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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

a) $\mathbb{E}(X)$
b) $\operatorname{Var}(X)$

Attempt to find answer: Let $X_i$ equal 1 if both balls of the $i^{th}$ withdrawn pairs are red, and let it equal 0 otherwise. Because

$\mathbb{E}[X_i]=\mathbb{P}[X_i=1]=\displaystyle\frac{r(r-1)}{2n(2n-1)}$

We have

$\mathbb{E}[X]=\displaystyle\sum_{i=1}^n \mathbb{E}[X_i]$

$\mathbb{E}[X]=\displaystyle\frac{r(r-1)}{4n-2}$

Question: Now how to compute its variance? I know

$\operatorname{Var}(X)=\displaystyle\sum_i\operatorname{Var}(X_i) + 2\displaystyle\sum_{i<j} \operatorname{Cov}(X_i,X_j)$

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    $\begingroup$ Please explain the distinction you appear to be drawing between "mean" and "expected mean" and between "variance" and "expected variance" in the title. $\endgroup$ – whuber Mar 25 '17 at 18:18
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    $\begingroup$ @whuber I have edited the title. $\endgroup$ – Dhamnekar Winod Mar 26 '17 at 2:33
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Answer to b)

Let us calculate $E[X_iX_j]$

$E[X_iX_j]=P[X_i=1,X_j=1]$

$E[X_iX_j]=P[X_i=1,X_j=1|X_i=1]$

$E[X_iX_j]=\frac{r(r-1)(r-2)(r-3)}{2n(2n-1)(2n-2)(2n-3)}$

For $Var(X)$ we use

$Var(X)=\displaystyle\sum_iVar(X_i) + 2\displaystyle\sum_{i<j}Cov(X_iX_j)$

$Var(X)=nVar(X_1)+ n(n-1)Cov(X_iX_j)$

Where because $X_i$ is a Bernoulli random variable

$Var(X_i)=E(X_1)(1-E[X_1])$

$Cov(X_1X_2)=\frac{r(r-1)(r-2)(r-3)}{2n(2n-1)(2n-2)(2n-3)}-(E[X_1])^2$

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