Question
An urn contains $n$ white and $m$ black balls, where
$n$ and $m$ are positive numbers.
If two balls are randomly withdrawn, what is the probability that they are the same color?
If a ball is randomly withdrawn and then
replaced before the second one is drawn, what
is the probability that the withdrawn balls are
the same color?
My Approach
- Probablity$$=\frac{\binom {n}{2}\,+\,\binom {m}{2}}{\binom {n+m}{2}}$$
Either choosed $2$ color from white $(n)$ or 2 from black $(m)$
2.Probablity$$=\frac{\binom {n}{1}\,*\,\binom {n-1}{1}\,+\,\binom {m}{1}\,*\,\binom {m-1}{1}}{\binom {n+m}{2}}$$
first choosed $1$ ball from black ,then $1$ from the remaining and did the same for
white ball .Am i correct?
Best Answer
For the first one: $$\large{\frac{\binom{m}{2}+\binom{n}{2}}{\binom{m+n}{2}}}$$ Explanation:
For the second one: $$\frac{m^2+n^2}{(m+n)^2}$$ Explanation: