If a random variable X follows Poisson distribution such that P(X=1)= 2P(X=2), then P(X=4) is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{24e}$
A Poisson distribution random variable X has the PMF -
$P(X=x)=\frac{e^{-λ}λ^x}{x!}$
$P(X=1)=2P(X=2)$
$\frac{e^{-λ}λ^1}{1!}=2×\frac{e^{-λ}λ^2}{2}$
$λ=1$
$P(X=4)=\frac{e^{-1}1^4}{4!}=\frac{1}{24e}$