Q76 / 120 · VITEEE 2011 · Mathematics
If $X$ is a poisson variate such that $P(X=1)$ $=P(X=2)$, then $P(X=4)$ is equal to
- A$\frac{1}{2 \mathrm{e}^{2}}$
- B$\frac{1}{3 \mathrm{e}^{2}}$
- Correct answerC$\frac{2}{3 \mathrm{e}^{2}}$
- D$\frac{1}{\mathrm{e}^{2}}$
Solution
Given : $\mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{X}=2)$
$$
\begin{array}{l}
\frac{\mathrm{e}^{-\lambda} \lambda}{1 !}=\frac{\mathrm{e}^{-\lambda} \lambda^{2}}{2 !} \\
\Rightarrow \lambda=2 \\
\therefore \mathrm{P}(\mathrm{X}=4)=\frac{\mathrm{e}^{-2} 2^{4}}{4 !}=\frac{2}{3 \mathrm{e}^{2}}
\end{array}
$$
Hence, the correct answer is option: 'C'.