The mean and variance of a binomial distribution are 4 and 3 respectively. Then the probability of getting exactly six successive in this distribution, is

${^{16}C}_6 \left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}$

Let X ∼ B(n, p) be  a binomial variate with mean 4 and variance 3. Then,

$np= 4,$ and $npq = 3 ⇒ q=\frac{3}{4}, p = \frac{1}{4}$ and $ n = 16 $

$∴ P(X=r) = {^{16}C}_r \left(\frac{1}{4}\right)^{r}\left(\frac{3}{4}\right)^{16-r}, r = 0, 1, 2, ..., 167$

$⇒ P(X=6)= {^{16}C}_6 \left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}$