Question

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.

Short answer

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

Step-by-step solution

Given Information

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

Explanation

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

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

Explanation

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.

Explanation

Assume that the seating is done at random. Then ,

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

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

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

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

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

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

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

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

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

Final Answer

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