Q81 / 160 · AP EAMCET 2013 · Mathematics
If $X$ is a poisson variate $P(X=1)=2 P(X=2)$, then $P(X=3)$ is equal to
- Correct answerA$\frac{e^{-1}}{6}$
- B$\frac{e^{-2}}{2}$
- C$\frac{e^{-1}}{2}$
- D$\frac{e^{-1}}{3}$
Solution
In poission distribution
$$
\begin{gathered}
P(X=x)=\frac{\lambda^x e^{-\lambda}}{x !} \\
P(X=1)=\frac{\lambda e^{-\lambda}}{1 !}=\lambda e^{-\lambda} \\
P(X=2)=\frac{\lambda^2 e^{-\lambda}}{2 !}=\frac{\lambda^2 e^{-\lambda}}{2} \\
P(X=1)=2 P(x=2) \\
\Rightarrow \quad \lambda e^{-\lambda}=2 \times \frac{\lambda^2 e^{-\lambda}}{2} \\
\Rightarrow \quad \lambda(\lambda-1)=0
\end{gathered}
$$
$$
\begin{aligned}
& \lambda \neq 0 \\
\therefore & \lambda=1
\end{aligned}
$$
Hence, $P(X=3)=\frac{(1)^3 e^{-1}}{3 !}=\frac{e^{-1}}{6} \quad(\because \lambda=1)$
Hence, the correct answer is option: 'A'.