Probabilty problem related to two cabinets containing gold and silver coins. Each of the two cabinets has 3 drawers. Cabinet I contains a gold coin in each drawer and cabinet II contains a gold coin in one of its drawers and a silver coin in the other. A cabinet is randomly selected ,one of its drawers is opened and a gold coin is found. Find the probability that there is a gold coin in the other drawer.
The answer given in book is 3/4

 A: I have no idea what $C$ is doing there. There is no need to specify the event that a cabinet is chosen, as that event is guaranteed to happen, and doesn't affect any probabilities.
Try using the events $L$ for picking the left cabinet and $G$ for picking a gold drawer. You want $$P(L\mid G)=\frac{P(L\cap G)}{P(G)}$$ $P(L\cap G)=P(L)$ is immediately seen to be $\frac12$, while $P(G)$ takes some calculation.
Alternately, as long as we are in land the land of assuming that $G$ occurs, there are $4$ equally likely possibilities. Three of them are in the all-gold cabinet and one isn't.
A: I assume the second cabinet also has an empty drawer to make three.  In that case, we are equally likely to select each drawer.  There are six possibilities, but we discard the two where we get the silver and nothing. leaving four.  Three of them have a gold coin in another drawer, so the chance is $\frac 34$.  
A more careful approach is that there is $\frac 12$ chance we choose the first cabinet.  In that case we surely choose a gold coin and have another in another drawer.  There is $\frac 16$ chance we choose the second cabinet and then the gold drawer.  Here we do not have another gold coin, so the chance we have another gold coin given that we found one in the first drawer is $$\frac {\frac 12}{\frac 12+\frac 16}=\frac 34$$
