# Multiple steps of branching probabilities

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.

• Have you used Bayes' Theorem? Have you heard of Bayes' Theorem? Are you correctly conditioning on the fact that the first two balls drawn are black? Commented Dec 7, 2022 at 13:07
• math.stackexchange.com/questions/2952508/probability-urns Commented Dec 7, 2022 at 13:10
• math.stackexchange.com/questions/2142363/… Commented Dec 7, 2022 at 13:12
• I did not use Bayes' Theorem, I figured I could just multiply the probability of having reached urn A by the probability of the next ball in that urn being black, do the same for all the other urns and sum the result Commented Dec 7, 2022 at 13:19
• @VadimChernetsov neither of those match the current scenario and emphasize the color of the first ball(s) drawn and how they influence our belief about the rest of the draws. @ Tsidia, this is a problem testing your knowledge and understanding of how and when to use Bayes' Theorem. Look at Bayes' Theorem. Understand it. Use it. Commented Dec 7, 2022 at 13:20

## 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.

• Mr. Moravitz, can you solve this problem through binoms? Commented Dec 7, 2022 at 13:36
• $\dfrac{\binom{6}{2}}{\binom{8}{2}}$ is the same as $\dfrac{6\cdot 5}{8\cdot 7}$. If you insist on using binomial coefficients where applicable, then you can certainly use binomial coefficients. Nothing I have written here prevents that. That being said, I expect it to be unnecessary. Commented Dec 7, 2022 at 13:38
• Mr. Moravitz, so what is the answer to this problem? I want to check with the mine result. Commented Dec 8, 2022 at 7:15