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~~
F.5 Maths M1 distribution
2015-05-10 8:04 pm
回答 (1)
2015-05-10 9:07 pm
✔ 最佳答案
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)λ⁰/0!]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) + λ⁴e^(-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) + λ⁴e^(-2λ)/4]
= λ²e^(-2λ) / [e^(-2λ) + λ²e^(-2λ) + λ⁴e^(-2λ)/4]
= λ² / (1 + λ² + λ⁴/4)
= 4λ² / (4 + 4λ² + λ⁴)
= 4λ² / (2 + λ²)²
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