If, for a binomial distribution, the number of trials is $9$, the variance is $2$ and the probability of success is greater than that of failure, find the probability of both success.

For a binomial distribution, variance $= npq = 2$ and $n = 9$. So, $9p(1 - p) = 2 \Rightarrow 9p - 9p^2 = 2 \Rightarrow 9p^2 - 9p + 2 = 0$. $\Rightarrow p = \dfrac{9 \pm \sqrt{81 - 72}}{18} = \dfrac{9 \pm 3}{18}$. Hence, $p = \dfrac{2}{3}$ or $\dfrac{1}{3}$. Since probability of success is greater than failure, $p = \dfrac{2}{3}$.