# Removing balls from an urn in pairs: expected number of pairs in which both are red

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.

Question:

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 $n(n-1)$.

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: