A group of $n$men and $n$ women is lined up at random.

(a) Find the expected number of men who have a woman next to them.

(b) Repeat part (a), but now assuming that the group is randomly seated at a round table.

A group of $n$ men and $n$ women is lined up at random.

Let $X_i$be such that:

$X_i=\left\{\begin{array}{l}1,ifmales\tan dingbesidesatleastonefemale\\ 0,otherwise\end{array}\right.$

$E\left[\sum _{i=1}^{2n}{X}_i\right]=\sum _{i=1}^{2n}E\left[X_i\right]$

$=\sum _{i=1}^{2n}P\left(X_i\right)$

$=1$

Now, $X_1=1$(and $X_{2n}=1$) only when $X_2$female (and $X_{2n-1}$female,)

$P\left(X_1\right)=\left\{\begin{array}{l}\mathrm{n}·\mathrm{n},\text{no of ways to select}1\text{male and}1\text{female}\\ 2\mathrm{n}(2\mathrm{n}/1),\text{no of ways to select}2\text{people of}2\mathrm{n}\end{array}\right.$

$P\left(X_{2n}=1\right)=P\left(X_1=1\right)$

$=\frac{n}{2(2n-1)}$

Now, $P\left(X_i=1\right)$for $iϵ[2,2n-1]$this will happen only when $(i-1{)}^{th},i^{th},(i+1{)}^{th}$people are:

• female-male-male
• male-male-female
• female-male-female

$P\left(X_i=1\right)=\frac{3n^2(n-1)}{2n(2n-1)(2n-2)}$

$=\frac{3n}{4(2n-1)}$

The expected number of men who have a woman next to them,

$=\frac{n}{2n-1}+\frac{3n(2n-2)}{4(2n-1)}$

$=\frac{3n^2-n}{4n-2}$

The expected number of men who have a woman next to them is$\raisebox{1ex}{$3n^2-n$}\!\left/ \!\raisebox{-1ex}{$$}\right.4n-2$.

A group of$n$men and$n$women is lined up at random.

The event $X_i$will be some as before just that the probabilities for $X_1=1$and $X_{2n}=1$,

Some other positions,

$P\left(X_i=1\right)=\frac{3n}{4(2n-1)}$

$2n\frac{3n}{4(2n-1)}=\frac{3n^2}{4n-2}$

The group is randomly selected at a round table is$\frac{3n^2}{4n-2}$.