Probability of finding a lost ball

Question

From a box with m white and n black balls, a ball has been lost. In order to figure out the content of the box, two balls have been taken from it. Both balls were white. Find the probability that the lost ball was white.

Answer 1

The pedestrian way:

\begin{align} P(\text{white lost})&=\frac m{m+n}\;,\\ P(\text{both white})&=\frac{\binom m2}{\binom{m+n}2}\;,\\ P(\text{both white}\mid\text{white lost})&=\frac{\binom{m-1}2}{\binom{m+n-1}2}\;. \end{align}

Thus

\begin{align} P(\text{white lost}\mid\text{both white})&=\frac{P(\text{both white}\mid\text{white lost})\,P(\text{white lost})}{P(\text{both white})}\\ &=\frac{\binom{m-1}2\binom{m+n}2}{\binom m2\binom{m+n-1}2}\frac m{m+n}\\ &=\frac{(m-2)(m+n)}{m(m+n-2)}\frac m{m+n}\\ &=\frac{m-2}{m+n-2}\;. \end{align}

The more elegant way:

Two white balls are known not to be lost. All other balls have the same probability of being lost.

Answer 2

This is an explanation for the "more elegant way" mentioned in joriki's answer.

There are $(m+n-2)$ balls remaining (including the lost one) after removing the two white balls.

The probability of any one of these balls being the lost ball is the same: $1/(m+n-2)$.

There are $(m-2)$ white balls remaining in this group, which gives the answer.