There are $4$ urns.
urn A has $2$ black balls and $6$ white balls
urn B has $4$ black balls and $4$ white balls
urn C has $6$ black balls and $2$ white balls
urn D has $8$ black balls
You choose an urn at random with equal probability, then draw $3$ balls from it, one at a time, without returning any back to the urn. What's the probability of drawing a black ball if $2$ black balls were drawn in the first two drawings?
I understand that I have to multiply probabilities in order to find one "branch" and then sum all the branches, but I keep getting wrong answers like $\frac{1}{2} \text{or} \frac{5}{14}$ and I can't figure out why.
Problem Setup:
Let events $A,B,C,D$ be the events that we happened to be drawing from urns $A,B,C,D$ respectively.
Let $X$ be the event that the first two balls drawn happened to be black.
Let $Y$ be the event that the third ball drawn happens to be black.
You are tasked with finding $\Pr(Y\mid X)$. To do this, recognize that $\Pr(Y\mid X) = \Pr((Y\cap A)\cup (Y\cap B)\cup (Y\cap C)\cup (Y\cap D)\mid X)$
$ = \Pr(Y\cap A\mid X)+\Pr(Y\cap B\mid X)+\Pr(Y\cap C\mid X)+\Pr(Y\cap D\mid X)$
$ = \Pr(Y\mid A\cap X)\Pr(A\mid X) + \dots +\Pr(Y\mid D\cap X)\Pr(D\mid X)$
The value of pieces like $\Pr(Y\mid C\cap X)$ can be found intuitively. This for instance, we are asking what the probability the third ball drawn will be black given we are pulling from urn C and two black balls have already been pulled. Well, in this scenario, there are $4$ black balls left after our earlier pulls out of $6$ balls in total giving $\Pr(Y\mid C\cap X) = \dfrac{4}{6}$
The rest of the pieces such as $\Pr(C\mid X)$ can be found from Bayes' Theorem. Recall that $\Pr(C\mid X) = \dfrac{\Pr(X\mid C)\Pr(C)}{\Pr(X)}$. Each of the pieces here should be readily obtainable. I leave the rest of the calculations and piecing of everything together to you.