Question

If $X$ is a Poisson variate such that $P\left(X=1\right)=P\left(X=2\right)$ then $P\left(X=4\right)$ is equal to

• A. $\frac{1}{2e^2}$
• B. $\frac{1}{3e^2}$
• C. $\frac{2}{3e^2}$
• D. $\frac{1}{e^2}$

Answer 1

The correct option is C

$\frac{2}{3e^2}$

Explanation for the correct option.

Step 1. Find the value of $\mu$.

For the Poisson variate $X$ and parameter $\mu$ $P\left(X=r\right)=\frac{e^{-\mu }{\mu }^r}{r!}$.

Now it is given that $P\left(X=1\right)=P\left(X=2\right)$, so

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

Step 2. Find the value of $P\left(X=4\right)$.

For $\mu =2$, the value of $P\left(X=4\right)$ is given as:

$\begin{array}{rcl}P\left(X=4\right)& =& \frac{e^{-2}{2}^4}{4!}\\ & =& \frac{e^{-2}\times 16}{24}\\ & =& \frac{2}{3e^2}\end{array}$

Hence, the correct option is C.