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

Answer: (A)

Required probability $= P\{GGB,\, GBG,\, BGG\}$
$= P(GGB) + P(GBG) + P(BGG)$
$= \frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{2}{7}\times \frac{2}{6}+\frac{3}{7}\times \frac{2}{6}$
$= 3\times \frac{1}{28}=\frac{3}{28}$