>>16790171 (OP)
Looking at this problem, it's a classic Bayesian probability puzzle (similar to the Monty Hall problem but with balls and boxes). The setup in part (1) is clear: three boxes with two balls each—GG, SS, and GS—and the boxes are indistinguishable except by their contents. Part (2) describes drawing a gold ball from a randomly selected box, and we need to find the probability that the remaining ball in that same box is silver. Part (3) reinforces that the selection is blind.
To solve this, we use Bayes' theorem to find the posterior probability that we selected the GS box given that the first ball drawn was gold, then determine the chance the second ball is silver conditional on that.
Step 1: Define the events and priors
- Let B_GG: Selected the GG box. P(B_GG) = 1/3.
- Let B_SS: Selected the SS box. P(B_SS) = 1/3.
- Let B_GS: Selected the GS box. P(B_GS) = 1/3.
- Let G_1: First ball drawn is gold.
The likelihood of drawing gold first from each box:
- P(G_1 | B_GG) = 1 (both balls gold).
- P(G_1 | B_SS) = 0 (no gold balls).
- P(G_1 | B_GS) = 1/2 (one gold out of two).
Step 2: Total probability of G_1
P(G_1) = P(G_1 | B_GG) P(B_GG) + P(G_1 | B_SS) P(B_SS) + P(G_1 | B_GS) P(B_GS) = (1)(1/3) + (0)(1/3) + (1/2)(1/3) = 1/3 + 0 + 1/6 = 1/2.
Step 3: Posterior probabilities P(B | G_1)
Using Bayes' theorem:
P(B_GG | G_1) = [P(G_1 | B_GG) P(B_GG)] / P(G_1) = [(1)(1/3)] / (1/2) = 2/3.
P(B_GS | G_1) = [P(G_1 | B_GS) P(B_GS)] / P(G_1) = [(1/2)(1/3)] / (1/2) = 1/3.
P(B_SS | G_1) = 0 (impossible to draw gold from SS).
Step 4: Probability second ball is silver given G_1
Let S_2: Second ball is silver.
- If B_GG, then P(S_2 | B_GG, G_1) = 0 (remaining ball is gold).
- If B_GS, then P(S_2 | B_GS, G_1) = 1 (drew the gold first, so silver remains).
- If B_SS, impossible.
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