A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, them the probability of getting exactly one red ball is

Probability of getting exactly one red $(R)$ ball $= {P_R} \cdot {P_{\bar R}} \cdot {P_{\bar R}} + {P_{\bar R}} \cdot {P_R} \cdot {P_{\bar A}} + {P_{\bar R}} \cdot {P_{\bar R}} \cdot {P_R}$

$= \frac{5}{8} \cdot \frac{3}{7}\frac{2}{6} + \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{6} + \frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}$

$= \frac{{15}}{{4 \cdot 7 \cdot 6}} + \frac{{15}}{{4 \cdot 7 \cdot 6}} + \frac{{15}}{{4 \cdot 7 \cdot 6}}$

$= \frac{5}{{56}} + \frac{5}{{56}} + \frac{5}{{56}} = \frac{{15}}{{56}}$