Probability that the withdrawn balls are the same color Question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.
  
  
*
  
*If two balls are randomly withdrawn, what is  the probability that they are the same color?
  
*If  a  ball  is  randomly  withdrawn  and  then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  is the probability that the withdrawn balls are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?


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.

