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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Aicharya · Accepted Answer

Given $X$ is a poisson variate such that $\begin{aligned} & \alpha=P(X=1)=P(X=2) \ & \frac{e^{-\lambda} \lambda}{1 !}=\frac{e^{-\lambda} \lambda^2}{2 !} \Rightarrow \lambda=2 \ & \alpha=P(X=1) \ & \quad=e^{-2} 	imes 2=\frac{2}{e^2}\end{aligned}$ $\begin{aligned} & 	herefore \quad P(X=4)=\frac{e^{-\lambda} \lambda^4}{4 !}=\frac{e^{-2}(2)^4}{24} \ & =\frac{e^{-2} 	imes 16}{24}=\frac{2}{3} e^{-2}=\frac{1}{3} \alpha\end{aligned}$ [from Eq. (i)]

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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