probability that the white balls are left in the urn I don´t understand the solution of next problem:
An urn contains n white balls and m black balls. The balls are withdrawn one at a time until only those of the same color are left. Show that with probability $$n\over m+n$$ they are all white
The hint is: imagine that the experiment continues until all the balls are removed, and consider the last ball withdrawn.
So if we take into account the hint, there are $(n+m)!$ outcomes of withdrawing all the balls from the urn (in order) and the event that the last ball removed is white has n(n+m-1)! possible outcomes hence the probability is $${n(n+m-1)!\over (n+m)!}= {n\over n+m}$$
The thing is that why do we have to consider the last ball withdrawn? why if the last ball drawn is white implies that all white balls are left in the urn? I don´t get it 
I was trying to do it like this: there are $(n+m)!$ outcomes of withdrawing all the balls and there are $m$ black balls wich wan be arrenged in $m!$ ways and the white balls can be arrenged in $n!$ ways so the probability is $$m!n!\over (n+m)!$$ but this is just my assumption.
I know this is a silly question but can you please explain me why do we have to consider that the last ball withdrawn is white? and why does this implies that all the white balls are the left in the urn? 
Is there another way to solve this problem? I really would appreciate your help :D
 A: Let experiment $E$ be the experiment in which balls are drawn until the instant when all the balls left have the same colour. 
Let $E^\ast*$ be the experiment in which we continue drawing one at a time until we have drawn all the balls.
When $E$ terminates, if the balls left are all white, then in $E^\ast$ the last drawn ball is white. 
When $E$ terminates, if the balls left are all black, then in $E^\ast$ the last drawn ball is black.
So when $E$ terminates, the balls left are all white if and only if in $E^\ast$ the last drawn ball is white. 
If we imagine the balls to be labelled, all sequences of drawing are equally likely. So the probability the last ball is white is the same as the probability the first ball is white, which is $\frac{n}{n+m}$.  
A: 
Please find the detailed solve using forward thinking method.


A: Following the hint: The last ball should be equally likely to be any of the original balls.  So the probability that the last ball is white is $\frac{n}{n+m}$.
The condition that when only one color remains that color is white is satisfied precisely if the last ball is white.
A: For the condition to be always satisfied, the last ball removed must be white, occurring at $m+n$ position. If we concentrate on the black balls then the last black ball can occupy any position $m+k$ where $k= \{0,1,2...n-1\}$. The rest of the $m-1 $ black balls can occupy the remaining $m+k-1$ positions. Since the conditions with each last black ball position are disjoint, then the probability is
$$
\begin{align}
&= \dfrac {\displaystyle\sum_{k=0}^{n-1}\dbinom {m+k-1}{m-1}}{\dbinom {m+n}{m}}
&= \dfrac {\dbinom {m+n-1}{n-1}}{\dbinom {m+n}{m}} \tag{Hockey stick identity}\\
&=\dfrac{n}{n+m}
\end{align}
$$
