Q80 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

If ' 3 ' is the variance of Poisson distribution, then $\mathrm{P}(1 < \mathrm{x} < 4)=$

  • A
    $\frac{123}{8} e^{-3}$
  • B
    $3 e^{-\sqrt{3}}$
  • Correct answer
    C
    $9 \mathrm{e}^{-3}$
  • D
    $\left(\frac{3+\sqrt{3}}{2}\right) e^{-3}$

Solution

$\mathrm{P}(1 < x < 4)=\mathrm{P}(x=2)+\mathrm{P}(x=3)$ $\begin{aligned} & \mathrm{P}(x=n)=\frac{\lambda^n \cdot e^{-\lambda}}{n !} \\ & \mathrm{P}(x=2)+\mathrm{P}(x=3)=\frac{(3)^2 \cdot e^{-3}}{2 !}+\frac{(3)^3 \cdot e^{-3}}{3 !}\end{aligned}$ $=\frac{3^2}{2}\left(e^{-3}+e^{-3}\right)=9 \cdot \mathrm{e}^{-3}$

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

Relevant Questions