Question

If $X$ is a Poisson variable with $P(X=0)=P(X=1)=k$, then show that $k=e^{-1}$.

Answer 1

$P(X=K)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$
Hence $P(X=0)=e^{-\lambda }$, $P(X=1)=\lambda .e^{-\lambda }$
Now $P\left(0\right)=P\left(1\right)$, hence
$e^{-\lambda }=e^{-\lambda }.\lambda$ or $e^{-\lambda }(\lambda -1)=0$
Since $e^{-\lambda }\ne 0$, we have $\lambda =1$.
Hence the $P\left(0\right)=P\left(1\right)=e^{-\lambda }=e^{-1}$