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P{X = n + kX > n} = P{ X = n + k} P{ X > n} p(1 − p) n + k −1 (1 − p ) n = p(1 − p)k−1 = If the first n trials are fall failures, then it is as if we are beginning anew at that time. 28. 29. The events {X > n} and {Y < r} are both equivalent to the event that there are fewer than r successes in the first n trials; hence, they are the same event. P{ X = k + 1} P{ X = k}  Np  N − np     k + 1 n − k − 1  =  Np  N − Np      k  n − k  = Chapter 4 ( Np − k )(n − k ) (k + 1)( N − Np − n + k + 1) 61 30.

I =1 ∞ = n − ∑ ( j + 1) n −1 − λ e λ j +1 / j ! j =0 = λ ∞ ∑ ( j + 1) n −1 − λ e λ j / j! j =0 = λE[( X + 1) n −1 ] Hence [X 3] = λE(X + 1)2] =λ ∞ ∑ (i + 1) e λ λ / i!  i =0 i =0  i =0  2 = λ[ E[ X ] + 2 E[ X ] + 1) = λ(Var(X) = E2[X] + 2E[X] + 1) = λ(λ + λ2 + 2λ + 1) = λ(λ2 + 3λ + 1) ∑ 20. ∑ ∑ Let S denote the number of heads that occur when all n coins are tossed, and note that S has a distribution that is approximately that of a Poisson random variable with mean λ. Then, because X is distributed as the conditional distribution of S given that S > 0, P{X = 1} = P{S = 1S > 0} = λe − λ P{S = 1} ≈ P{S > 0} 1 − e − λ 21.

No—they are conditionally independent given the coin selected. 89. 2)3 97 10 = 10 . 8) 15 10 . 8) 2 33 10 . 8) 2 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 P(N1 ∪ N2) = P(N1) + P(N2) − P(N1N2) = (1 − p1)n + (1 − p2)n − (1 − p1 − p2)n Hence, the probability that both outcomes occur at least once is 1 − (1 − p1)n − (1 − p2)n + (p0)n. Chapter 3 39 Theoretical Exercises 1.