Q50 / 160 · TS EAMCET 2010 · Mathematics

Suppose that a random variable $X$ follows Poisson distribution. If $P(X=1)=P(X=2)$ then $P(X=5)$ is equal to

  • A
    $\frac{2}{3} e^{-2}$
  • B
    $\frac{3}{4} e^{-2}$
  • Correct answer
    C
    $\frac{4}{15} e^{-2}$
  • D
    $\frac{7}{8} e^{-2}$

Solution

Let $\lambda$ be the mean of the poisson variate $x$. Then, $P(X=r)=\frac{\lambda^r e^{-\lambda}}{r !}, r=0,1,2, \ldots$ Now, $P(X=1)=P(X=2) \Rightarrow \frac{\lambda e^{-\lambda}}{1 !}=\frac{\lambda^2 e^{-\lambda}}{2 !}$ $\Rightarrow \quad \lambda=2$ Hence, $\quad P(X=5)=\frac{\lambda^5 e^{-\lambda}}{5 !}=\frac{2^5 e^{-2}}{120}$ $=\frac{32 \cdot e^{-2}}{120}=\frac{4 e^{-2}}{15}$

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

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