If a random variable X follows Poisson distribution such that P(X=1)= 2P(X=2), then P(X=4) is :

$\frac{1}{24e}$

The correct answer is Option (2) → $\frac{1}{24e}$

A Poisson distribution random variable X has the PMF -

$P(X=x)=\frac{e^{-λ}λ^x}{x!}$

$P(X=1)=2P(X=2)$

$\frac{e^{-λ}λ^1}{1!}=2×\frac{e^{-λ}λ^2}{2}$

$λ=1$

$P(X=4)=\frac{e^{-1}1^4}{4!}=\frac{1}{24e}$