If eight coins are tossed together, then the probability of getting exactly 3 heads is

We know that, probability distribution $P(X = r){ = ^n}{C_r}{(p)^r}{q^{n - r}}$ Here $n = 8,r = 3,p = \frac{1}{2}$ and $q = \frac{1}{2}$

$\therefore$ Required probability${ = ^8}{C_3}{\left( {\frac{1}{2}} \right)^3}{\left( {\frac{1}{2}} \right)^{8 - 3}} = \frac{{8!}}{{5!3!}}{\left( {\frac{1}{2}} \right)^8}$

$= \frac{{8 \cdot 7 \cdot 6}}{{3 \cdot 2}} \cdot \frac{1}{{16 \cdot 16}} = \frac{7}{{32}}$