If X is a Poisson variate and $\mathrm{P}\left(\mathrm{X}=1\right)=2\mathrm{P}(\mathrm{X}=2) \mathrm{then} \mathrm{P}(\mathrm{X}=3)=$
 

Given X follows poisson distribution 

Given that $\mathrm{P} \left(X=1\right)=2 \mathrm{P}(X=2)$

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

Now $\mathrm{P}(X=3)=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^3}{3!}$

                         $$\begin{gathered} =\frac{{\mathrm{e}}^{-1}\times {\left(1\right)}^3}{6} \\ =\frac{{\mathrm{e}}^{-1}}{6} \end{gathered}$$