If a die is thrown 5 times, then find the probability that an odd number will come up exactly three times.

Here, $n = 5,p = \left( {\frac{1}{6} + \frac{1}{6} + \frac{1}{6}} \right) = \frac{1}{2}$

and $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$

Also, $r = 3$

$\therefore$ $P(X = r){ = ^n}{C_r}{(p)^r}{(q)^{n - r}}{ = ^5}{C_3}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^{5 - 3}}$

$= \frac{{5!}}{{3!2!}} \cdot \frac{1}{8} \cdot \frac{1}{4} = \frac{{10}}{{32}} = \frac{5}{{16}}$