Probability

**alakazam** · February 17th 2009, 11:36 PM

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?

**Grandad** · February 18th 2009, 01:57 AM

Hello alakazam

Originally Posted by alakazam

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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