# hw9 - MATH 471 HW 9 Solutions Pengsheng Ji November 7 2006...

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MATH 471 HW 9 Solutions Pengsheng Ji November 7, 2006 1 (a) Let N be the number of customers arriving in the store during a typical hour. It has a Poisson distribution with parameter λ . Let K be the number of female customers arriving in a given hour. We know that, conditioning on N = n , the distribution of K is Binomial ( n,p ). By the law of total probability, the probability of K = k is given by P ( K = k ) = X n = k P ( K = k,N = n ) = X n = k P ( K = k | N = n ) P ( N = n ) = X n = k n ! k !( n - k )! p k (1 - p ) n - k · e - λ λ n n ! = e - λ p k λ k k ! X n = k [ λ (1 - p )] n - k ( n - k )! . Actually the summation is the Taylor series of e λ (1 - p ) . Therefore, P ( K = k ) = e - λ p k λ k k ! · e λ (1 - p ) = e - λp ( λp ) k k ! , which means K has a Poisson distribution. (b) Let A= ¨ No females arriving ¨ , B= ¨ No males arriving ¨ , q=1-p. Then, P ( A ) = P ( K = 0) = e - λp . Similarly to Part(a), the number of males in a typical hour has a Poisson distribution with parameter

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hw9 - MATH 471 HW 9 Solutions Pengsheng Ji November 7 2006...

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