Q81 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

If a random variable $\mathrm{X}$ follows a poisson distribution such that $P(X=1)=3 P(X=2)$, then $P(x=3)=$

  • Correct answer
    A
    $\frac{4}{81} e^{-\frac{2}{3}}$
  • B
    $\frac{2}{81} e^{-\frac{2}{3}}$
  • C
    $\frac{2}{27} e^{-\frac{2}{3}}$
  • D
    $\frac{4}{81} e^{-\frac{1}{3}}$

Solution

In Poisson distribution $$ \begin{aligned} & P(X=r)=\frac{\lambda^p e^{-\lambda}}{r !} \text {, where } \lambda=n p \\ & P(X=1)=3 P(X=4 \\ & \frac{\lambda e^{-\lambda}}{1 !}=\frac{3 \lambda^2 e^{-\lambda}}{2 !} \\ & \Rightarrow \quad \lambda=2 / 3 \\ & \therefore \quad P(X=3)=\frac{\lambda^3 e^{-\lambda}}{3 !}=\frac{\left(\frac{2}{3}\right)^3 e^{-2 / 3}}{3 !}=\frac{4}{81} e^{-2 / 3} \\ & \end{aligned} $$

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

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