Q80 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

If $X$ is a Poisson random variate with mean 3, then $P(|X-3| < 2)=$

  • A
    $\frac{9}{2 e^3}$
  • Correct answer
    B
    $\frac{99}{8 e^3}$
  • C
    $\frac{3}{2 e^3}$
  • D
    $\frac{1}{3 e^3}$

Solution

For poisson distribution, $P(X=k)=\frac{\lambda^k \cdot e^{-\lambda}}{k !}$ $$ \begin{aligned} & \text { Here, } \lambda=3,|X-3| < 2 \\ & \Rightarrow \quad 1 < X < 5 \\ & \therefore \quad X=2,3,4 \\ & \therefore \quad P(\mid X-3) < 2)=P(X=2)+P(X=3)+P(X=4) \\ & \quad=e^{-3}\left[\frac{9}{2}+\frac{27}{6}+\frac{81}{24}\right]=\frac{99}{8 e^3} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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