Q93 / 225 · JEE Main 2003 · 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

$\left.\begin{array}{c}\mathrm{np}=4 \\ \mathrm{npq}=2\end{array}\right\} \Rightarrow \mathrm{q}=\frac{1}{2} \cdot \mathrm{p}=\frac{1}{2}, \mathrm{n}=8$ $\mathrm{p}(\mathrm{X}=1)={ }^8 \mathrm{C}_1\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^7=8 \cdot \frac{1}{2^8}=\frac{1}{2^5}=\frac{1}{32}$

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

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