Mean and variance of a binomial distribution are 6 and 2 respectively. The probability of 2 successes will be

$\frac{16}{3^7}$

The correct answer is Option (2) → $\frac{16}{3^7}$

Given:

Mean: $μ = np = 6$

Variance: $σ^2 = np(1-p) = 2$

From mean: $np = 6 \Rightarrow n = \frac{6}{p}$

From variance: $np(1-p) = 2 \Rightarrow 6(1-p) = 2 \Rightarrow 1-p = \frac{1}{3} \Rightarrow p = \frac{2}{3}$

Then $n = \frac{6}{2/3} = 9$

$P(X=2) = \frac{9!}{2!7!} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^7$

$\frac{9\cdot 8}{2} \cdot \frac{4}{9} \cdot \frac{1}{2187} = 36 \cdot \frac{4}{9} \cdot \frac{1}{2187}$

$36 \cdot \frac{4}{9} = 16$, so $16/2187$

Answer: $\frac{16}{2187}$