The mean and the variance of a binomial distribution are 4 and 2 respectively.

Question

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

Accepted Answer

Given that mean = 4 = np = 4 and variance = 2

npq = 2 $$\Rightarrow$$ 4q = 2

$$ \Rightarrow q = {1 \over 2}$$

$$\therefore$$ $$p = 1 - q = 1 - {1 \over 2} = {1 \over 2}$$

Also, n = 8

Probability of 2 successes $= P(X = 2) = {}^8{C_2}{p^2}{q^6}$

$$ = {{8!} \over {2! \times 6!}} \times {\left( {{1 \over 2}} \right)^2} \times {\left( {{1 \over 2}} \right)^6} = 28 \times {1 \over {{2^8}}}$$

$$ = {{28} \over {256}}$$

Solutions · 1