If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$, then $P(X=4)$ is equal to

If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$, then $P(X=4)$ is equal to
  1. $\frac{1}{2 e^2}$
  2. $\frac{1}{3 e^2}$
  3. $\frac{2}{3 e^2}$
  4. $\frac{1}{e^2}$

Solution

Given that, $P(X=1)=P(X=2)$ $ \begin{array}{rlrl} & \therefore & \frac{e^{-\lambda} \lambda^1}{1 !} & =\frac{e^{-\lambda} \lambda^2}{2 !} \\ & \Rightarrow & \lambda & =2 \\ & \therefore & P(X=4) & =\frac{e^{-2}(2)^4}{4 !} \\ & =\frac{e^{-2} \times 16}{24}=\frac{2}{3 e^2} \end{array} $

Asked in: AP EAMCET 2008

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