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MTH 202 : Probability and Statistics Homework 2 6th January, 2017 (1) A man has n keys exactly one of which fits the lock. He tries the keys one at a time, at each trial choosing at random from the keys that were not tried earlier. Find the probability that the r-th key tried is the correct key. [Ref : Exercise-3, Hoel, Port, Stone, Page-46] Solution : The total number of keys tried up to the r-th attempt (without replacement) is n(n − 1) . . . (n − r + 2)(n − r + 1) Now if the r-th key tried is the correct one, all the attempts made up to (r − 1)-th attempt has to be a failure. Thus, keeping the correct key aside, the number of attempts (out of the remaining (n − 1) keys) made up to the (r − 1)-th attempt is (n − 1) . . . (n − r + 2)(n − r + 1)(1) with only one choice left for the r-th attempt. Hence the required probability is 1 (n − 1) . . . (n − r + 2)(n − r + 1) = . n(n − 1) . . . (n − r + 2)(n − r + 1) n (2) If Sam and Peter are among n men who are arranged at random in a line, what is the probability that exactly k men stand between them? [Ref : Exercise-7, Hoel, Port, Stone, Page-46] Solution : Keeping Sam and Peter and k men m1 , m2 , . . . , mk out of the remaining n − 2 we treat the block : ∆ := Sam, m1 , m2 , . . . , mk , Peter as a single component and distribute the remaining n − k − 2 men in left and right segment. Keeping the block ∆ fixed and treating it as a object there are (n − k − 1)! ways to permute these. Now notice that : (i) there are n−2 ways of choosing these k men, k (ii) there are 2! ways one can keep Sam and Peter at the ends 1 2 of these blocks (either left or right end), (iii) once k men are chosen, there are k! ways of permuting them within a block. Hence out of the total n! ways to permute all people there are 2 n−2 k!(n − k − 1)! many correct alignments. Hence the rek quired probability is 2 n−2 k!(n − k − 1)! 2(n − k − 1) k = n! n(n − 1) (3) Suppose n balls are distributed in n boxes. (i) What is the probability that exactly one box is empty? (ii) Given that box 1 is empty, what is the probability that only one box is empty? (iii) Given that only one box is empty, what is the probability that box 1 is empty? [Ref : Exercise-10, Hoel, Port, Stone, Page-47] Proof : Note that here both the balls and the boxes are distinguishable. For (i) let Ai denote the event that exactly box number i is empty. Then as computed in Example 9 (where we computed the probability of exactly box number 1 being empty; the computation would be same for any particular box), we have that n (n − 1)! P (Ai ) = 2 (1 ≤ i ≤ n) nn Now the required probability is n n n n [ X (n − 1)! n! 2 2 P ( Ai ) = P (Ai ) = n × = nn nn i=1 i=1 since the events A1 , . . . , An are mutually disjoint. (ii) Let B1 denote the event that the box number 1 is empty. While B1 happen the remaining boxes numbered 2, 3, . . . , n could be arbitrarily filled. Hence we see here that A1 ⊆ B1 while Ai ∩ B1 = ∅ for any i 6= 1. Also while box number 1 is empty, there are n balls each have a possible (n − 1) destinations, which mean there are (n − 1)n possible ways. Hence P (B1 ) = (n − 1)n nn 3 Now the required conditional probability is S T n n [ (n − 1)! P (( ni=1 Ai ) B1 ) P (A1 ) 2 P ( Ai |B1 ) = = = P (B1 ) P (B1 ) (n − 1)n i=1 (iii) Here we compute the conditional probability TS n [ P (B1 ( ni=1 Ai )) P (A1 ) 1 Sn Sn = = P (B1 | Ai ) = P ( i=1 Ai ) P ( i=1 Ai ) n i=1 Sn as we have noticed in part (i) that P ( i=1 Ai ) = nP (A1 ). (4) Compute the probability of each of the following poker hands of five cards : (i) Straight flush (five cards of the same suit in a sequence), (ii) Straight (five cards in a sequence, regardless of the suit), (iii) One pair (face values of the form (w, w, x, y, z) where w, x, y and z are distinct). [Ref : Exercise-14, Hoel, Port, Stone, Page-47] Solution : (i) Straight flush : five cards of the same suit in a sequence (see Fig 1) Figure 1 choices for a poker Recall that there are always a total of 52 5 hand of five cards. Now note that there are 10 possible sequences that can be made out of a fixed suit : [A, 2, 3, 4, 5], [2, 3, 4, 5, 6], . . . , [9, 10, J, Q, K], [10, J, Q, K, A] which is possible because an ace ”A” can be treated as both the least and the highest value in a sequence. Since there are 4 4 suits, the total number of such sequence is 4 × 10 and hence the required probability is 40/ 52 5 (ii) Straight : five cards in a sequence, regardless of the suit (see Fig 2) Figure 2 Keeping the same symbols A, 2, 3, . . . , 10, J, Q, K as before there are again 10 possible sequences out of these symbols. Now each of these symbols can be chosen arbitrarily from the four possible suits. This give a required count 45 × 10. The required probability is thus 45 × 10 52 5 (iii) One pair : face values of the form (w, w, x, y, z) where w, x, y and z are distinct. Recall that a face value mean the symbols A, 2, 3, . . . , 10, J, Q, K. Now we can choose one face value among the 13 for the pair and choose any two of the four suits with the fixed face value. 