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The variance of Poisson distribution is 1. Then the probability that Poisson random variable X, takes the value equal to 2 is : |
$\frac{1}{2e^2}$ $\frac{1}{2^2e}$ $\frac{e}{2}$ $\frac{1}{2e}$ |
$\frac{1}{2e}$ |
The correct answer is Option (4) → $\frac{1}{2e}$ The probability mass function (PMF) is, $P(X=k)=\frac{λ^ke^{-λ}}{k!}$ Substitute $λ=1,k=2$ $P(X=2)=\frac{1^2e^{-1}}{2!}=\frac{1}{2e}$ |
