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Urgent probability questions

  1. Sep 27, 2007 #1
    Urgent probability questions!!!!

    Please help!

    How can 5 black and 5 yellow balls be put into two urns to maximize the probability that a yellow ball is drawn from a randomly chosen urn?

    I got:
    P(draw yellow)=P(draw yellow intersect urn1)+P(draw yellow intersect urn2)
    =P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2)

    where y=number of yelow balls in urn 1 and k=number of black balls in urn 2
    and y, k<=5

    Then, I think I should plug in k=1, 2, 3..., differentiate the above, and set it to zero. Find critical value--but I never get a solution! I get something like -9=0....

    Also, if there are b black balls and y yellow balls in an urn, and a person takes one out, replaces it, and adds in another of the same color, prove that at the nth trial, the probability of picking a black ball is b/b+y
  2. jcsd
  3. Sep 28, 2007 #2


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    Let F = P(draw yellow|urn1)P(urn1) + P(draw yellow|urn2)P(urn2) = (y1/n1 + y2/n2)/2.
    You also have
    y1 + y2 = 5
    n1 + n2 = 10.
    Therefore you can write y2 = 5 - y1 and n2 = 10 - n1 and substitute in to F.
    Now, how do you simultaneously set y1 and n1 to maximize F?
  4. Sep 28, 2007 #3

    D H

    Staff: Mentor

    This doesn't work for a couple of reasons:
    • The solution (y1=2.5, n1=5) is not an integral solution.
    • More importantly, the solution is a saddle point. Its neither a minimum nor a maximum. If you had six of each color rather than five, the derivative-based solution is acheivable but is not maximal.

    The resultant resultant probability for the derivative-based approach is 50%. There are many realizable arrangements that yield a 50% probability, for example
    • One of each color in one jug, the rest in the other jug
    • Two of each color in one jug, the rest in the other jug
    • One yellow and four black balls in one jug, four yellow and one black in the other jug.

    There are several arrangements that do better than 50%. For example, two yellow and four black balls in one jug, three yellow balls and one black ball in the other yields a probability of 1/3 for jug A, 3/4 for jug B, or 13/24 for a random jug. There are ways to do even better. If you think about the problem for a bit one arrangement should pop out as special. Of course, you could brute-force it; there are only 21 combinations to worry about.
  5. Sep 28, 2007 #4


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    I was wrong; y1=3, n1=5 is a saddle point. Thanks for catching it.
    Last edited: Sep 28, 2007
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