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

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) (3/28) (B) (2/21) (C) (1/28) (D) (167/168)

Tardigrade Admin · Accepted Answer

Required probability = P GGB, GBG, BGG = P(GGB) + P(GBG) + P(BGG) = (3/8)× (2/7)× (2/6)+(3/8)× (2/7)× (2/6)+(3/7)× (2/6) = 3× (1/28)=(3/28)

Q. 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

$\frac{3}{28}$

$\frac{2}{21}$

$\frac{1}{28}$

$\frac{167}{168}$

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

Q. 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

283​

212​

281​

168167​