Question

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

Answer 1

Obtain the required probability applying the concept of total probability.
It is given that, the box contains $3$ orange balls, $3$ green balls and $2$ blue balls, i.e., total $8$ balls.
Let $G=\text{The event that the ball drawn is a green ball}$
and $B=\text{The event that the ball drawn is a blue ball}$.

The required probability is
$P\left(2\text{green balls and one blue ball}\right)$
$=P(G,G,B)+P(G,B,G)+P(B,G,G)$
$=\frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}+\frac{2}{8}\times \frac{3}{7}\times \frac{2}{6}$
$=\frac{1}{28}+\frac{1}{28}+\frac{1}{28}$
$=\frac{3}{28}$

Hence, the correct option is a.