Statistics I - Bayes's Theorem A desk has three drawers. The first contains two gold coins, the second has two silver coins and the third has one gold and one silver coin. A drawer is selected at random and a coin is drawn at random from the drawer. Suppose that the coin selected was silver. Use Bayes's Theorem to find the probability that the other coin in that drawer is gold. 
 A: You select a drawer and select a coin from that drawer.
Let $A$ denote the event that the coin that you select is silver.
Let $B$ denote the event that the other coin in the drawer is gold.
You are equally likely to pick any coin, so
$$P(A) = \frac{3\text{ silver coins}}{6 \text{ coins}} = \frac 12$$
There are 6 different ways you could have selected a coin. Only one of them results in you selecting a silver coin AND the other coin in the drawer being gold, so
$$P(A \cap B) = \frac{1\text{ desired outcome}}{6 \text{ total outcomes}} = \frac 16$$
By the definition of conditional probability,
$$P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{1/6}{1/2} = \frac 13$$

As I mentioned in my comment, Bayes's Theorem is useful if you want to know $P(B\mid A)$ but you only know $P(A\mid B)$. Here, we don't know $P(A \mid B)$ either, so Bayes's Theorem doesn't really help.
A: Let the choice of picking a drawer amongst 3 represent 3 events called $gg, gs, ss$ for both gold, one gold one silver and 2 silver drawer respectively. These probabilities are $\frac{1}{3}$ each.
From the $2g$ drawer, probability of picking a silver coin $p(s|gg) = 0$.
Similarly, $p(s|gs) = \frac{1}{2}$ and $p(s|ss) = 1$.
The only way the other coin can be gold is if you've picked the $gs$ drawer. So you're trying to find $p(gs|s)$.
By Bayes' theorem,
$$p(gs|s) = \frac{p(s|gs)\cdot p(gs)}{p(s|gs)\cdot p(gs) + p(s|gg)\cdot p(gg) + p(s|ss)\cdot p(ss)} \\ = \frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}\cdot\frac{1}{3} + 0 \cdot \frac{1}{3} + 1\cdot\frac{1}{3}} \\ = \frac{1}{3}$$
