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Math Help - Probability problem

  1. #1
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    Probability problem

    There are 2 boxes with n pencils each one.

    Randomly we remove a pencil of these boxes, and is not import of what box we remove the pencil.

    In a inesperd moment, a box of them this empty one...

    Calculate the probability that in the other box (nonempty) they are k pencils.
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  2. #2
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    Lets restructure the wording.
    Suppose the two boxes are labeled H & T, both with n pencils.
    We flip a coin, if it is heads we take one pencil from box H and tails we take one from T; then continue.
    If 1\le k\le n we will calculate the probability that when T is first empty there will be exactly k remaining in box H.
    Because this is a symmetric problem we will just double that to answer the question.
    To have the above setup we need the last flip to be a tail.
    Before that we need to have had (n-1) tails come up and (n-k) heads to have shown. This is a total of (2n-k-1) flips occurring in any order. The one more flip on which tails, then box T is empty and box H contains k pencils.
    What is the probability of that?
    \binom {2n-k-1} {n-1}\left( {\frac{1}{2}} \right)^{2n - k}.
    Those (n-1) tails may occur anywhere among the first (2n-k-1) flips, then one more tail.
    To get either box we double that.
    P(X=k)=\binom {2n-k-1} {n-1}\left( {\frac{1}{2}} \right)^{2n - k-1}.
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