Q.7.12 - A First Course in Probability

Question

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. 

Step-by-Step Solution

Answer

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

b). The group is randomly seated at a round table is $\raisebox{1ex}{$3n^2$}\!\left/ \!\raisebox{-1ex}{$4n-2$}\right.$.

Step 1: Given Information (Part a)

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

Step 2: Explanation (Part a)

Let $X_i$ be such that:

$X_i=\left\{\begin{array}{l}1, if male s\tan ding besides atleast one female\\ 0, otherwise \end{array}\right.$

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

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

$=1$

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

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

Step 3: Explanation (Part a)

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

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

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,

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

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

Step 4: Final Answer (part a)

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

Step 5: Given Information (Part b)

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

Step 6: Explanation (Part b)

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

Some other positions,

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

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

Step 7: Final Answer (Part b)

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