Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 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.3/8

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}

Answers (1)

Given-

There are total 8 balls in box.

Therefore, P(G)=\frac{3}{8}  , Probability of green ball

P(B)=\frac{2}{8}  , Probability of blue ball
The probability of drawing 2 green balls and one blue ball is
P(E)=P(G).P(G).P(B)+P(B).P(G).P(G)+P(G).P(B).P(G)

\\ P(E)=\left(\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}\right)+\left(\frac{2}{8} \times \frac{3}{7} \times \frac{2}{6}\right)+\left(\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}\right) \\ P(E)=\frac{1}{28}+\frac{1}{28}+\frac{1}{28} \\ P(E)=\frac{3}{28} \\ $ Hence, $\quad P(E)=\frac{3}{28}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question