# Conditional Probability (Where did I go wrong?)

Four identical balls are numbered $1, 2, 3$ and $4$ and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.

For the first question:

What is the probability that the sum of the numbers on the two balls is $5$?

Here, there is no problem: we could take $\{1,4\}$ , $\{4,1\}$ , $\{2,3\}$ , $\{3,2\}$. Each of those four possibilities has a probability $\frac{1}{4}\cdot \frac{1}{3}=\frac{1}{12}$. So let $F$ be the sample space above, such that $$Pr\left( F \right)=4\cdot \frac{1}{12}=\frac{1}{3}$$

Now here's where I ran into problems:

Given that the sum of the numbers on the two balls is $5$, what is the probability that the second ball drawn is numbered $1$?

What I did was this: $$Pr\left( 1|F \right)=\frac{Pr\left( 1\cap F \right)}{Pr\left( F \right)}=\frac{Pr\left( 1\; ,\; 4 \right)+Pr\left( 4\; ,\; 1 \right)}{Pr\left( F \right)}=\frac{\frac{1}{12}+\frac{1}{12}}{\frac{1}{3}}=\frac{1}{2}$$

The answer is meant to be $\frac{1}{4}$. Where did I go wrong?