#### Question

A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.

#### Solution

\[\text{ We have, } \]

\[p = \text{ probability of getting an even number on a toss } = \frac{3}{6} = \frac{1}{2} \text{ and } \]

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

\[\text{ Let X denote a success of getting an even number on a toss . Then } , \]

\[\text{ X follows binomial distribution with parameters n = 2 and } p = \frac{1}{2}\]

\[ \therefore Var\left( X \right) = \text{ npq } = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}\]

Is there an error in this question or solution?

Solution A Die is Tossed Twice. a 'Success' is Getting an Even Number on a Toss. Find the Variance of Number of Successes. Concept: Random Variables and Its Probability Distributions.