Question

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:

Hint:

In problems involving conditional probability, Bayes' Theorem allows you to calculate the probability of an event given other known probabilities. Make sure to compute the total probability of the outcome across all possible scenarios.

Answer 1

Correct Answer: 0.25

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.