Using Baye's Rule with two possible outcomes

Question

I have a problem that I cannot figure out how to apply Baye's rule to. The question is:

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) a silver coin;
(b) a gold coin.

For b, I defined each chest as $A_{1},...,A_{4}$ and defined $B :=$ a drawer contains a gold coin. Then I used Baye's rule on $P(A_1|B)$.

Question a, though, is different. I need to find $P(A_2 or A_3 or A_4)$. I am not asking for someone to solve the problem, just to tell me what $A$ would be in $P(A|B)$.

Answer 1

$P(A|B) = P(A \cap B) / P(B) $

A = second draw has silver coin

B = first draw has gold coin

$P(A \cap B) $= P(chose 1 or 2) P(1st coin gold) P(2nd coin silver) = (1/2) x(1/2) x 1 = 1/4

[note above:for gold-silver you have to choose chest 1 or 2, with probability 2/4 = 1/2 - you would then have to choose a gold coin (probability 1/2) then a silver coin (probability then 1 because there is only a silver coin)]

P(B) = 4/8 = 1/2

$P(A|B) = (1/4) / (1/2) = 1/2$

that's born out by considering that you chose with equal probability one of 4 gold coins, in two of the cases the other coin has to be silver, and in two of the cases it has to be gold

ii) A = second draw has gold coin B = first draw has gold coin

this can only happen if you choose chest 3 $P(A \cap B) $= P(choose 3) = (1/4)

$P(A|B) = (1/4) / (1/2) = 1/2$

also it has to be the complement of the probability found in (i)

Answer 2

For both questions you have chosen either chest number one, two or three (as these are the only ones containing gold coins). So we actually don't care about chest number four.

In the three interesting chests, there are four gold coins and two silver coins. Try looking at each scenario when finding the desired probabilities. If this is too vague, please let me know (and I'll be able to respond in a couple of hours).