# Finding the number of red balls drawn before the first black ball is chosen

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 They are withdrawn from the urn, one at a time and without replacement.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).

• Note that you not only have to count the number of possibilities of drawing $k$ red balls, but you have to multiply the resulting probabilities with the number $k$ ($0\le k \le n$) – Peter Nov 2 '14 at 20:19
• Probably, some addition theorem with binomials leads to the desired result. – Peter Nov 2 '14 at 20:21
• What I would like to know is, why doesn't the expression for $E(X_i)$ somehow involve the value of $i$? – David K Nov 2 '14 at 20:28
• @David K: before the draw, each of the red balls is equivalent to the other red balls. And the numbering of the red balls has to take place before the draw for the question to make sense. So $E[X_i]$ cannot depend on $i$. – Henry Nov 3 '14 at 0:21
• It appears that I misread the phrase "if $i$ red ball is taken before any black ball". I read it as "if $i$ red balls are taken before any black ball". If it means "if red ball number $i$ is taken before any black ball" then indeed $E(X_i)$ would be independent of $i$. It also fits the given solution, so I agree your reading is correct. – David K Nov 3 '14 at 2:37

## 1 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 $\sum_i E[X_i] = \sum_i \frac{1}{1+m} = \frac{n}{1+m}$.

• Ohhh. That makes sense. You are right the ambiguity of the indicator was giving me trouble. – Undergradstudent Nov 3 '14 at 1:31