Probability that the withdrawn balls are the same color

Question

An urn contains $n$ white and $m$ black balls, where $n$ and $m$ are positive numbers.

1. If two balls are randomly withdrawn, what is the probability that they are the same color?

2. If a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color?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

1. 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?

1 Answer

For the first one: $$\large{\frac{\binom{m}{2}+\binom{n}{2}}{\binom{m+n}{2}}}$$ Explanation:

• $\large{\binom{m+n}{2}}$ ways of selecting two balls.
• $\large{\binom{m}{2}}$ ways of selecting two black balls.
• $\large{\binom{n}{2}}$ ways of selecting two white balls.

For the second one: $$\frac{m^2+n^2}{(m+n)^2}$$ Explanation:

• $(m+n)^2$ ways of selecting two balls, one at a time, with replacement.
• $m^2$ ways of selecting two black balls.
• $n^2$ ways of selecting two white balls.