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It seems easy, but I don't get it

  1. Aug 27, 2010 #1
    Hi everybody.
    Once more, I need your help.

    The following is a problem that appears in Rohatgi's "An introduction to Probability and Statistics":

    "Each of n urns contains four white and six black balls, while another urn contains five white and five black balls. An urn is chosen at random from the (n+1) urns, and two balls are drawn from it, both being black. The probability that five white and three black balls remain in the chosen urn is 1/7. Find n."

    My first guess would be that n=6, but the solution given in the book is n=4.

    Someone can tell me if I am wrong or there is a mistake in the book ?
    Thanks.
     
  2. jcsd
  3. Aug 28, 2010 #2
    My guess is the following one:

    The probability that five white and three black balls remain in the chosen urn is zero for the n urns that had initially only four white balls.
    For the urn that had initially 5 white and 5 black balls, the probability of having 5 white and 3 black balls after having drawn 2 black balls is 1.
    The probability that this former is the chosen urn is 1/(n+1).
    So n must be equal to 6.

    Am I wrong or there is a mistake in the book?
    I would appreciate your hints.
    Thanks
     
  4. Aug 28, 2010 #3
    Let's say E is the event that the urn with 5 white and 5 balls is chosen, and B is the number of black balls drawn. You are told that

    [tex]P(E \;|\; B=2) = 1/7[/tex].

    By definition, [tex]P(E \;|\; B=2) = \frac{P(E \; \cap \; B=2)}{P(B=2)}[/tex].

    Take it from there.
     
  5. Aug 30, 2010 #4
    Thanks for your answer, awkward.
    This was my first attempt to the problem, but I wrote it in the following terms:

    Let's say A is the event that five white and three black balls remain in the chosen urn, B is the number of black balls drawn and Cm is the event that the urn m has been chosen.
    Then we are told that

    [tex]
    P(A \;|\; B=2) = \frac{P(A \; \cap \; B=2)}{P(B=2)}
    [/tex]

    where
    [tex]
    P(A \; \cap \; B=2) = \sum_{m=1}^{n+1}p(C_{m})p(A\; \cap \; B=2 \;|\; C_{m}) = \sum_{m=1}^{n+1} p(A\; \cap \; B=2 \cap \; C_{m})
    [/tex]

    and my problem was that I did not apply correctly the multiplication rule, that is

    [tex]
    p(A\; \cap \; B=2 \cap \; C_{m}) = p(C_{m}) p(B=2 \;|\; C_{m}) p(A \;|\; B=2 \cap \; C_{m})
    [/tex]

    If I apply it correctly, then I obtain the result n=4, the same that appears in the book, and indeed the same if I apply your hint.
    Thanks a lot.
     
    Last edited: Aug 30, 2010
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