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.

  • $\begingroup$ 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? $\endgroup$
    – JMoravitz
    Commented Dec 7, 2022 at 13:07
  • $\begingroup$ math.stackexchange.com/questions/2952508/probability-urns $\endgroup$ Commented Dec 7, 2022 at 13:10
  • $\begingroup$ math.stackexchange.com/questions/2142363/… $\endgroup$ Commented Dec 7, 2022 at 13:12
  • $\begingroup$ 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 $\endgroup$
    – Tsidia
    Commented Dec 7, 2022 at 13:19
  • $\begingroup$ @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. $\endgroup$
    – JMoravitz
    Commented Dec 7, 2022 at 13:20

1 Answer 1


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.

  • $\begingroup$ Mr. Moravitz, can you solve this problem through binoms? $\endgroup$ Commented Dec 7, 2022 at 13:36
  • $\begingroup$ $\dfrac{\binom{6}{2}}{\binom{8}{2}}$ 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. $\endgroup$
    – JMoravitz
    Commented Dec 7, 2022 at 13:38
  • $\begingroup$ Mr. Moravitz, so what is the answer to this problem? I want to check with the mine result. $\endgroup$ Commented Dec 8, 2022 at 7:15

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