Find the probability of getting 5 exactly twice in 7 throws of a dice.

.: Let $p$ be the probability of success of getting $5 = \cfrac{1}{6}$

$\Rightarrow q = 1 - p = 1 - \cfrac{1}{6} = \cfrac{5}{6}$

Let $X$ has a binomial distribution with $n = 7,p = \cfrac{1}{6},q = \cfrac{5}{6}$

$\therefore$ $P$ (getting 5 exactly twice ) $= P(X = 2){ = ^7}{C_2}{q^5}{p^2}$

$= \cfrac{{7 \times 6}}{{2 \times 1}} \times {\left( {\cfrac{5}{6}} \right)^5}{\left( {\cfrac{1}{6}} \right)^2} = \cfrac{{21 \times {5^5}}}{{{6^7}}} = \cfrac{7}{{12}}{\left( {\cfrac{5}{6}} \right)^5}$