Choosing without replacement I was asked this problem by a friend and I feel like its so simple but I can't seem to figure out how to adjust for the decreasing sample size. 
Problem: bag with m red balls and n white balls. You choose 1 at a time and don't put it back in the bag. What's the probability you choose a red ball on the kth round?
Attempt: I tried to break it down into the 1, 2, ..., k rounds themselves in hopes I could visualize it. 
So on the first time the probability should just be m/m+n, the second time would be m-1/m+n-1 so would the kth trial just be m-k/m+n-k? This doesn't seem right to me as the different coloured balls don't have the equal chance of being chosen so I can't just use relative probability but I'm not sure how else to approach it. I've also tried matching it to a probability distribution but I also got stuck there. 
Any help is appreciated!
 A: To start you need to think about how many ways you actually sample a red ball on the kth draw, and then compare this to how many total configurations that you can have without imposing any restrictions. Let's address the second part now.
The total number of ways that you can either draw "m" red and "n" white balls one by one, until you have drawn them all is (n+m)!/(n!*m!), where "!" means factorial. The reason I am dividing by m! and n! is because I assuming that the balls are indistinguishable. For example, consider m=1 and n=2. Now take the order that I draw red ball 1 (i.e. r1) and then white balls 1 and 2, respectively. This is r1, w1, w2. However, I could also have r1, w2, w1. But if I do not care that about specific ball labels this is really the same thing, that is r,w,w. So I would need to divide each configuration by 1!*2!. 
Now if we fix the kth draw to be red, then this means we can let the other (n+m-1) balls be drawn however we want. Similar to the first part this can be done in (n+m-1)!/(n!*(m-1)!) ways. Therefore the answer is:
$$\frac {\frac{(m+n-1)!}{(m-1)!n!}}{\frac{(m+n)!}{m!n!}} = \frac{m}{m+n}$$
Hope this helps!
