Question:
An urn contains $n + m$ balls, of which n are red and m are black. They are withdrawn from the urn, one at a time and without replacement. Let $X$ be the number of red balls removed before the first black ball is chosen. Find the expected value $\mathbb{E}[X]$.
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In the solution, of this problem , from here,
Finding the number of red balls drawn before the first black ball is chosen
It has been said that, there are $m+1$ special black balls. But, from the problem description we know that, there are m black balls.
i) Why are $m+1$ balls considered? In a post I read that it has to do with some ordering, but I am not sure what ordering is being mentioned there.
[ $X_i$ is $1$ if i-th red ball is picked before any black ball]
Also,
ii) Why is the probability of picking the $i$'th red ball $\mathbb{P}\{X_i=1\}= 1/(m+1) $
iii) Can we pick the $i$-th ball before picking $(i-1)$ th ball ?
If that's true then, if i-th ball is removed then should that not imply that, all the other i-1 balls before it have already been removed ?
In that case, should the probability, $\mathbb{P}\{Xi=1\}$, not be the following?
$$\frac{n-i+1}{m+n-i+1}$$
Since we are picking one red ball from the urn and not returning it? And, if we pick the $i$-th ball then, have the $1$ st to $(i-1)$th balls are already been removed from the urn?
Let $R$ be any one of the red balls, and imagine that we temporarily paint $R$ black, so that there are $n-1$ red balls and $m+1$ black balls in the urn. What is the probability that $R$ is the first black ball picked? There are $(m+1)!$ equally likely different orders in which the $m+1$ black balls can be picked, and $R$ is the first ball in $m!$ of those orders, so the probability that $R$ is the first black ball picked is $\frac{m!}{(m+1)!}=\frac1{m+1}$. Note that it doesn’t matter in what order or exactly when in the process the $n-1$ red balls are picked: at the moment we’re interested only in the order in which the $m+1$ black balls are picked.
Now just pretend that we temporarily repaint $R$ black: the same argument shows that the probability that $R$ is drawn before any of the $m$ black balls is $\frac1{m+1}$. The point is that this depends only on the order in which these $m+1$ balls are picked, not on anything that happens to the other $n-1$ red balls, and each of these $m+1$ balls is equally likely to be picked first.
Now label the $n$ red balls $R_1$ through $R_n$.
For $k=1,\ldots,n$ let
$$X_k=\begin{cases} 1,&\text{if }R_k\text{ is chosen before all of the black balls}\\ 0,&\text{if at least one black ball is chosen before }R_k\;. \end{cases}$$
We’ve just seen that $\Bbb P[X_k=1]=\frac1{m+1}$ for each $k=1,\ldots,n$, and of course it follows that $\Bbb E[X_k]=\frac1{m+1}$ for each $k=1,\ldots,n$ and hence by linearity of expectation that
$$\Bbb[X]=\sum_{k=1}^n\Bbb E[X_k]=\frac{n}{m+1}\;,$$
where $X=\sum_{k=1}^nX_k$, the number of red balls drawn before the first black ball is drawn.