Suppose in Self-Test Problem $7.3$that the $20$people are to be seated at seven tables, three of which have $4$ seats and four of which have $2$ seats. If the people are randomly seated, find the expected value of the number of married couples that are seated at the same table.

The $20$people are to be seated at seven tables, three of which have $4$seats and four of which have $2$ seats.

Let X represents the number of married couples that are seated at the same table, and let's define indicator variables I_j as:

$I_j=\{\begin{array}{ll}1,& \text{if}{E}_j\text{occurs}\\ 0,& \text{if}{E}_j\text{does not occur}\end{array}$

$\text{whereby}{E}_j,j=1,2,\dots ,10\text{, denote the event:}$

$E_j="j\text{th married couple is at the same table "}.$

Then,

$X=\sum _{j=1}^{10}{I}_j$

and therefore the expected number of married couples that are seated at the same table is

$E\left[X\right]=E\left[\sum _{j=1}^{10}{I}_j\right]=\sum _{j=1}^{10}E\left[I_j\right]=\sum _{j=1}^{10}P\left\{E_j\right\}\cdot (\ast )$

Consider the next events:

${W_j}^i=$" Woman from j th married couples is at i th table"

${M^i}_j=$" Man from j th married couples is at i th table"

whereby, without loss of generality, we assume that the $1$st, $2$nd and $3$rd tables consist of $4$seats and the $4$th, $5$th, $6$th and $7$ th tables consist of $2$ seats.

Assume that the seating is done at random. Then ,

$P\left\{E_j\right\}=P\{"j\text{th married couple is at 1st table"}\}$

$+\cdots +P\{"j\text{th married couple is at}7\text{th table}"\}=$

$P\left\{W_j^1\right\}P\{M_j^1\mid W_j^1\}+P\left\{W_j^2\right\}P\{M_j^2\mid W_j^2\}+P\left\{W_j^3\right\}P\{M_j^3\mid W_j^3\}+$

$P\left\{W_j^4\right\}P\{M_j^4\mid W_j^4\}+P\left\{W_j^5\right\}P\{M_j^5\mid W_j^5\}+P\left\{W_j^6\right\}P\{M_j^6\mid W_j^6\}+$

$P\left\{W_j^7\right\}P\{M_j^7\mid W_j^7\}=$

$\frac{4}{20}\left(\frac{3}{19}\right)+\frac{4}{20}\left(\frac{3}{19}\right)+\frac{4}{20}\left(\frac{3}{19}\right)+$

$\frac{2}{20}\left(\frac{1}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)=\frac{11}{95}$

$\text{and therefore according to}(*)\text{we get:}$

$E\left[X\right]=10\left(\frac{11}{95}\right)=\frac{\mathbf{22}}{\mathbf{19}}$

If the people are randomly seated, the expected value of the number of married couples that are seated at the same table is$\frac{22}{19}$.