4 These would give 13 × choices. Note here that the number 2 4 comes because keeping the face value fixed the suits can be 2 permuted within them, e.g. in Fig. 3 (King of hearts, King of spades) and (King of spades, King of hearts) would represent the same hand. Now out of the remaining 12 face values 3 are to be chosen and in each of the coordinates (x, y, z) a arbitrary suit can occur. Since the face values are all different here, these can be chosen in 12 × 43 ways. Hence the required probability 3 5 Figure 3 is 13 × 4 2 × 12 3 × 43 52 5 (5) Cards are dealt from an ordinary deck of playing cards one at a time until the first king appears. Find the probability that this occurs with the n-th card dealt. [Ref : Exercise-16, Hoel, Port, Stone, Page-48] Solution : Since there are 52 cards are they are dealt without replacement, the total number of choices up to the n-th stage are (52)n = 52(52 − 1) . . . (52 − n + 1) (see the notation in Page-29, Hoel, Port, Stone). Now since the event require the King to appear at n-th stage and there are a total of 4 kings in the deck, the first (n − 1) times the cards has to come from the remaining 52 − 4 = 48 choices, and the last choice could be any of the 4 kings. Hence the required probability is (48)n−1 × 4 (52)n (6) Find the probability that a poker hand of 5 cards will contain no card smaller than 7, given that at least 1 card over 10, where aces are treated as high cards. [Ref : Exercise-24, Hoel, Port, Stone, Page-48] Solution : The cards which at least 7 are the eight face values given as : 7, 8, 9, 10, J, Q, K, A. Considering the four suits, there 6 are total 8 × 4 = 32 such cards. Similarly the cards over 10 are J, Q, K, A and there are 4 × 4 = 16 such cards. To estimate the probability we set C to be the event that no card is smaller than 7 and D represent the event that at least one card is over 10. We compute P (C|D) = P (C ∩ D) P (D) To compute P (D) notice that out of the total 52 choices, there 5 36 would be 5 hands representing all five cards come from the face values 2, 3, . . . , 10 (9 cards, 4 suits and hence 4 × 9 = 36 such choices). By removing these choices from the total we compute P (D) as 52 − 36 5 5 P (D) = 52 5 Similarly while computing P (C ∩ D) we notice that out of the total 32 choices representing all cards have face values at 5 least 7, there are 16 choices representing no card are over 10, 5 i.e. these are among 7, 8, 9, 10 (4 cards, 4 suits and hence total 4 × 4 = 16 choices). By removing these from the total we have P (C ∩ D) as 32 16 − P (C ∩ D) = 5 52 5 5 Hence the required probability is 32 − 5 P (C|D) = 52 − 5 16 5 36 5 (7) If you hold three tickets to a lottery for which n tickets were sold and 5 prizes are to be given, what is the probability that you will win at least 1 prize? [Ref : Exercise-25, Hoel, Port, Stone, Page-48] Solution : We can compute the probability of no prizes won. Since there are a total of n tickets sold, the total number of ways to purchase 3 tickets out of them is n3 . Now as there are 5 prizes among these (n ≥ 5), the number of ways of getting no 7 prize is n−5 (assuming the prizes to ticket numbers are already 3 decided). Hence the probability of winning at least one prize is n−5 = 1 − P (winning no prize) = 1 − n3 3 Note that n−5 3 n 3 n−3 5 n 5 = which means in an alternate way we can solve the same : there are a total of n5 ways to assign the prizes to the tickets and ensuring that no prizes to be won, there are n−3 ways to assign 5 the prizes to the tickets which are not purchased. (8) A box of 10 washers contain 5 defective ones. What is the probability that two washers selected at random (without replacement) from the box are both good? [Similar to : Exercise-26, Hoel, Port, Stone, Page-48] Solution : The total number of ways to select two washers (without replacement) is 10 and the number of ways to choose 2 5 two goodones out of 5 are 2 . Hence the required probability is 52 / 10 . 2