Q81 / 125 · BITSAT 2021 · Mathematics

The mean and variance of a random variable $\mathrm{X}$ having binomial distribution are 4 and 2 respectively, then $\mathrm{P}(\mathrm{X}=1)$ is

  • A
    $\frac{1}{4}$
  • Correct answer
    B
    $\frac{1}{32}$
  • C
    $\frac{1}{16}$
  • D
    $\frac{1}{8}$

Solution

$\begin{aligned} n p &=4 \\ n p q &=2 \\ \Rightarrow q &=\frac{1}{2}, p=\frac{1}{2}, n=8 \\ \mathrm{P}(\mathrm{X}&=1)={ }^{8} \mathrm{C}_{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{7}=8 \cdot \frac{1}{2^{8}}=\frac{1}{2^{5}}=\frac{1}{32} \end{aligned}$

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

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