Is the answer B?

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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?