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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Probability

BITSAT 2021 · Mathematics · Probability

The mean and variance of a random variable X having binomial...

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

Options

  1. A14\frac{1}{4}
  2. B132\frac{1}{32}
  3. C116\frac{1}{16}
  4. D18\frac{1}{8}

Correct answer

B. 132\frac{1}{32}

Step-by-step solution

np=4npq=2⇒q=12,p=12,n=8P(X=1)=8C1(12)(12)7=8⋅128=125=132\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}
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