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If $X$ is a poisson variate such that $\alpha=P(X=1)=P(X=2)$, then $P(X=4)$ is equal to
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Verified Answer
The correct answer is:
$\frac{\alpha}{3}$
Given $X$ is a poisson variate such that
$\begin{aligned} & \alpha=P(X=1)=P(X=2) \\ & \frac{e^{-\lambda} \lambda}{1 !}=\frac{e^{-\lambda} \lambda^2}{2 !} \Rightarrow \lambda=2 \\ & \alpha=P(X=1) \\ & \quad=e^{-2} \times 2=\frac{2}{e^2}\end{aligned}$
$\begin{aligned} & \therefore \quad P(X=4)=\frac{e^{-\lambda} \lambda^4}{4 !}=\frac{e^{-2}(2)^4}{24} \\ & =\frac{e^{-2} \times 16}{24}=\frac{2}{3} e^{-2}=\frac{1}{3} \alpha\end{aligned}$
[from Eq. (i)]
$\begin{aligned} & \alpha=P(X=1)=P(X=2) \\ & \frac{e^{-\lambda} \lambda}{1 !}=\frac{e^{-\lambda} \lambda^2}{2 !} \Rightarrow \lambda=2 \\ & \alpha=P(X=1) \\ & \quad=e^{-2} \times 2=\frac{2}{e^2}\end{aligned}$
$\begin{aligned} & \therefore \quad P(X=4)=\frac{e^{-\lambda} \lambda^4}{4 !}=\frac{e^{-2}(2)^4}{24} \\ & =\frac{e^{-2} \times 16}{24}=\frac{2}{3} e^{-2}=\frac{1}{3} \alpha\end{aligned}$
[from Eq. (i)]
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