Bag B₁ contains 6 white and 4 blue balls, Bag B₂ contains 4 white and 6 blue balls, and Bag B₃ 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 B₂, is :

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$\frac{1}{3}$

$\frac{4}{15}$

$\frac{2}{3}$

$\frac{2}{5}$

answer is B.

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Detailed Solution

E₁ : Bag B₁ is selected
B₁ B₂ B₃
6W 4B 4W 6B 5W 5B
E₂ : bag B₂ is selected
E₃: Bag B₃ is selected
A : Drawn ball is white
We have to find $P (\frac{E_{2}}{A})$

$\begin{array}{l} P (\frac{E_{2}}{A}) = \frac{P (E_{2}) P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})} \\ = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}} \\ = \frac{4}{15} \end{array}$