Question

The mean and variance of a binomial distribution are \[\frac{4}{3}\] and \[\frac{8}{9}\] respectively. Find P (X ≥ 1).

 

 

Answer 1

\[\text{ Mean }  \left( np \right) = \frac{4}{3}\text{ and Variance} \left( npq \right) = \frac{8}{9}\]

\[ \therefore q = \frac{2}{3}\] 

\[\text{ and } p = 1 - \frac{2}{3} = \frac{1}{3}\]

\[\text{ Therefore, } n = \frac{\text{ Mean} }{p} = 4\]

\[\text{ Hence, the distribution is given by } \]

\[P(X = r) = ^{4}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{4 - r} , r = 0, 1, 2, 3, 4\]

\[P(X \geq 1) = 1 - [P(X = 0)] \]

\[ = 1 - \left[ \left( \frac{2}{3} \right)^4 \right]\]

\[ = \frac{81 - 16}{81}\]

\[ = \frac{65}{81}\]