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 The balls are withdrawn one at a time until only those of the same color are left.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The hint is: imagine that the experiment continues until all the balls are removed, and consider the last ball withdrawn.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The thing is that why do we have to consider the last ball withdrawn?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I know this is a silly question but can you please explain me why do we have to consider that the last ball withdrawn is white?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

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 $\frac{n}{n+m}$.

• thank you very much andre I finally understand it!! :D really appreciate your help Aug 21, 2014 at 2:25
• You are welcome. The fact that the (unconditional) probability that the $k$-th drawn ball is white is $\frac{n}{n+m}$ is, for a while, intuitively unreasonable. Then at some stage it switches from unreasonable to intuitively obvious! Aug 21, 2014 at 2:30
• @Andre Nicolas I have a much better answer with forward approach. This one is really confusing. May I post? Jan 20, 2018 at 2:17
• I would like to add that converting expt E to E* ensures that all outcomes are equally likely. To clarify, you can define an outcome in $E$ as a sequence that terminates prematurely, i.e. not necessarily $n$ labelled balls long, whereas outcomes in $E*$ are guaranteed to be $n$-permutations. In expt E, an outcome $O_j$ with j balls left and an outcome $O_k$ with k balls left are not equally likely. In particular, $P(O_k)=\frac{k!}{n!}$. Jan 31, 2021 at 9:51
• Additionally, the extension from $E$ to $E*$ "preserves" the values of probability function applied to events. Since the event comprising outcomes with white balls left in $E$ "maps" to the event comprising outcomes with last ball white in $E*$ (not a bijection, however!), their probabilities are also the same Jan 31, 2021 at 9:58

Please find the detailed solve using forward thinking method.

• @user128422. Here in place of n white and m black, we took m white and n black Jan 20, 2018 at 2:27

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.

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}