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Math Help - Probability - balls in baskets

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    Probability - balls in baskets

    Suppose we have n white balls and n black balls and we place them randomly into two baskets such that each basket has n balls. Suppose now we randomly pick one ball from each basket and swap them. What is the probability that we originally had i white balls in the first basket and now have j white balls in the first basket?
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  2. #2
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    Probability

    Hello alakazam
    Quote Originally Posted by alakazam View Post
    Suppose we have n white balls and n black balls and we place them randomly into two baskets such that each basket has n balls. Suppose now we randomly pick one ball from each basket and swap them. What is the probability that we originally had i white balls in the first basket and now have j white balls in the first basket?
    The number of ways of selecting the n balls to go in the first basket is \binom{2n}{n}.

    The number of ways of selecting i white balls from n, and (n-i) black balls from n to go in the first basket is \binom{n}{i} \times \binom{n}{n-i}=\binom{n}{i}^2.

    So the probability that there are initially i white balls in the first basket is

    \frac{\binom{n}{i}^2}{\binom{2n}{n}}

    At the commencement of the second stage, there are i white and (n-i) black balls in the first basket, and (n-i) white and i black in the second. Then there are three possibilities:

    j = i+1, where a black ball is removed from the first basket, and a white from the second. The probability of this is

    \frac{n-i}{n}\times\frac{n-i}{n}= \frac{(n-i)^2}{n^2}

    j = i, where either a black is removed from the first and a white from the second, or the other way round. The probability of this is

    \frac{n-i}{n}\times\frac{i}{n} + \frac{i}{n}\times\frac{n-i}{n}= \frac{2i(n-i)}{n^2}

    j = i -1, where a white is removed from the first and a black from the second. The probability of this is

    \frac{i}{n}\times\frac{i}{n}= \frac{i^2}{n^2}

    The overall probabilities therefore are:

    j = i+1: \frac{\binom{n}{i}^2}{\binom{2n}{n}}\times \frac{(n-i)^2}{n^2}

    j = i:\frac{\binom{n}{i}^2}{\binom{2n}{n}}\times \frac{2i(n-i)}{n^2}

    j = i-1:\frac{\binom{n}{i}^2}{\binom{2n}{n}}\times \frac{i^2}{n^2}

    Grandad
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