Suppose we have four chests each having two drawers. Chests 1 and 2
have a gold coin in one drawer and a silver coin in the other drawer.
Chest 3 has two gold coins and chest 4 has two silver coins. A chest is
selected at random and a drawer opened. It is found to contain a gold
coin. 

**Find the probability that the other drawer has a silver coin.**

I'm not sure how to solve this problem, here is what I tried:

(1)

$A:$select a chest that contains at least one gold coin.

$B$: select a chest that contains at least one silver coin.

So $P(A) = \frac{3}{4}$ and $P(A\cap B)=\frac{1}{2}$ 

Then $P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{2}{3}$

(2)

$A\_{1}, A\_{2},A\_{3},A\_{4}:$ selecting a chest.

$B:$ finding a silver coin 

Then $P(A\_{i})=\frac{1}{4}$ for $i\in\{1,2,3,4\}$

$P(B\mid A\_{1})=\frac{1}{2}$,
$P(B\mid A\_{2})=\frac{1}{2}$,
$P(B\mid A\_{3})=0$,
$P(B\mid A\_{4})=1$

So I need to calculate: $P(A\_{1}\mid B) \cup P(A\_{2}\mid B)$

$P(A\_{1}\mid B) \cup P(A\_{2}\mid B)=P(A\_{1}\mid B)+P(A\_{2}\mid B)$

$=\frac{P(A\_{1})P(B\mid A\_{1})}{\sum\_{k=1}^{4} P(A\_{k})P(B\mid A\_{k})}+\frac{P(A\_{2})P(B\mid A\_{2})}{\sum\_{k=1}^{4} P(A\_{k})P(B\mid A\_{k})}=\frac{1}{2}$

I don't know if any of this is correct, any help thanks