If $X$ is a Poisson random variate with mean 3, then $P(|X-3| < 2)=$
If $X$ is a Poisson random variate with mean 3, then $P(|X-3| < 2)=$
- $\frac{9}{2 e^3}$
- $\frac{99}{8 e^3}$
- $\frac{3}{2 e^3}$
- $\frac{1}{3 e^3}$
Solution
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}
$
Asked in: AP EAMCET 2022 (07 Jul Shift 2)
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