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 $P(A|B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{\frac12\cdot\frac18}{\frac12\cdot(1-\frac78)+(1-\frac12)\cdot1}=\cfrac19$.
Given that the first seven drawers were empty either $A\cup B$ where:

* $A$: The letter is in the eighth drawer: $\mathsf P(A)=\tfrac 12 \tfrac 18$
* $B$: The letter is in no drawer: $\mathsf P(B)=\tfrac 12$
	+ $A,B$ are disjoint

So by the definition of conditional probability

$$\mathsf P(A\mid A\cup B) ~=~ \dfrac{\color{grey}{\tfrac 12 \tfrac {\color{black}{1}}8}}{\color{grey}{\tfrac 12 \tfrac {\color{black}{1}}8+\tfrac 12 \tfrac {\color{black}{8}}8}}~=~\dfrac 19$$

$\blacksquare$