Bayes theorem in a simple example Having this situation
Two urns with the number of balls in there pictured above..
and two events:
$A$ = urn is 1
$B$ = ball is white
I know that $P(A) = \frac{1}{2}$, $P(\text{not }A) = \frac{1}{2}$, $P(B \mid A) = \frac{2}{3}$, $P(B\mid\text{not }A) = \frac{3}{4}$
but if I try to verify the Bayes Theorem with $P(B/A)$, I get troubles..
$$
P(B\mid A) = \dfrac{P(A\mid B)P(B)}{P(A)}
$$
$P(B\mid A)$ is $\frac{2}{3}$, $P(A)$ is $\frac{1}{2}$, $P(B)$ I think is $\frac{5}{7}$, but how about $P(A\mid B)$?
Is it meaningful asking for the probability that I choose urn 1 knowing that I extracted a white ball? I think not but I'm unsure... am I asking the probability that the urn was the first known the ball extracted was white?
 A: I am just discussing the situation and hope that will answer your question.
First of all, I quite disagree with your contention that $P(B)$ is $5/7$. Because, to calculate this, you need to make two cases: the urn chosen was 1 and the urn chosen was not 1, and in each case, consider the probability of getting a white ball.
So, $P(B)$ is $$P(B)=P(A).P(B|A)+ P(\bar A).P(B|\bar A)$$ So, it turns out that $P(B)=\frac{17}{24}$. So, Now you can calculate, $P(A|B)$ using Bayes' Rule, 
$$P(A|B)=\frac{P(B|A).P(A)}{P(B)}$$ Here, from the information you have provided yourself, we get $P(A|B)=\frac{8}{17}$. So, now see that, the expression you have written down is true.
So, the probability you are asking about, the probability that urn 1 was chosen given a white ball was drawn, is computable and is reasonable to ask. But, note that computation of this fact needs Bayes' Rule.
A: I can't comment, but the question seems a bit loose.  To say it back, you have two urns, $X$ and $Y$, where $X$ has two white balls and one black one and $Y$ has three white balls and one black one.
Now you define events: $A$ is picking urn $X$ and $B$ is getting a white ball.
You are now going to pick an urn uniformly at random and then pick a ball uniformly from the selected urn.
At this point it should be clear that $P(B)$ is not $5/7$.  It is actually: $(1/2)(2/3) + (1/2)(3/4) = 17/24$.  Now use Bayes rule to get $P(A|B)$, which is otherwise not obvious.
