#### 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

2/3

4/5

7/8

15/16

#### Solution

15/16

Mean =2 and variance =1

\[\Rightarrow np = 2 \text{ and npq } = 1\]

\[ \Rightarrow q = \frac{1}{2} \]

\[ \Rightarrow p = 1 - \frac{1}{2} = \frac{1}{2} \]

\[n = \frac{\text{ Mean} }{p}\]

\[ \Rightarrow n = 4\]

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

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

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

\[ = 1 - \frac{1}{2^4}\]

\[ = \frac{15}{16}\]

Is there an error in this question or solution?

Solution 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 (A) 2/3 (B) 4/5 (C) 7/8 (D) 15/16 Concept: Bernoulli Trials and Binomial Distribution.