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If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$, then $P(X=4)$ is equal to
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2186 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{3 e^2}$
Given that, $P(X=1)=P(X=2)$
$$
\begin{array}{rlrl}
& \therefore & \frac{e^{-\lambda} \lambda^1}{1 !} & =\frac{e^{-\lambda} \lambda^2}{2 !} \\
& \Rightarrow & \lambda & =2 \\
& \therefore & P(X=4) & =\frac{e^{-2}(2)^4}{4 !} \\
& =\frac{e^{-2} \times 16}{24}=\frac{2}{3 e^2}
\end{array}
$$
$$
\begin{array}{rlrl}
& \therefore & \frac{e^{-\lambda} \lambda^1}{1 !} & =\frac{e^{-\lambda} \lambda^2}{2 !} \\
& \Rightarrow & \lambda & =2 \\
& \therefore & P(X=4) & =\frac{e^{-2}(2)^4}{4 !} \\
& =\frac{e^{-2} \times 16}{24}=\frac{2}{3 e^2}
\end{array}
$$
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