A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

$\frac{3}{28}$

The correct answer is Option (1) → $\frac{3}{28}$ ##

Let $G$ and $B$ represents green and blue balls respectively.

$∴\text{Probability of drawing 2 green balls and one blue ball}$

$= P(G) \cdot P(G) \cdot P(B) + P(B) \cdot P(G) \cdot P(G) + P(G) \cdot P(B) \cdot P(G)$

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

$= \frac{1}{28} + \frac{1}{28} + \frac{1}{28} = \frac{3}{28}$