Here is the problem:
Each of 2 cabinets identical in appearance has 2 drawers. One of the cabinets has a silver coin in each drawer. The other cabinet has a silver coin in one drawer and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is revealed. What is the probability that the other drawer of the cabinet holds a silver coin?
And here's what I did:
Let $C1$ = event that cabinet 1 is chosen
Let $C2$ = event that cabinet 2 is chosen
Let $S$ = event that a silver coin is chosen
$P(C1|S) = \frac{P(S|C1)P(C1)}{(P(S|C1)P(C1)+P(S|C2)P(C2)}$
$P(C1|S) = \frac{1\times0.5}{1\times0.5 + 0.5\times0.5}$
$P(C1|S) = \frac{2}{3}$
However, I'm not entirely sure if this is correct, is it possible to get feedback/hints?
Best Answer
Looks fine. That is how you apply Bayes' Rule and the Law of Total Probability.
Another approach: You could have picked any drawer without bias, due to the drawer selection method. So the silver coin you choose could have been equally likely to be any from the three silver coins, and two of these coins are in a cabinet containing another silver coin. Thus verifying your result of $2/3$.
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