Find the probability of throwing at most 2 sixes in 6 throws of a single dice.

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

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

$\therefore \quad P(X = r){ = ^n}{C_r}{(q)^{n - r}}{p^r} \therefore$

Required probability $= P$ (getting atmost 2 sixes)

$= P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$= (1){\left( {\cfrac{5}{6}} \right)^6}(1) + 6{\left( {\cfrac{5}{6}} \right)^5}\left( {\cfrac{1}{6}} \right) + 15{\left( {\cfrac{5}{6}} \right)^4}{\left( {\cfrac{1}{6}} \right)^2}$

$= {\left( {\cfrac{5}{6}} \right)^4}\left( {\cfrac{{25}}{{36}} + \cfrac{5}{6} + \cfrac{5}{{12}}} \right) = {\left( {\cfrac{5}{6}} \right)^4}\left( {\cfrac{{25 + 30 + 15}}{{36}}} \right) = \cfrac{{35}}{{18}}{\left( {\cfrac{5}{6}} \right)^4}$