There are three boxes containing white balls and black balls. Box-1 contains 2 black and 1 white ball. Box-2 contains 1 black and 2 white balls. Box-3 contains 3 black and 3 white balls. In a random experiment, one of these boxes is selected, where the probability of choosing Box-1 is $ \frac{1}{2} $, Box-2 is $ \frac{1}{3} $, and Box-3 is $ \frac{1}{6} $. A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is:

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There are three boxes containing white balls and black balls.   Box-1 contains 2 black and 1 white ball.  Box-2 contains 1 black and 2 white balls.  Box-3 contains 3 black and 3 white balls.  In a random experiment, one of these boxes is selected, where the probability of choosing Box-1 is $ \frac{1}{2} $, Box-2 is $ \frac{1}{3} $, and Box-3 is $ \frac{1}{6} $. A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is:

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Let $ B_1 $, $ B_2 $, and $ B_3 $ be the events of selecting Box-1, Box-2, and Box-3, respectively. Let $ W $ be the event of drawing a white ball. Given: $ P(B_1) = \frac{1}{2} $ $ P(B_2) = \frac{1}{6} $ $ P(B_3) = \frac{1}{3} $ Conditional probabilities: $ P(W|B_1) = \frac{1}{3} $ (1 white ball out of 3 total) $ P(W|B_2) = \frac{2}{3} $ (2 white balls out of 3 total) $ P(W|B_3) = \frac{3}{6} = \frac{1}{2} $ (3 white balls out of 6 total) We need to find $ P(B_2|W) $. Using Bayes' Theorem: $$ P(B_2|W) = \frac{P(W|B_2)P(B_2)}{P(W)} $$ First, calculate $ P(W) $ using the Law of Total Probability: $$ P(W) = P(W|B_1)P(B_1) + P(W|B_2)P(B_2) + P(W|B_3)P(B_3) $$ $$ P(W) = \left(\frac{1}{3} \times \frac{1}{2}\right) + \left(\frac{2}{3} \times \frac{1}{6}\right) + \left(\frac{1}{2} \times \frac{1}{3}\right) $$ $$ P(W) = \frac{1}{6} + \frac{2}{18} + \frac{1}{6} = \frac{1}{6} + \frac{1}{9} + \frac{1}{6} = \frac{3}{18} + \frac{2}{18} + \frac{3}{18} = \frac{8}{18} = \frac{4}{9} $$ Now, calculate $ P(B_2|W) $: $$ P(B_2|W) = \frac{\left(\frac{2}{3} \times \frac{1}{6}\right)}{\frac{4}{9}} = \frac{\frac{2}{18}}{\frac{4}{9}} = \frac{\frac{1}{9}}{\frac{4}{9}} = \frac{1}{9} \times \frac{9}{4} = \frac{1}{4} = 0.25 $$ Answer: The probability that the white ball is drawn from Box-2 is 0.25.

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