If X has a Poisson distribution such that 3 P(X = 2) = P(X = 4), then P(X = 1) is:

$6 e^{-6}$

The correct answer is Option (1) → $6 e^{-6}$

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

$3P(X=2)=P(X=4)$

$3 \cdot \frac{e^{-\lambda}\lambda^2}{2!} = \frac{e^{-\lambda}\lambda^4}{4!}$

$3 \cdot \frac{\lambda^2}{2} = \frac{\lambda^4}{24}$

$36\lambda^2 = \lambda^4$

$\lambda^2(\lambda^2 - 36)=0 \Rightarrow \lambda=6$

$P(X=1)=\frac{e^{-6}6}{1!}=6e^{-6}$

$P(X=1)=6e^{-6}$