Having problems understanding the correct answer on a classic urn probability question. There are two urns, one urn $U_1$ containing $3$ black balls $B$ and $6$ white balls $B^c$, while the other urn $U_2$,  contains $100$ white balls. An urn is selected uniformly at random and then a ball is drawn uniformly at random from the chosen urn. Suppose that the first drawn ball was white and returned to its original urn. What is the probability that another ball drawn from that same urn, will be black?
The answer: $\frac{2}{15}$ makes no sense to me. Even when I see the answer described (so I may need this explained like I am $5$).
I first assumed $P(B^c, U_1) = \frac{1}{3}$ and $P(B|U{_1}) = \frac{1}{3}$. Then the answer would be $\frac{1}{3}x\frac{1}{3}=\frac{1}{9}$.
Second, I reasoned that $P(B^c | U_{1,2}) = \frac{5}{6}$, and drawing from $U_1$ or $P(U_1) = \frac{1}{2}$. So $P(B^c, U{_1}) = \frac{5}{12}$, and $P(B|U{_1}) = \frac{1}{3}$. Then the answer would be $\frac{5}{12}x\frac{1}{3} = \frac{5}{36}$
Answer: you saw that the probability $\frac{5}{6}$ of event $B_1^c$ (first drawn ball was white), is composed of $P(B_1^c, U_1) = \frac{1}{2}$ while $P(B_1^c, U_1^c) = \frac{1}{3}$.
So, by the definition of the conditional probability $P(U_1|B_1^c) = \left(\frac{1}{2}\right)/\left(\frac{5}{6}\right)= \frac{3}{5}$.
This is precisely the probability of using the urn $U_1$ in the experiment described in part (c), so repeating the analysis of part (a) we conclude that the event $B$ of getting black ball when using same urn twice, has the following probability conditioned on $B_1^c$,
$$\begin{align*}
P(B|B_1^c) &=P(B|U_1,B_1^c)P(U_1|B_1^c)+P(B|U_1^c,B_1^c)P(U_1^c|B_1^c) \\
&= 0\times\frac{3}{5}+\frac{1}{3}\times\frac{2}{5} \\
&= \frac{2}{15}.
\end{align*}$$
 A: If you are looking for an intuitive explanation, you can proceed this way.  Suppose that the experiment is run $90$ times.  $45$ times we choose urn $2$, and the first ball will be white.  $45$ times we choose urn $1$.  $30$ times the first ball is white, and $15$ times the first ball is black, but that didn't happen!  So we have $75$ possibilities to consider, $30$ where we chose urn $1$ and $45$ where we chose urn $2$.  The second ball can only be black if we chose urn $1$.  $10$ times it is black and $20$ times it is white.  Therefore, the probability that the second ball is balck, given that the first is white is $$\frac{10}{75}=\frac2{15}$$
You can see that this conforms to the definition of conditional probability.  $\frac{10}{90}$ is the probability that the first ball is white and the second black, and $\frac{75}{90}$ is the probability that the first ball is white.       
A: It's a little hard to follow your question because you don't define clearly the events. We have three random experiments here: 


*

*Selection of the urn (let $U_1$ be -not a urn label but- the event that urn $1$ is selected; and let $U_2= U_1^c$)

*Selection of first ball: let $W_1$ be the event that the first ball is white.

*Selection of second ball: let $B_2$ be the event that the second ball is black.
We want $$P(B_2|W_1) = \frac{P(B_2,W_1) }{P(W_1)}$$
Then, using total probability $P(X)=P(X|Y)P(Y) + P(X|Y^c)P(Y^c)$ we have:
$$P(W_1)=P(W_1 | U_1) P(U_1) + P(W_1 | U_2) P(U_2) =\frac{6}{9} \frac12 + 1 \frac12 = \frac56$$
and
$$P(B_2,W_1)=P(B_2,W_1 | U_1) P(U_1) + P(B_2,W_1 | U_2) P(U_2) =\frac{6}{9} \frac{3}{9}\frac12 + 0 \frac12 = \frac19$$
Then  $$P(B_2|W_1) = \frac19/\frac56=\frac{2}{15}$$
