Q.7.34 - A First Course in Probability

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.

Step-by-Step Solution

Answer

a) The expected number is $\frac{20}{19}$.

b) The variance of the number of wives who are seated next to their husbands are $\frac{360}{361}$.

Step 1: Given Information (Part a)

10 married couples are randomly seated at a round table.

The expected number $=?$

Step 2: Explanation (Part a)

Let $X_i$ is the number of couples sits next to each other.

Let $X$ denotes the wives are seated next to their husbands is $X=\sum _{1-i}^{10}{X}_i$

a) Suppose the seating as the first seat wife $i$.

The couple $i$ are seated together.

Clearly there are 2 places to seat husband $i$ and $18!$Ways of seating.

If there is no restriction, there are $19 !$Ways of seating the other $19$people.

$E\left[X\right]=\sum _{i=1}^{10}iP\left(X_i\right)$

$=10\times \frac{2\times 18!}{19!}$

$=\frac{20}{19}$

Step 3: Final Answer (Part a)

Hence, the expected number is $\frac{20}{19}.$

Step 1: Given Information (Part b)

10 married couples are randomly seated at a round table.

The variance of the number of wives who are seated next to their husbands = ? 

Step 2: Explanation (Part b)

Note that $X_i$ are not independent

$V\left[X\right]=Cov\left(\sum _{i=1}^{10}{X}_i,\sum _{j=1}^{10}{X}_j\right)$

$=\sum _{j-1}^{10}\sum _{i-1}^{10}Cov\left(X_i,X_j\right)$

$=\sum _{i=1}^{10}Cov\left(X_i,X_j\right)+\sum _{i=1j}^{10}\sum _{j,j\ne i}^{10}Cov\left(X_i,X_j\right)$

$=\left(10\times \frac{2}{19}\times \frac{17}{19}\right)+10\times 9Cov\left(X_i,X_j\right)$

$\because$All of the $Cov\left(X_i,X_j\right)$ with $i\ne j$ are equal

Step 3: Explanation (Part b)

Calculate the variance of the number,

$\begin{array}{r}\mathrm{Cov}\left(X_1,X_2\right)=E\left(X_1{X}_2\right)-E\left(X_1\right)E\left(X_2\right)\end{array}$

$\begin{array}{r}=E\left(X_1{X}_2\right)-\frac{4}{{19}^2}\end{array}$

$\begin{array}{r}=P\left\{\text{ Couple }1\text{ is seated together and couple }2\text{ is seated together }\right\}-\frac{4}{{19}^2}\end{array}$

$\begin{array}{r}=\frac{4\times 17!}{19!}-\frac{4}{{19}^2}\end{array}$

$=\frac{4\times 19-4\times 18}{18\times {19}^2}$

$=\frac{2}{9\times {19}^2}$

And,

$\begin{array}{r}V\left[X\right]=\left(10\times \frac{2}{19}\times \frac{17}{19}\right)+\left(10\times 9\mathrm{Cov}\left(X_i,X_j\right)\right)\end{array}$

$=\left(10\times \frac{2}{19}\times \frac{17}{19}\right)+\left(10\times 9\left(\frac{2}{9\times {19}^2}\right)\right)$

$=\frac{360}{361}$

Step 4: Final Answer

The variance of the number of wives who are seated next to their husbands are  $\frac{360}{361}.$