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If X is a random poission variate such that 2P(X=0)+P(X=2)=2P(X=1) then E(X)=

A
4
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B
3
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C
2
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D
1
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Solution

The correct option is D 2
Poisson's distribution implies
P(X=n)=λneλn!
Where mean=variance=λ
Hence 2.P(x=0)+P(x=2)=2P(x=1)
2λ0eλ0!+λ2eλ2!=2λ1eλ1!

2eλ+λ2eλ2!=2λeλ

2λeλ2eλ=λ2eλ2!
4λ4=λ2
λ24λ+4=0
(λ2)2=0
λ=2
Hence mean=variance=2

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