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 They are withdrawn from the urn, one at a time and without replacement.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 of $X$.
The problem begins by stating calculating the expected value of $X_{i}$
Where $X_{i} = 1$ if i red ball is taken before any black ball is chosen
$X_{i}=0$ otherwise
The $E[X_{i}]$ is given to be
$\frac{{m+n\choose 1+m}m!(n-1)!}{(m+n)!}$
which simplies to $\frac{1}{m+1}$
Then summing over the $X_{i}$ the $E[X]$ is given to be $\frac{n}{1+m}$
I know that $m+n$ is the total number of ways you can chose. m! is how the m balls can be premutated and $n-1$ is the number of way you can order the remaining red balls. I'm having trouble understanding why you choose (1+m).
Best Answer
The description of your indicator $X_i$ is slightly ambiguous and that may be causing some confusion.
If you imagine that the red balls are numbered $1,2,3,\ldots,n$ before they are drawn, then the probability that red ball $i$ is drawn before any of the $m$ black balls (ignoring when the other $n-1$ red balls are drawn) is $\frac{1}{1+m}$ by considering possible orderings of $1+m$ balls. $E[X_i]$ is equal to this probability.
Now, by linearity of expectation, the expected total number of red balls drawn before any black balls is $\sum_i E[X_i] = \sum_i \frac{1}{1+m} = \frac{n}{1+m}$.