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Probability with urns and balls

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data
    If n balls are distributed into k urns, what is the probability that the last urn contains j balls?

    2. Relevant equations

    3. The attempt at a solution

    not really sure how to do it. This is only a review question, not HW, but with that being known, I would appreciate an explanation more than an answer.
    Last edited: Sep 25, 2010
  2. jcsd
  3. Sep 25, 2010 #2
    Re: probability

    There are a few ways to approach this problem. That being said..

    To find to number of possible outcomes (i.e. ways to fill the urns), I would start by choosing a fixed number of balls that must be in the same urn, then finding out how many ways the others can be distributed.
    So initially, let's say all n balls must be in one urn (clearly only one way to choose them). There are k urns to put them in, so this can be done in k ways.

    Now, (n-1) balls must be in one urn. [tex]{n\choose n-1}[/tex]
    so n ways to choose these n-1 balls. These can go in any urn, so k(n) ways to fill one urn with n-1 balls. The extra ball may go in any urn except the one already occupied, so it has k-1 urns to choose from. There are k(n)*(k-1) ways to fill the urns in this way.

    Requiring that (n-2) balls be in one urn is similar, but the extra 2 balls may both go in any of the k-1 remaining urns, so they can be distributed in [tex] (k-1)^{2}[/tex] ways.

    BTW: [tex]n\choose k[/tex] [tex]=\frac{n!}{k!(n-k)!}[/tex], if you aren't familiar with the notation. It's same thing as imposing an equivalence class so that order doesn't matter. I.E. (a, b) is the same thing as (b, a), which just means that you want the set {a, b}.

    Does this make sense?
  4. Sep 25, 2010 #3
    Re: probability

    how would i calculate the total number of possible ways to put n balls in k urns?
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