If X is a Poisson variable such that P(X = 1) = P(X = 2), then P(X =…

The correct answer is Option (1) → $\frac{4}{3 e^2}$

$P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!}$

$P(X=1)=\frac{e^{-\lambda}\lambda}{1!},\;\; P(X=2)=\frac{e^{-\lambda}\lambda^2}{2!}$

$\frac{e^{-\lambda}\lambda}{1} = \frac{e^{-\lambda}\lambda^2}{2}$

$\lambda = \frac{\lambda^2}{2}$

$2\lambda = \lambda^2$

$\lambda = 2$

$P(X=3)=\frac{e^{-2}2^3}{3!} = \frac{8e^{-2}}{6} = \frac{4}{3}e^{-2}$

$P(X=3)=\frac{4}{3}e^{-2}$