I am sorry that I cannot make the title more clear. The following is from Sheldon M. Ross: Introduction to Probability Models (11th Edition). I am able to reach the desired answer but actually I don't fully understand it. Please help.


An urn contains $2n$ balls, of which $r$ are red. The balls are randomly removed in $n$ successive pairs. Let $X$ denote the number of pairs in which both balls are red. Find $E[X]$ and Var$(X)$.

I then denote $X_i$ be a random variable which equals to $1$ if the $i$-th pair which $2$ red balls and $0$ if otherwise. Then, we have $$ E[X] = \sum_{i=1}^n {E[X_i]} = \sum_{i=1}^n{\frac{r(r-1)}{2n(2n-1)}}=\frac{r(r-1)}{2(2n-1)}$$ which is the correct answer.

But then I am not satisfied by the answer: why all $E[X_i], i=1,...,n$, are of the same value? Say, I am calculating $E[X_2]$. Then I should have $$E[X_2] = (1)P(X_2 = 1) +(0)P(X_2 = 0).$$ But shouldn't $P(X_2 = 1)$ depend on the outcome of the first pair?

On the other hand, I calculate Var$(X)$. The suggested solution gives the following: $$Var(X) = \sum_{i=1}^n Var(X_i) + 2 \sum \sum_{i<j} Cov(X_i,X_j)=(n)Var(X_1)+(n)(n-1)Cov(X_1,X_2).$$ I am not sure about the last equality, especially the existence of $n(n-1)$.

Thanks in advance!

  • 3
    $\begingroup$ Because each pair is made of two uniformly chosen balls, whether this is the first pair to be removed or any other pair. $\endgroup$
    – Did
    Oct 5, 2014 at 12:43

1 Answer 1


As Didier pointed out in a comment, you can see that the probabilities are the same for each pair by noting that each pair contains two uniformly chosen balls, that is, since the balls are by definition identical (except for the colour, which doesn't influence their selection), in a given draw every pair of balls is equally likely to be drawn, and this determines all probabilities related to a single draw.

Regarding the variance, your calculation is correct (the factor $\binom n2$ is the number of pairs in the summation over the covariances), but I find it easier to calculate variances using expectation values, like this:

\begin{align} \operatorname{Var}(X)&=E[X^2]-E[X]^2\\ &=nE[X_1^2]+n(n-1)E[X_1X_2]-(nE[X_1])^2\\ &=nE[X_1]+n(n-1)E[X_1X_2]-(nE[X_1])^2\\ &=n\frac{\binom r2}{\binom{2n}2}+n(n-1)\frac{\binom r4}{\binom{2n}4}-n^2\left(\frac{\binom r2}{\binom{2n}2}\right)^2\;. \end{align}

  • $\begingroup$ How did you go from first line to the second? I mean $E[X^2] = E[(\sum_{i=1}^n X_i)^2]$. I don't fully understand why all the $X_i$ are identical, I mean if you picked red balls then it should the future $X_i$. But even taking that as granted, how does $X_1$ and $X_2$ appear? $\endgroup$
    – Aditya P
    Sep 24, 2019 at 14:08
  • $\begingroup$ @Aditya: Yes, if you picked red balls, it influences the future $X_i$, but that just means that the conditional quantities (probabilities and expectations) are different. The unconditional quantities (when we haven't picked any red balls yet) are the same. I tried to explain this in the first paragraph; if it's still not clear, please indicate what part of that explanation is unclear. About $X_1$ and $X_2$: The squared sum contains $n$ terms of the form $X_i^2$ and $n(n-1)$ terms of the form $X_iX_j$ with $i\ne j$, and these are all equal in expectation to $X_1^2$ and $X_1X_2$, respectively. $\endgroup$
    – joriki
    Sep 28, 2019 at 11:53

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