F.5 Maths M1 distribution－wszspj2的部落格

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F.5 Maths M1 distribution

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The number of customers arriving at any one counter in a supermarket in a minute follows the distribution Po(λ) and is independent of the others. Given that in a particular minute the numbers of customers arriving at two counters are equal and each number is not more than 2, show that the probability... 顯示更多 The number of customers arriving at any one counter in a supermarket in a minute follows the distribution Po(λ) and is independent of the others. Given that in a particular minute the numbers of customers arriving at two counters are equal and each number is not more than 2, show that the probability that there is 1 customer arriving at each counter is (4λ^2) / [(λ^2)+2]^2. In general, the probability of the Poisson distribution should have the term 'e^-λ', but why in this question the probability does not have it? And how to prove it? Please help. Thank you~~

最佳解答:

Let X ~ Po(λ) be the number of customers arriving the first counter. Let Y ~ Po(λ) be the number of customers arriving the second counter. It is given that X and Y are independent. Now, the question asks for Pr(X = 1 and Y =1 | X = Y and X, Y ≤ 2) = Pr(X = 1 and Y =1 and X = Y and X, Y ≤ 2)/Pr(X = Y and X, Y ≤ 2) = Pr(X = 1 and Y =1)/Pr(X = Y and X, Y ≤ 2) = Pr(X = 1)Pr(Y =1)/Pr(X = Y and X, Y ≤ 2) Consider Pr(X = 1) = e^(-λ)λ1/1! = λe^(-λ) Pr(Y = 1) = e^(-λ)λ1/1! = λe^(-λ) The denominator is Pr(X = Y and X, Y ≤ 2) = Pr(X = 0 and Y = 0) + Pr(X = 1 and Y = 1) + Pr(X = 2 and Y = 2) = Pr(X = 0)Pr(Y = 0) + Pr(X = 1)Pr(Y = 1) + Pr(X = 2)Pr(Y = 2) = [Pr(X = 0)]2 + [Pr(X = 1)]2 + [Pr(X = 2)]2 = [e^(-λ)λ?/0!]2 + [e^(-λ)λ1/1!]2 + [e^(-λ)λ2/2!]2 = [e^(-λ)]2 + [λe^(-λ)]2 + [λ2e^(-λ)/2]2 = e^(-2λ) + λ2e^(-2λ) + λ?e^(-2λ)/4 Therefore, the required probability is Pr(X = 1)Pr(Y =1)/Pr(X = Y and X, Y ≤ 2) = [λe^(-λ)]2 / [e^(-2λ) + λ2e^(-2λ) + λ?e^(-2λ)/4] = λ2e^(-2λ) / [e^(-2λ) + λ2e^(-2λ) + λ?e^(-2λ)/4] = λ2 / (1 + λ2 + λ?/4) = 4λ2 / (4 + 4λ2 + λ?) = 4λ2 / (2 + λ2)2