If X is a Poisson variable representing number of successes in 50 trials such that 

$2\mathrm{P}\left(\mathrm{X}=1\right)=5\mathrm{P}\left(\mathrm{X}=5\right)+2\mathrm{P}\left(\mathrm{X}=3\right)$ then the probability of getting success in one trail is

Given X follows Poisson distribution 

Now $\mathrm{\lambda }=np$

         $\mathrm{\lambda }=50p$

Given that $2\mathrm{P}(\mathrm{x}=1)=5\mathrm{P}(\mathrm{x}=5)+2\mathrm{P}(\mathrm{x}=3)$

$$\begin{gathered} \Rightarrow \frac{2.{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^1}{1!}=\frac{5.{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^5}{5!}+\frac{2.{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^3}{3!} \\ \Rightarrow {\mathrm{e}}^{-\mathrm{\lambda }}\left(2\mathrm{\lambda }\right)={\mathrm{e}}^{-\mathrm{\lambda }} \left[\frac{5.{\mathrm{\lambda }}^5}{5\times 4!}+\frac{2.{\mathrm{\lambda }}^3}{3\times 2\times 1}\right] \\ \Rightarrow 2\mathrm{\lambda }=\frac{{\mathrm{\lambda }}^5}{24}+\frac{{\mathrm{\lambda }}^3}{3} \\ \Rightarrow 2\mathrm{\lambda }=\mathrm{\lambda }\left(\frac{{\mathrm{\lambda }}^4+8{\mathrm{\lambda }}^2}{24}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow 2\times 24={\mathrm{\lambda }}^4+8{\mathrm{\lambda }}^2 \\ \Rightarrow {\mathrm{\lambda }}^4+8{\mathrm{\lambda }}^2-48=0 \\ \Rightarrow {\mathrm{\lambda }}^4+12{\mathrm{\lambda }}^2-4{\mathrm{\lambda }}^2-48=0 \\ \Rightarrow {\mathrm{\lambda }}^2({\mathrm{\lambda }}^2+12)-4({\mathrm{\lambda }}^2+12)=0 \\ \Rightarrow ({\mathrm{\lambda }}^2-4) ({\mathrm{\lambda }}^2+12)=0 \\ \Rightarrow {\mathrm{\lambda }}^2=4 (\therefore {\mathrm{\lambda }}^2+12\ne 0) \\ \Rightarrow \mathrm{\lambda }=\sqrt{4}=2 \end{gathered}$$

Since $\mathrm{\lambda }=50p \Rightarrow p=\frac{2}{50}=0.04$

$\therefore$ Probability of getting Success in one trail =0.04