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

$\frac{15}{56}$

The correct answer is Option (3) → $\frac{15}{56}$ ##

Let $R$ represents red ball.

Probability of getting exactly one red ball

$= P(R) \cdot P(\bar{R}) \cdot P(\bar{R}) + P(\bar{R}) \cdot P(R) \cdot P(\bar{R}) + P(\bar{R}) \cdot P(\bar{R}) \cdot P(R)$

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

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

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