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

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$\frac{2}{21}$

$\frac{167}{168}$

$\frac{3}{28}$

$\frac{1}{28}$

answer is A.

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Detailed Solution

Since, probability of drawing two green balls and one blue ball

$= \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6} + \frac{2}{8} \times \frac{3}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6} = \frac{1}{28} + \frac{1}{28} + \frac{1}{28} = \frac{3}{28}$ .

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