Question

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

Options

• 2/3

• 4/5

• 7/8

• 15/16

   

Answer 1

15/16

Mean =2 and variance =1

\[\Rightarrow np = 2 \text{ and npq }  = 1\]
\[ \Rightarrow q = \frac{1}{2} \]
\[ \Rightarrow p = 1 - \frac{1}{2} = \frac{1}{2} \]
\[n = \frac{\text{ Mean} }{p}\]
\[ \Rightarrow n = 4\]
\[\text{ Hence, the distribution is given by } \]
\[P\left( X = r \right) =^{4}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{4 - r} , r = 0, 1, 2, 3, 4\]
\[ \therefore P(X \geq 1) = 1 - P(X = 0) \]
\[ = 1 - \frac{1}{2^4}\]
\[ = \frac{15}{16}\]