Question

Mark the correct alternative in the following 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

Options

• \[ \frac{167}{168}\]

• \[ \frac{1}{28}\]

• \[ \frac{2}{21}\]

• \[\frac{3}{28}\]

Answer 1

\[\text{ Let } : \]
\[\text{ O be the event of drawing a orange ball, } \]
\[\text{ G be the event of drawing a green ball and } \]
\[\text{ B be the event of drawing a blue ball} \]
\[\text{ We have, } \]
\[n\left( O \right) = 3, n\left( G \right) = 3 \text{ and }  n\left( B \right) = 2\]
\[ \text{ Also, total balls } = 8\]
\[ \text{ Now } , \]
\[P\left( \text{ drawing 2 green balls and one blue ball } \right) = P\left( GGB \right) + P\left( GBG \right) + P\left( BGG \right)\]
\[ = P\left( G \right) \times P\left( G|G \right) \times P\left( B|GG \right) + P\left( G \right) \times P\left( B|G \right) \times P\left( G|GB \right) + P\left( B \right) \times P\left( G|B \right) \times P\left( G|BG \right)\]
\[ = \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6} + \frac{2}{8} \times \frac{3}{7} \times \frac{2}{6}\]
\[ = \frac{1}{28} + \frac{1}{28} + \frac{1}{28}\]
\[ = \frac{3}{28}\]