Puzzle: another probability problem

If you have an urn who already contains either a black or white ball, and you add a white ball to it, and then you subtract a white ball from it, then what is the probability of having a white ball left?

[update: adding the fact that the original ball is either black or white]


Comments (10)

  1. Eusebio Rufian-Zilbermann says:

    the same probability as the ball being white at the beginning 😉

  2. Bruce says:

    Depends on whether the first ball, the one already in the urn, is white or not.

    Lets see if I remember any math…

    If it is, then the answer is 1 – always

    If it is not, then the answer is 0 – never

    If the probability of the first ball being white is A, then the probability of the remaining ball being white is also A.

    If you’re playing stupid puzzle tricks on David Letterman, then one might ask "what else is in the urn, in addition to a ball of indeterminate color?" Answers could include things like other balls, a ball-making machine, or a clown.

  3. AdiOltean says:

    Sorry, I meant tosay that the original ball in the urn can be either black or white.

  4. I think we can make this into a Monty Hall problem.

    Start with an urn that contains two white balls and a black ball. The black ball wins a prize. Take one ball and hold it in your hand without looking at it.

    Now, we have the same setup as Adi’s problem right after you drop the white ball in.

    The host removes one white ball from the urn. You either keep the ball in hand or switch to the ball in the urn. The urn ball has 2/3 probability of being black and 1/3 probability of being white.

  5. Ian Horwill says:

    I agree with Eusebio and Bruce, but I had to think about it 🙂

    There are 3 possible siutuations (A is the probability of the original ball being white):

    1. The original ball is white, and that’s the one you pick out (prob = A x 0.5)

    2. The original ball is white, but you pick the 2nd ball back out (prob = A x 0.5)

    3. The original ball is black, but you pick the 2nd ball back out (prob = 1 – A)

    The probability that the remaining ball is white = prob (1 or 2) = A x 0.5 + A x 0.5 = A

  6. AndrewSeven says:

    Isn’t the wording a little off?

    You have an urn that contains a ball that is either black or white.

    You add a white ball to the url.

    You remove a (random) ball from the urn and get a white ball.

    What is the probability of having a white ball left?

  7. Rob says:

    Yeah, with the original wording, the probability of #FFFFFF would be infinitesimally small (one choice of an infinite spectrum).

  8. Arturo says:

    What Nicholas says is incorrect. His initial setup (2 white balls, 1 black ball, remove one ball without looking at the color…) After this setup, the probability of having two white balls left is only 1/3, when the right setup of the problem should be 1/2.

    We have 4 scenarios with 1/4 chance of happening,

    1. B1 is black, B2 is white. We retrieve B1.

    2. B1 is black, B2 is white. We retrieve B2.

    3. B1 is white, B2 is white. We retrieve B1.

    4. B1 is white, B2 is white. We retrieve B2.

    Scenario #1 didn’t happen. We have 3 scenarios left, 2 where the remaining ball is white.

    Therefore, the probability of having a white ball left in the urn is 2/3.

  9. molly says:

    the probability of one while ball left is 2/3

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