Step-by-step Solution:

We are given three identical boxes \( B_1, B_2, B_3 \) containing two balls each: - \( B_1 \) contains two golden balls. - \( B_2 \) contains two silver balls. - \( B_3 \) contains one golden and one silver ball. We need to determine the conditional probabilities that a golden ball drawn comes from \( B_1 \), \( B_2 \), or \( B_3 \), respectively. \[Step 1:\] Define Events - Let \( G \) be the event that a golden ball is drawn. - Let \( B_i \) be the event that the ball is drawn from box \( B_i \). - The probability of choosing any box is equal, so: \[ P(B_1) = P(B_2) = P(B_3) = \frac{1}{3} \] - The probability of drawing a golden ball from each box is: \[ P(G | B_1) = 1 \quad \text{(since both balls are golden in \( B_1 \))} \] \[ P(G | B_2) = 0 \quad \text{(since \( B_2 \) has only silver balls)} \] \[ P(G | B_3) = \frac{1}{2} \quad \text{(since \( B_3 \) has one golden and one silver ball)} \] Step 2: Use Bayes’ Theorem The total probability of drawing a golden ball, using the Law of Total Probability, is: \[ P(G) = P(G | B_1) P(B_1) + P(G | B_2) P(B_2) + P(G | B_3) P(B_3) \] Substituting values: \[ P(G) = (1 \times \frac{1}{3}) + (0 \times \frac{1}{3}) + \left(\frac{1}{2} \times \frac{1}{3} \right) \] \[ = \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{2} \] Now, using Bayes’ Theorem, the conditional probability that the golden ball came from box \( B_i \) is: \[ P(B_i | G) = \frac{P(G | B_i) P(B_i)}{P(G)} \] For \( B_1 \): \[ P(B_1 | G) = \frac{(1 \times \frac{1}{3})}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} \] For \( B_2 \): \[ P(B_2 | G) = \frac{(0 \times \frac{1}{3})}{\frac{1}{2}} = 0 \] For \( B_3 \): \[ P(B_3 | G) = \frac{\left(\frac{1}{2} \times \frac{1}{3} \right)}{\frac{1}{2}} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} \] Final Answer: The conditional probabilities are: \[ P(B_1 | G) = 1, \quad P(B_2 | G) = 0, \quad P(B_3 | G) = \frac{1}{2} \] Thus, the correct answer is: \[ \boxed{1, 0, \frac{1}{2}} \]