive formula from the addendum to the notes for a coin-tossing scenario, we have that 1 1 bk (i) bk 1 (i 1) bk 1 (0) for i 0,1 2 2 with boundary conditions b0 (i ) 0 for i 0,1 and bk (2) 1 . Then the probability that we have no runs of at least 2 consecutive heads in a sequence of k tosses of a fair coin is just a k 1 bk (0) . The following R function will be used to calculate bk (0) for k = 5, 10, and 25 by calculating b0 (0), b0 (1), b0 (2), b1 (0), b1 (1), , bk 1 (1), bk 1 (2), bk (0) . bmat <- matrix(rep(0, 26*3), nrow = 26) # bmat will store the b_k(i)’s. The 1, ..., 26 rows of bmat correspond to # k=0,...,25 and the 1,...3 columns of bmat correspond to i=0,1,2. bmat[,3] <- 1 # set b_k(2)=1 for(k in 2:26){ bmat[k,1:2] <- .5*bmat[(k-1),2:3]+.5*bmat[(k-1),1] # Apply recursive formula } To get a 5 , a10 , and a 25 from the matrix, we just need to input 1 – bmat[c(5, 10, 25)+1, 1]: > 1 - bmat[c(5,10,25)+1, 1] # gives us a_5, a_10, and a_25, respectively [1] 0.406250000 0.140625000 0.005853713 So, as we can see, the probability of not seeing at least 1 run of 2 heads or longer in a sequence of k tosses decreases fairly rapidly as k increases. 4. Ross 4.4 p. 187 We have 5 men and 5 women who are ranked according to their scores on an exam, where no two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a woman. Clearly, since 6 is the lowest possible rank attainable by the highest-scoring female, we must have P( X 7) P( X 8) P( X 9) P( X 10) 0 . For X 1 (female is highest-ranking scorer), we have 5 possible choices out of 10 for the top spot that satisfy this requirement; hence P( X 1) 1 / 2 . For X 2 (female is 2nd-highest scorer), we have 5 possible choices for the top male, then 5 possible choices for the female who ranked 2nd overall, and then any arrangement of the remaining 8 individuals is acceptable (out of 10! possible arrangements of 10 individuals); hence, P( X 2) (5 5 8!) / 10! 5 / 18 0.278 . For X 3 (female is 3rd-highest scorer), acceptable configurations yield (5)(4)=20 possible choices for the top 2 males, 5 possible choices for the female who ranked 3rd overall, and 7! different arrangement of the remaining 7 individuals (out of a total of 10! possible arrangements of 10 individuals); hence, P( X 3) (5 4 5 7!) / 10! 5 / 36 0.139 . For X 4 (female is 4th-highest scorer), acceptable configurations yield (5)(4)(3)=60 possible choices for the top 3 males, 5 choices for the female who ranked 4th overall, and 6! different arrangements of the remaining 6 individuals (out of a total of 10! possible arrangements of 10 individuals); this gives us P( X 4) (5 4 3 5 6!) / 10! 5 / 84 0.060 . For X 5 (female is 5th-highest scorer), satisfactory arrangements have (5)(4)(3)(2)=120 possible choices for the top 4 males, 5 choices for the female who ranked 5th overall, and 5! different configurations of the remaining 5 individuals, thus giving us a probability of P( X 5) (5 4 3 2 5 5!) / 10! 5 / 252 0.020 . For X 6 (female is 6th-highest scorer), we basically have 5! arrangements of the males in the top 5 positions, and another 5! arrangements of the females in the bottom 5 positions; hence, the probability of all 5 males outperforming the top female is P( X 6) (5!5!) / 10! 1 / 252 0.004 . 