another problem with covariance

Question

If 10 married couples are randomly seated at a round table, compute

(a) the expected number and

(b) the variance of the number of wives who are seated next to their husbands.

(a)Let $X_i$ be a random variable that is 1 if the i-th wife is seated next to her husband, 0 otherwise.

$X=\sum_{i=1}^{10} X_i$= number of wives seated next to their husband.

$E[X]=\sum_{i=1}^{10} E[X_i]=10* \frac{2}{19}=\frac{20}{19}$ with $E[X_i]=\frac{2*(20-2)}{(20-1)(20-2)}=\frac{2}{19}$

(b) $$\begin{align} Var (X) &=Var (\sum_{i=1}^NX_i)\\ &=\sum_{i=1}^N Var (X_i)+2*\sum_{}^ {}\sum_{i

Now I don't know how to calculate $E[X_i X_j]$

for $i

$$\begin{align}E[X_i X_j] &= P(X_i=1,X_j=1)\\ &=P(X_i)P(X_j=1|X_i=1)\\ &=\frac{2}{19}*???\end{align}$$

in particular how to calculate $P(X_j=1|X_i=1)$ the probability that the j-th wife has her husband next to her if the i-th wife has her husband next to her

I've tried and this is my attempt: $$\frac{2*(17-2)}{(17-1)*(17-2)}+\frac{1}{17}=\frac{2}{16}+\frac{1}{17}$$ but the solution in the book is $\frac{2}{18}$

Answer 1

$$P(X_j=1 \mid X_i=1) = \frac{17 \cdot 2}{18 \cdot 17} = \frac{1}{9}.$$

After the $i$th couple sits together, there are $18$ remaining seats which you can think of as a row (with two ends). There are $17$ pairs of adjacent seats to choose from, and $2$ ways for the couple to sit in any given pair. There are $18 \cdot 17$ total ways for them to sit in two of the remaining $18$ seats.