If $X$ is a poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$, then the mean $X$ is

If $X$ is a poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$, then the mean $X$ is
  1. 1
  2. 2
  3. $\frac{3}{2}$
  4. $\frac{1}{2}$

Solution

Given, $ 3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0) $ $X$ is poisson distribution this gives $ P(X=x)=\frac{e^{-\lambda} \lambda^x}{x !}, \quad[x=0,1,2, \ldots] $ Now, $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$ $ \begin{aligned} & \Rightarrow \quad \frac{3 e^{-\lambda} \lambda^4}{4 !}=\frac{1}{2} \frac{e^{-\lambda} \lambda^2}{2 !}+\frac{e^{-\lambda} \lambda^0}{0 !} \\ & \Rightarrow \quad \frac{3 \lambda^4}{4 !}=\frac{\lambda^2}{4}+1 \\ & \Rightarrow \quad \frac{\lambda^4}{8}=\frac{\lambda^2}{4}+1 \text { or } \lambda^4-2 \lambda^2-8=0 \end{aligned} $ Let $\lambda^2=u$, then $u^2-2 u-8=0$ Gives, $u=4,-2$ \begin{array}{ll} \therefore & \lambda^2=4 \\ \Rightarrow & \lambda=2 \end{array} $\left[\because \lambda^2 \neq-2\right]$ \text { Mean of } X=2 $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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