If $X$ is a poisson variate such that $P(X=2)=P(X=3)$, then $e^3, P(X=4)$ is

If $X$ is a poisson variate such that $P(X=2)=P(X=3)$, then $e^3, P(X=4)$ is
  1. $\left(\frac{3}{2}\right)^3$
  2. $\frac{3}{2}$
  3. $\frac{e^{-3} \cdot 3^4}{4 !}$
  4. $\frac{e^3 \cdot 3^4}{4 !}$

Solution

$X$ is poisson variate $ \begin{aligned} & \text { and } P(X=2)=P(X=3) \\ & \Rightarrow \frac{e^{-\lambda} \cdot \lambda^2}{2 !}=\frac{e^{-\lambda} \lambda^3}{3 !} \quad\left[\because P(X=r)=\frac{e^{-\lambda} \cdot \lambda^r}{r !}\right] \\ & \Rightarrow \quad \frac{\lambda^2}{2}=\frac{\lambda^3}{6} \Rightarrow \frac{6}{2}=\lambda \\ & \Rightarrow \quad \lambda=3 \\ & \therefore \quad P(X=4)=\frac{e^{-\lambda} \cdot \lambda^4}{4 !}=\frac{e^{-3} \cdot(3)^4}{4 !} \\ & =\frac{e^{-3} \cdot 3 \times 3 \times 3 \times 3}{4 \times 3 \times 2 \times 1}=e^{-3}\left(\frac{27}{8}\right)=e^{-3} \cdot\left(\frac{3}{2}\right)^3 \\ & \therefore e^3 P(X=4)=e^3 \times e^{-3}\left(\frac{3}{2}\right)^3 \\ & =\left(\frac{3}{2}\right)^3 \\ & \end{aligned} $

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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