Q144 / 225 · JEE Main 2004 · Mathematics

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

A
$\frac{37}{256}$
B
$\frac{219}{256}$
C
$\frac{128}{256}$
Correct answer
D
$\frac{28}{256}$

Solution

Given that $n \mathrm{p}=4, \mathrm{n} p \mathrm{q}=2 \Rightarrow \mathrm{q}=1 / 2 \Rightarrow \mathrm{p}=1 / 2, \mathrm{n}=8 \Rightarrow \mathrm{p}(\mathrm{x}=2)={ }^8 \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{28}{256}$

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

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