5. Ross 4.14 p. 188 So we have 5 numbers randomly distributed to players numbered 1 through 5; players compare numbers in pairs, with the higher number winning. Let X denote the number of times player 1 is a winner. Then the probabilities P( X k ) are as follows: P( X 0) P (1 loses to 2) 1 / 2 P( X 1) P(of 1, 2, 3 : 3 has largest,then1, then 2) (1 / 3)(1 / 2) 1 / 6 P( X 2) P(of 1, 2, 3, 4 : 4 has largestand 1 has next largest) (1 / 4)(1 / 3) 1 / 12 P( X 3) P (of 1, 2, 3, 4, 5 : 5 has largest then1) (1 / 5)(1 / 4) 1 / 20 P( X 4) P(1 has largest) 1 / 5 6. i) Ross 4.21 p. 190 a) It would seem reasonable to believe that E (X ) should be larger than E (Y ) . Note that each of the bus drivers is equally likely to be selected, whereas the student selected is more likely to have come from a bus with larger capacity. In other words, X has greater weight placed on the larger busses, and Y does not. b) For X, we know that P( X k ) k / 148 , where k can take on values of 40, 33, 25, and 50. Hence, E ( X ) (40)(40 / 148) (33)(33 / 148) (25)(25 / 148) (50)(50 / 148) 39.284 . On the other hand, E (Y ) (40)(1 / 4) (33)(1 / 4) (25)(1 / 4) (50)(1 / 4) 37 . ii) Ross 4.37 p. 192 We now want Var(X ) and Var(Y ) for the random variables from (i). To determine their values, let us first calculate E ( X 2 ) an E (Y 2 ) : E ( X 2 ) (40) 2 (40 / 148) (33) 2 (33 / 148) (25) 2 (25 / 148) (50) 2 (50 / 148) 1625.419 E (Y 2 ) (40) 2 (1 / 4) (33) 2 (1 / 4) (25) 2 (1 / 4) (50) 2 (1 / 4) 1453.5 This gives us the variances of X and Y: Var( X ) E ( X 2 ) E ( X ) 2 1625.419 (39.284) 2 82.186 . Var(Y ) E (Y 2 ) E (Y ) 2 1453.5 (37) 2 84.5 7. Ross 4.32 p. 192 a) Let T be a random variable denoting the number of tests that are required for a group of 10 people. Then we have that T 1 if all 10 people turn up negative T 11 if at least 1 people turns up positive. Hence, P(T 1) (1 0.9)10 1 P(T 11) , and this gives us as the expected number of tests necessary for each group E (T ) (1)( 0.9)10 (11)[1 (0.9)10 ] 11 (10 )( 0.9)10 7.513 b) Let p be the probability that an individual has the disease. Then the expected number of tests required for each group of 10 people under the group testing scheme is E (T ) 11 (10 )(1 p)10 compared to 10 tests required for each group of 10 people under a “test each individual” scheme. Hence, let us find that value of p for which E (T ) 10 : 11 (10)(1 p )10 10 1 (10)(1 p )10 0.1 (1 p )10 0.794 1 p p 0.206 Noting from (a) that p 0.1 yielded an expected number of tests less than 10, we can conclude that the group testing scheme is better when p 0.206 and consequently, it is better to test each individual separately when p 0.206 . 8. Ross 4.35 p. 192 a) Since we are drawing without replacement, we can see that (letting R denote drawing a red marble and B denote drawing a blue marble) P( RR) P( BB) 2 / 9 and P( RB) P( BR) 5 / 18 . Hence, if we let X be the random variable representing your winnings playing this marble game, then P( X 1.1) 4 / 9 1 P( X 1) . This gives us an expected value for X of E ( X ) (1.10)(4 / 9) (1.00)(5 / 9) 0.6 / 9 0.067 . In other words, you would expect to lose about 7 cents every time you play this game. b) To compute the variance, we can use the fact that Var ( X ) E ( X 2 ) E ( X ) 2 , which only requires to determine what E ( X 2 ) is, since we have E (X ) from part (a): E ( X 2 ) (1.1) 2 (4 / 9) (1.0) 2 (5 / 9) (1.21)( 4 / 9) (1.0)(5 / 9) 1.093 . Hence, the variance of the amount you win is just Var ( X ) 1.093 (0.067 ) 2 1.089 .

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