Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold? 1/3 2/3 3/4 4/5 5/6

Is the answer B? ----- Let A, B, and C be the three boxes. G represents gold, and S represents silver. So, A: G,G B: S,S C: G,S The probability of selecting any of the threee boxes is 1/3. P(A)=P(B)=P(C)=1/3 Probability of selecting the gold coin from each of the three boxes, P(G,A)=1 (both coins are gold) P(G,B)=0 (no gold coins) P(G,C)=1/2 (one of the two coins is gold) The question asks for the probability of the other coin being gold, which essentially means the probability of drawing a gold coin from bag A and this cannot happen unless you picked bag A. So, P(A,G) (NOT P(G,A) because we need to pick bag A first), => [{P(A)*P(G,A)}/{P(A)*P(G,A)+P(A)*P(G,B)+P(C)*P(G,C)}] => (1/3*1)/[(1/3*1)+(1/3*0)+(1/3*1/2)] => (1/3)/(1/3+1/6) => (1/3)/(1/2) => 2/3 I only did this because I know the formula for conditional probability. Does the GMAT test conditional probabilities? Also, is there another way of doing it? I was tempted to just say 1/3, but then the probability of selecting the second gold coin is conditional on the fact that you select bag A first. Is my logic correct?