Baye's Theorem ball selection problem Urn $1$ has $5$ red ball and 6 black balls. Urn $2$ has $7$ red balls and $6$ black balls. If I combine the balls from both urns in a bag and draw a red ball, what is the probability it will come from Urn $1$?
I'm having a hard time getting figuring out what the numbers are that I need to plug into Bayes Theorem, any hints/help is appreciated.
So far I know I have:
$$P(R|U_1)= \frac{P(R,U_1)}{P(U_1)}$$
But I'm not sure where to go from there.  Sorry if formatting isn't right.
 A: Note that in general, $P(A|B)$ is the probability of $A$ given $B$.
You are given that the ball selected is red, so you do not want $P(\text{red} | \text{urn 1})$. Rather what you want is $P(\text{urn 1} | \text{red})$. 
$$\begin{align*}
P(\text{urn 1} | \text{red})
&=\frac{P(\text{urn 1} \cap \text{red})}{P(\text{red})}\\\\
&=\frac{\frac{5}{24}}{\frac{12}{24}}\\\\
&=\frac{5}{12}
\end{align*}$$
where $\frac{5}{24}$ comes from there being $11+13=24$ total balls, and $5$ of them being red and from urn $1$ and $\frac{12}{24}$ comes from there being $24$ balls and $5+7=12$ of them being red.
If you'd like to expand it in terms of $P(\text{red} | \text{urn 1})$ you could do
$$\begin{align*}
P(\text{urn 1} | \text{red})
&=\frac{P(\text{urn 1} \cap \text{red})}{P(\text{red})}\\\\
&=\frac{P(\text{red} | \text{urn 1})\cdot P(\text{urn 1})}{P(\text{red})}\\\\
&=\frac{\frac{5}{11}\cdot \frac{11}{24}}{\frac{12}{24}}\\\\
&= \frac{5}{12}
\end{align*}$$
where $P(\text{red}|\text{urn 1}) = \frac{5}{11}$ because urn $1$ has $5$ reds balls and $11$ total balls.
