# Balls of Two Colors

An urn contains $w$ white balls and $b$ black balls $(w \gt 0)$ and $b \gt 0).$ The balls are thoroughly mixed and two are drawn, one after the other, *without* replacement. Let $W_{i}$ and $B_{i}$ denote the respective outcomes 'white on the *i*th draw' and 'black on the $i$th draw,' for $i = 1, 2.$

Prove that $\displaystyle P(W_{2}) = P(W_{1}) = \frac{w}{w+b}.$ (Which clearly implies a similar identity for $B_1$ and $B_{2}.)$

Furthermore, $\displaystyle P(W_{i}) = \frac{w}{w+b},$ for any $i$ not exceeding the total of white number of balls $w.$

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Copyright © 1996-2017 Alexander Bogomolny

We must remember the formula for the **total** probability:

$P(w_{2}) = P(w_{2}|w_{1})\cdot P(w_{1}) + P(w_{2}|b_{1})\cdot P(b_{1}),$

from which

$\begin{align}\displaystyle P(w_{2}) &= \frac{w - 1}{b + w - 1}\cdot \frac{w}{b + w} + \frac{w}{b + w - 1}\cdot \frac{b}{b + w} \\ &=\frac{w\cdot (w - 1 + b)}{(b + w - 1)\cdot (b + w)}, \end{align}$

which is simplified to $\displaystyle\frac{w}{w+b}.$

The wonderful thing here is that the initial probability $\displaystyle\frac{w}{w+b}$ is being maintained even though the balls are not returned to the urn.

### References

- Ruma Falk,
*Understanding Probability and Statistics*, A K Peters, 1993 - Paul Nahin,
*Will You Be Alive In 10 Years From Now?*, Princeton University Press, 2013 (Introduction)

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