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

(a) $\frac{3}{28}$
(b) $\frac{2}{21}$
(c) $\frac{1}{28}$
(d) $\frac{167}{168}$

Answer 1

(a) Probability of drawing 2 green balls and one blue ball
$=P_G.P_G.P_B+P_B.P_G.P_G+P_G.P_G.P_G=\frac{3}{8}.\frac{2}{7}.\frac{2}{6}+\frac{2}{8}.\frac{3}{7}.\frac{2}{6}+\frac{3}{8}.\frac{2}{7}.\frac{2}{6}=\frac{1}{28}+\frac{1}{28}+\frac{1}{28}=\frac{3}{28}$