There are three boxes containing white balls and black balls.
Box-1 contains 2 black and 1 white balls.
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 \(\frac{1}{6}\), and Box-3 is \(\frac{1}{3}\). 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 ________(Round off to two decimal places)
Answer
0.25
Explanation
Let \(W\) be the event that the ball is white. The chance of white from each box is \(\frac{1}{3}\), \(\frac{2}{3}\) and \(\frac{3}{6} = \frac{1}{2}\). By the law of total probability,
\[P(W) = \frac{1}{2}\cdot\frac{1}{3} + \frac{1}{6}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{2} = \frac{3}{18} + \frac{2}{18} + \frac{3}{18} = \frac{4}{9}.\]
By Bayes' theorem,
\[P(\text{Box-2} \mid W) = \frac{\frac{1}{6}\cdot\frac{2}{3}}{\frac{4}{9}} = \frac{1/9}{4/9} = \frac{1}{4} = 0.25.\]
Box-2 has the highest share of white balls, but it is the least likely box to be chosen, which keeps its posterior low.
Answer: 0.25