Bag contains 6 white and 4 blue balls, Bag contains 4 white and 6 blue balls, and Bag contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is dra

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Solution

Q.
Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from bag $B_{2}$ , is : [2025]

1 $\frac{2}{5}$

2 $\frac{4}{15}$

3 $\frac{2}{3}$

4 $\frac{1}{3}$

Ans.
(2)

Consider the events, $E_{1}$ : Bag $B_{1}$ is selected, $E_{2}$ : Bag $B_{2}$ is selected, $E_{3}$ : Bag $B_{3}$ is selected

A : Ball drawn is white. Then $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$

$P (A / E_{1}) = \frac{6}{10}, P (A / E_{2}) = \frac{4}{10}, P (A / E_{3}) = \frac{5}{10}$

Required probability

$P (\frac{E_{2}}{A}) = \frac{P (E_{2}) P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})}$

$= \frac{\frac{1}{3} \times \frac{4}{10}}{(\frac{1}{3}) (\frac{6}{10}) + (\frac{1}{3}) (\frac{4}{10}) + (\frac{1}{3}) (\frac{5}{10})} = \frac{\frac{4}{30}}{\frac{6}{30} + \frac{4}{30} + \frac{5}{30}} = \frac{4}{15}$

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