# Expected number of red balls removed from an urn before the first black ball

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 They are withdrawn from the urn one at a time and without replacement.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. We are interested in determining the $E[X]$. To obtain this quantity, number the red balls from 1 to n. Define the random variables $X_i$ $(i=1,2...,n)$ by

$$X_i = \begin{cases} 1 \quad & : \text{if red ball labeled (i) is taken before any black ball is chosen}\\ 0 \quad & : \text{otherwise} \end{cases}$$

Express $X$ in terms of $X_i$s and find $E[X]$.

### My Attempt:

Now we can express $X$ as $X = X_1+X_2+...+X_n$

To find the $X_i's$, what I think is it not geometric, but it is more like this $$\begin{array}{c|c|c|c} X & 1 & 2 & \ldots \\ \hline P(X) & \frac{m}{n+m} & (\frac{n}{n+m})(\frac{m}{(n+m)-1}) & \ldots \end{array}$$ Now to determine the random variables, I tried to think about this step-by step. For $X_1$, it means that is the red ball labeled 1 is picked before any black ball is chosen, and so on for $X_2, X_3,..,X_N$. What I think is to determine the random variables, it is like a permutation going on since the balls are numbered, but it's not. There is only 1 ball in the urn which is red and is labeled #1. There is only 1 ball which is red and is labeled #2, etc.

The question is that am I thinking about this the right way? Again, I really apologize about the organization. It is my first time on this site and I am trying to get familiar with the programming. Thank you for all of your help!

Think of all the balls as distinguishable, and imagine taking out the balls one at a time until they are all gone. Then all sequences of balls are equally likely. Temporarily, let us call Red Ball $i$ and all the black balls special. Each of the $m+1$ special balls is equally likely to be the first special ball drawn. It follows that the probability Red Ball $i$ is drawn before any black is $\frac{1}{m+1}$.

The Bernoulli random variable $X_i$ takes on value $1$ with probability $\frac{1}{m+1}$, and therefore $E(X_i)=\frac{1}{m+1}$.

By the linearity of expectation we then have $E(X)=E(X_1)+\cdots+E(X_n)=\frac{n}{m+1}$.

Remark: In the second part of the post, you were going after the distribution of the random variable $X$. So let us do that. There are $\binom{n+m}{m}$ equally likely ways to choose the positions of the $m$ black balls. Now we count the number of ways to choose positions of the black balls if the first $k$ positions are to be black and the next one red. That leaves $n+m-k-1$ positions, and $m-k$ black. Positions for them can be chosen in $\binom{n+m-k-1}{m-k}$ ways, or equivalently $\binom{n+m-k-1}{n-1}$ ways. It follows that $$\Pr(X=k)=\frac{\binom{n+m-k-1}{n-1}}{\binom{m+n}{m}}\tag{1}$$ for $k=0,1,\dots, m$.

Now we could use (1) to write down an expression for $E(X)$. With some fooling around with binomial coefficients, we could after a while simplify this to $\frac{n}{m+1}$. However, that is a painful way to find $E(X)$. The method of Indicator Random Variables described in the first part of the OP is far easier.

• Okay. The question I have is that in the urn, there are m+n balls. For the random variables, $X_i$, the probability is $\frac{1}{m+1}$. I still do not understand why that is? Thank you for all of your help. – user211962 Feb 1 '15 at 15:20
• OK, let's do it in a lengthy way. (It is done in a short way in the answer.) We want $\Pr(X_i=1)$, the probability Red Ball $i$ is chosen before any black. There are $(n+m)!$ permutations of the balls. We now count the favourables. Call Red Ball $i$ and the blacks special. There are $m+1$ special balls. There are $\binom{n+m}{m+1}$ equally likely ways to choose the locations reserved for special balls (we have not yet chosen which ball goes where). [to be continued] – André Nicolas Feb 1 '15 at 15:56
• [continued] For every such choice, there is $1$ way to have Red Ball $i$ before all the black, and then the black can be arranged in $m!$ ways, and the $n-1$ non-special balls can be arranged in $(n-1)!$ ways, for a total of $\binom{n+m}{m+1}m!(n-1)!$ "favourable" ways. This is $\frac{(n+m)!}{(n-1)!(m+1)!}\cdot m!(n-1)!$. Divide by $(n+m)!$ and simplify. We get $\frac{1}{m+1}$. – André Nicolas Feb 1 '15 at 16:00