Question: You have a chest of $8$ drawers. With probability $1/2$, you put a letter in one of the drawers. With probability $1/2$, you don't put a letter in any drawer. I open the first $7$ drawers, all are empty. What is the probability there is a letter in the $8th$ drawer? I thought the answer was $1/16$ but apparently its $1/9$ can any one explain why?

Answer: 
Let $A$ denote the event in which there is a letter in the $8$th drawer.

Let $B$ denote the event in which the first $7$ drawers are all empty.

Then $P(A|B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{\frac12\cdot\frac18}{\frac12\cdot(1-\frac78)+(1-\frac12)\cdot1}=\cfrac19$.
Alternative way:

Think of it as a chest of $2\times8=16$ drawers and put a letter in one of the drawers. 

After opening $7$ empty drawers there are $9$ drawers left and in exactly one of them there is a letter.