A group of $n = 10$ men exchange their gloves randomly (altogether there are 20 gloves). Let $X_i$ be the random variable which attains the value 1 if the $i-th$ man got at least one of his gloves back and it attains the value 0 otherwise. Set $p = E(X_i)$ and $σ^2 = Var(X_i)$.
Observe that $p = \frac{37}{190}$ and that $Var (X_i) = \frac{5661}{190^2}$. Let $b$ be the probability that for $i\ne j$ both the $i-th$ and $j-th$ men got at least one of their gloves back.
Show that $b = \frac {1157}{{20 \choose 2}{18 \choose 2}}$.
1.) Let $X$ be the total number of men who received at least one of their gloves. Find $E(X)$.
2.) Calculate $Var(X).$ You may write it as an expression in $p$, $σ^2$ and $b.$
At first I just thought that $E(X)=np=10\frac{37}{190}$ and then $Var(X)=npq=10\cdot 37\cdot\frac{153}{190^2}$ but now I am not sure about that, as the events do intersect: if someone takes gloves of someone else, then someone else has nothing to choose from. However, I do not know how to count these. Could you help me, please? Thanks
Your calculations of $p$, $\sigma^2$ and $\mathbb EX$ are okay.
Covariance is bilinear, so that by symmetry:
$$\mathsf{Var}(X)=\mathsf{Cov}(\sum_{i=1}^{10}X_i,\sum_{j=1}^{10}X_j)=\sum_{i=1}^{10}\sum_{j=1}^{10}\mathsf{Cov}(X_i,X_j)=10\mathsf{Var}(X_1)+90\mathsf{Cov}(X_1,X_2)=$$$$10\sigma^2+90(\mathbb EX_1X_2-\mathbb EX_1\mathbb EX_2)=10\sigma^{10}+90(P(X_1=X_2=1)-P(X_1=1)P(X_2=1))=$$$$10\sigma^2+90(b-p^2)$$
Here $b=P(X_1=X_2=1)$ so it remains to find $P(X_1=X_2=1)$
edit:
Let $Z_{1}$ denote the number of gloves taken by person 1 that belong to person 1.
Let $Y_{1}$ denote the number of gloves taken by person 1 that belong to person 2.
Then:
$$P\left(X_{1}=X_{2}=1\right)=$$$$P\left(X_{2}=1\mid Z_{1}=2\right)P\left(Z_{1}=2\right)+$$$$P\left(X_{2}=1\mid Z_{1}=1\wedge Y_{1}=0\right)P\left(Z_{1}=1\wedge Y_{1}=0\right)+$$$$P\left(X_{2}=1\mid Z_{1}=1\wedge Y_{1}=1\right)P\left(Z_{1}=1\wedge Y_{1}=1\right)$$
This with
$P\left(X_{2}=1\mid Z_{1}=2\right)=1-P\left(X_{2}=0\mid Z_{1}=2\right)=1-\frac{16}{18}\frac{15}{17}$
$P\left(X_{2}=1\mid Z_{1}=1\wedge Y_{1}=0\right)=1-P\left(X_{2}=0\mid Z_{1}=1\wedge Y_{1}=0\right)=1-\frac{16}{18}\frac{15}{17}$
$P\left(X_{2}=1\mid Z_{1}=1\wedge Y_{1}=1\right)=1-P\left(X_{2}=0\mid Z_{1}=1\wedge Y_{1}=1\right)=1-\frac{17}{18}\frac{16}{17}$
$P\left(Z_{1}=2\right)=\frac{2}{20}\frac{1}{19}$
$P\left(Z_{1}=1\wedge Y_{1}=0\right)=\frac{2}{20}\frac{16}{19}+\frac{16}{20}\frac{2}{19}$
$P\left(Z_{1}=1\wedge Y_{1}=1\right)=\frac{2}{20}\frac{2}{19}+\frac{2}{20}\frac{2}{19}$