Probability problem solution verification 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?
 A: It looks good to me. You should make a tree diagram for these types of problems so you can easily compute the probabilities, assuming these events are mutually exclusive.
A: 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 $2/3$.
$\begin{array}{c}
&&&\nearrow & \tfrac 12 & \circ
\\
&\nearrow
& \tfrac 12 & \searrow & \tfrac 12 & \circ
\\[1ex]
& \searrow
 & \tfrac 12 & \nearrow & \tfrac 12 & \circ
\\
&&& \searrow & \tfrac 12 & \bullet
\end{array}$ 
A: IF you want a better understanding on how to solve this question, here is a link to the book that I used last semester in my probability math course. Head to page 141 and it should explain how conditional probability works.
https://math.dartmouth.edu/~prob/prob/prob.pdf
A: Actually, after thinking about it hard, I don't think any of this is right. The question is asking the probability of a silver coin being in the other drawer, given that you already chose a silver coin.
I'm thinking of solving it as follows:
P(choosing silver coin|silver coin already chosen) $ = P(S|C1)P(C1) + P(S|C2)P(C2)$
$= (1)(0.5) + (0)(0.5) = \frac{1}{2}$
Since we don't know which cabinet we have chosen, we would have to calculate the total probability, because the probability that a silver coin is chosen really depends on which cabinet you have chosen. I think this reasoning makes more sense. What do you think?
