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If $X$ is a Poisson random variate with mean 3, then $P(|X-3| < 2)=$
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Verified Answer
The correct answer is:
$\frac{99}{8 e^3}$
For poisson distribution, $P(X=k)=\frac{\lambda^k \cdot e^{-\lambda}}{k !}$
$$
\begin{aligned}
& \text { Here, } \lambda=3,|X-3| < 2 \\
& \Rightarrow \quad 1 < X < 5 \\
& \therefore \quad X=2,3,4 \\
& \therefore \quad P(\mid X-3) < 2)=P(X=2)+P(X=3)+P(X=4) \\
& \quad=e^{-3}\left[\frac{9}{2}+\frac{27}{6}+\frac{81}{24}\right]=\frac{99}{8 e^3}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Here, } \lambda=3,|X-3| < 2 \\
& \Rightarrow \quad 1 < X < 5 \\
& \therefore \quad X=2,3,4 \\
& \therefore \quad P(\mid X-3) < 2)=P(X=2)+P(X=3)+P(X=4) \\
& \quad=e^{-3}\left[\frac{9}{2}+\frac{27}{6}+\frac{81}{24}\right]=\frac{99}{8 e^3}
\end{aligned}
$$
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