#### Question

If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is

##### Options

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

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

\[^{12}{}{C}_6 \left( \frac{1}{20} \right) \left( \frac{3}{4} \right)^6\]

\[^{12}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^6\]

#### Solution

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

Mean (*np*) = 4 and Variance (*npq*) = 3

\[\therefore q = \frac{3}{4}\]

\[ \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}\text{ and } n = 16\]

\[\text{ Let X denotes the number of successes in 16 trials . Then, } \]

\[P(X = r) = ^{16}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{16 - r} \]

\[ \Rightarrow P(X = 6) = \text{ Probability (getting exactly 6 successes } )\]

\[ = 16 C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10} \]