A pair of dice is thrown 200 times. If getting a sum of 9 is considered a success, then find the mean and the variance of the number of successes.

Mean = $\frac{200}{9}$, Variance = $\frac{1600}{81}$

The correct answer is Option (2) → Mean = $\frac{200}{9}$, Variance = $\frac{1600}{81}$

When a pair of dice is thrown, the sample space has 36 equally likely outcomes. Outcomes favourable to sum of 9 are (5, 4), (4, 5), (6, 3), (3, 6) only.

∴ $p$ = probability (sum is 9)

so $q=1-\frac{1}{9}=\frac{8}{9},n=200$

Hence, mean = $np = 200(\frac{1}{9})=\frac{200}{9}$ and

variance = $npq=200(\frac{1}{9})(\frac{8}{9})=\frac{1600}{81}$