# If X is a Poisson variate such that $P(X=1)=P(x=2)$ the $P(x=4)$ is equal to

$\begin {array} {1 1} (A)\;\large\frac{1}{2e^2} & \quad (B)\;\large\frac{1}{3e^2} \\ (C)\;\large\frac{2}{3e^2} & \quad (D)\;\large\frac{1}{e^2} \end {array}$

Given that $P(x=1)=P(x=2)$
$= \large\frac{e^{-\lambda}\lambda'}{1!} = \large\frac{e^{-\lambda}\lambda^2}{2!}$
$\lambda = 2$
$P(x=4) = \large\frac{e^{-2}(2)^4}{4!}$
$= \large\frac{e^{-2} \times \not{16}2}{\not{24}3}$
$= \large\frac{2}{3e^2}$
Ans : (C)