Q106 / 160 · TS EAMCET 2012 · Mathematics

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

  • A
    $2 \alpha$
  • Correct answer
    B
    $\frac{\alpha}{3}$
  • C
    $\alpha e^{-2}$
  • D
    $\alpha e^2$

Solution

Given $X$ is a poisson variate such that $\begin{aligned} & \alpha=P(X=1)=P(X=2) \\ & \frac{e^{-\lambda} \lambda}{1 !}=\frac{e^{-\lambda} \lambda^2}{2 !} \Rightarrow \lambda=2 \\ & \alpha=P(X=1) \\ & \quad=e^{-2} \times 2=\frac{2}{e^2}\end{aligned}$ $\begin{aligned} & \therefore \quad P(X=4)=\frac{e^{-\lambda} \lambda^4}{4 !}=\frac{e^{-2}(2)^4}{24} \\ & =\frac{e^{-2} \times 16}{24}=\frac{2}{3} e^{-2}=\frac{1}{3} \alpha\end{aligned}$ [from Eq. (i)]

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

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