If X is a poisson variate such that P(X = 1) = P(X = 2), then P(X = 4) = 
 

Given X follows poisson distribution.

Given that $\mathrm{P}(\mathrm{x}=1)=\mathrm{P}(\mathrm{x}=2)$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^1}{1!}=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^2}{2!} \\ \Rightarrow \mathrm{\lambda }=\frac{{\mathrm{\lambda }}^2}{2} \\ \Rightarrow \mathrm{\lambda }=2 \end{gathered}$$

Now  $\mathrm{P}(X=4)=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^4}{4!}$

                          $$\begin{gathered} =\frac{{\mathrm{e}}^{-2}.2^4}{24} \\ =\frac{2}{3{\mathrm{e}}^2} \end{gathered}$$