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Thread: probability problem

  1. #1
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    probability problem

    n balls are randomly distributed among three boxes. Find the probability of
    (a) exactly two empty boxes;
    (b) exactly one empty box
    (c) no empty boxes.

    so the answers are (a) 1/3^(n-1)
    (b) [(2^n)-2]/3^(n-1)
    (c) [3^(n-1)-2^n+1]/3^(n-1)

    I don't understand why the denominator is 3^(n-1) and why the numerator for (b) is 2^n-2?
    shouldn't it be 3^n instead?
    Last edited by elmomleo; Nov 5th 2017 at 08:52 AM.
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  2. #2
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    Re: probability problem

    The probability that all balls will be in the leftmost box is 1ŕ3^n. The middle is 1/3^n. The right is 1/3^n. Add that up and you get 3/3^n.

    For (b) choose 1 box to be empty (3 ways). The n balls have 2^n different ways of being placed in the two boxes, except we cannot have all of the balls in the left box or all in the right box. So, (2^n)-2 ways so that at least one ball is in each box.
    Finally, product principle: 3*((2^n)-2)/3^n
    Last edited by SlipEternal; Nov 5th 2017 at 09:01 AM.
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