Given three identical boxes I,II and III,each | vedclass

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(A) Let $E_1, E_2$,and $E_3$ be the events that boxes $I, II$,and $III$ are chosen,respectively. Since the boxes are chosen at random,$P(E_1) = P(E_2)

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$2/3$

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(A) Let $E_1, E_2$,and $E_3$ be the events that boxes $I, II$,and $III$ are chosen,respectively. Since the boxes are chosen at random,$P(E_1) = P(E_2) = P(E_3) = 1/3$. Let $A$ be the event that the coin drawn is gold. Then,the conditional probabilities are: $P(A|E_1) = 2/2 = 1$ (both coins are gold in box $I$). $P(A|E_2) = 0/2 = 0$ (no gold coins in box $II$). $P(A|E_3) = 1/2$ (one gold coin in box $III$). We want to find the probability that the other coin is also gold,which means we must have chosen box $I$. By Bayes' theorem,$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$. Substituting the values: $P(E_1|A) = \frac{(1/3 	imes 1)}{(1/3 	imes 1) + (1/3 	imes 0) + (1/3 	imes 1/2)} = \frac{1/3}{1/3 + 1/6} = \frac{1/3}{3/6} = \frac{1/3}{1/2} = 2/3$.

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