Q80 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

If $\mathrm{X}$ is a Poisson variate satisfying the condition $3 P(x=2)=P(x=4)$ then $P(x-6)=$

  • A
    $\frac{162}{5 e^6}$
  • B
    $\frac{108}{5 \mathrm{e}^6}$
  • Correct answer
    C
    $\frac{324}{5 \mathrm{e}^6}$
  • D
    $\frac{648}{5 e^6}$

Solution

$\begin{aligned} & \text {} 3 P(x=2)=P(x=4) \\ & \because P(x)=\frac{e^{-\lambda} \lambda^x}{x !} \\ & \Rightarrow \frac{3 \cdot e^{-\lambda} \lambda^2}{2 !}=\frac{e^{-\lambda} \lambda^4}{4 !} \\ & \Rightarrow \lambda^2=\frac{4 ! \times 3}{2 !}=36 \\ & \Rightarrow \lambda=6 \\ & \therefore \quad P(x=6)=\frac{e^{-6}(6)^6}{6 !}=\frac{324}{5 e^6} .\end{aligned}$

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

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