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MATH 5010–001 SUMMER 2003 SOLUTIONS TO ASSIGNMENT 5 Problems from pp. 171–180 28. Let X denote the number of defectives in the sample. The possible values of X are 0, 1, 2, 3. The probabilities are: 16 P {X = 0} = P {X = 1} = P {X = 2} = P {X = 3} = Therefore, E(X) = 1 · 16 4 2 1 20 3 3 20 3 16 4 2 1 20 3 16 4 1 2 20 3 4 3 20 . 3 +2· 16 4 1 2 20 3 +3· 4 3 20 . 3 30. Let Hj denote the event that the jth toss results in heads, and Tj the event that the jth toss results in tails. Then, n 1 n P {X = 2 } = P {H1 ∩ · · · ∩ Hn−1 ∩ Tn } = = 2−n . 2 Therefore, E(X) = ∞ X 2n P {X = 2n } = n=1 n X 2n 2−n = +∞, n=1 as asserted. 46. Let G denote the event that the randomly-selected defendant is guilty; we know that P (G) = 0.65. Conditionally on G, the number X of jurors who vote “innocent” is binomial with parameters n = 12 and p1 = 0.2. Conditionally on G c , X is binomial with parameters n = 12 and p2 = 0.9. We are after P {X ≥ 9}. P {X ≥ 9} = P {X ≥ 9 | G}P (G) + P {X ≥ 9 | G c }P (G c ) = P {X ≥ 9 | G}0.65 + P {X ≥ 9 | G c }0.35 12 12 X X 12 12 j 12−j = 0.65 0.2 0.8 0.9j 0.112−j . + 0.35 j j j=9 j=9 1 Theoretical Problem from p. 183 32. There are 2n − 1 nonempty subsets. On the other hand, for any j = 1, . . . , n, the number of subsets that have exactly j elements is nj . Therefore, P {X = j} = Thus, n j 2n − 1 , n j n X j = 1, . . . , n. n 1 X n! E(X) = = n j· n j· 2 −1 2 − 1 j=1 j!(n − j)! j=1 n 1 X n! h i = n 2 − 1 j=1 (j − 1)! (n − 1) − (j − 1) ! n n−1 n X n−1 n X n−1 = n = n k 2 − 1 j=1 j − 1 2 −1 k=0 = n2n−1 , 2n − 1 Pn−1 = (1 + 1)n−1 = 2n−1 . Divide because the binomial theorem asserts that k=0 n−1 k the numerator and the denominator by 2n−1 to obtain the first statement. Similarly, n n n X 1 X n! j E(X 2 ) = = n j2 · n j· 2 −1 2 −1 (j − 1)!(n − j)! j=1 = 2n 1 −1 1 = n 2 −1 j=1 n X (j − 1 + 1) · j=1 n X n! (j − 1)!(n − j)! n 1 X n! n! + n (j − 1) · (j − 1)!(n − j)! 2 − 1 j=1 (j − 1)!(n − j)! j=1 n = n 1 X (n − 1)! n X n! h i + (j − 1) · 2n − 1 j=2 (j − 1)!(n − j)! 2n − 1 j=1 (j − 1)! (n − 1) − (j − 1) ! n n n(n − 1) X (n − 2)! n X n−1 h i + n = n j −1 2 −1 (j − 2)! (n − 2)(j − 2) ! 2 − 1 j=2 j=1 n n−1 n(n − 1) X n − 2 n X n−1 = n + n k 2 − 1 j=2 j − 2 2 −1 = n(n − 1) 2n − 1 n−2 X `=0 k=0 n−2 n2n−1 n(n − 1)2n−2 + n2n−1 + n = . ` 2 −1 2n − 1 2 Now, n(n − 1)2n−2 + n2n−1 = n2 2n−2 + n(2n−1 − 2n−2 ) = n2 2n−2 + n2n−2 (2 − 1) = (n2 + n)2n−2 . Because n2 + n = n(n + 1), this yields, n(n + 1)2n−2 Var(X) = − 2n − 1 n2n−1 2n − 1 2 = n(n + 1)2n−2 (2n − 1) − n2 22n−2 . (2n − 1)2 The numerator can be simplified a bit by first observing that n(n + 1)2n−2 (2n − 1) = n(n + 1)22n−2 −n(n+1)2n−2 . Therefore, n(n+1)2n−2 (2n −1)−n2 22n−2 = n22n−2 −n(n+1)2n−2 , and so n22n−2 − n(n + 1)2n−2 Var(X) = , (2n − 1)2 as claimed. To finish, note that as n → ∞, (2n − 1)2 ∼ 22n . Therefore, n(n + 1)2n−2 = 0. n→∞ (2n − 1)2 lim Consequently, Var(X) ∼ n2n−2 n = n2−2 = . 2n 2 4 3