1. **Problem statement:** We have two bags, $B_1$ with 2 white and 1 black ball, and $B_2$ with 1 white and 2 black balls. One bag is chosen at random, then two balls are drawn from it. (i) Find the probability that the two balls drawn from $B_2$ are one white and one black. (ii) Given that the two balls drawn are one white and one black, find the probability they came from $B_2$. 2. **Formulas and rules:** - Probability of an event = $\frac{\text{favorable outcomes}}{\text{total outcomes}}$. - For combinations, number of ways to choose $k$ items from $n$ is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. - Conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$. 3. **Step (i): Probability of one white and one black from $B_2$:** - Total balls in $B_2$: 3 (1 white, 2 black). - Total ways to choose 2 balls: $\binom{3}{2} = 3$. - Favorable ways (one white, one black): choose 1 white from 1 and 1 black from 2: $\binom{1}{1} \times \binom{2}{1} = 1 \times 2 = 2$. - Probability: $$P = \frac{2}{3}$$ 4. **Step (ii): Probability that balls came from $B_2$ given one white and one black drawn:** - Probability of choosing each bag: $\frac{1}{2}$. - Probability of drawing one white and one black from $B_1$: - $B_1$ has 2 white, 1 black. - Total ways: $\binom{3}{2} = 3$. - Favorable ways: choose 1 white from 2 and 1 black from 1: $\binom{2}{1} \times \binom{1}{1} = 2 \times 1 = 2$. - Probability: $$P = \frac{2}{3}$$ - Total probability of drawing one white and one black: $$P(\text{one white, one black}) = P(B_1) \times P(\text{one white, one black}|B_1) + P(B_2) \times P(\text{one white, one black}|B_2) = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} + \frac{2}{6} = \frac{4}{6} = \frac{2}{3}$$ - Using Bayes' theorem: $$P(B_2 | \text{one white, one black}) = \frac{P(B_2) \times P(\text{one white, one black}|B_2)}{P(\text{one white, one black})} = \frac{\frac{1}{2} \times \frac{2}{3}}{\frac{2}{3}} = \frac{1}{2}$$ **Final answers:** (i) Probability = $$\frac{2}{3}$$ (ii) Probability = $$\frac{1}{2}$$