Question

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 respectively of the number of successes.

• A. $\frac{400}{9}$, $\frac{1600}{81}$
• B. $\frac{1600}{81}$, $\frac{400}{9}$
• C. $\frac{1600}{81}$, $\frac{200}{9}$
• D. $\frac{200}{9}$, $\frac{1600}{81}$

Answer 1

Answer: (D)

$\frac{200}{9}$, $\frac{1600}{81}$

Answer 2

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. So $p =$ probability (sum is $9$) $= \frac{4}{36} = \frac{1}{9}$, $\therefore q = 1-\frac{1}{9}= \frac{8}{9}$, $n = 200$ Hence, mean $= np= 200\left(\frac{1}{9}\right)= \frac{200}{9}$ and variance $= npq = 200\left(\frac{1}{9}\right)\left(\frac{8}{9}\right) = \frac{1600}{81}$