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
Best Answer
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.