Q. 7.11 - A First Course in Probability

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.

Step-by-Step Solution

Answer

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

Step 1: Given Information

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

Step 2: Explanation

The indicator variables, whereby,

$E_j,j=1,2,\dots ,10$ denotes the events,

$E_j="j$  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

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

Step 3: Explanation

$W_j^i="$ woman from $j$  married couple is at $i$  table " ,

 $M_j^i="$ man from $j$ married couple is at $i$  table "

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

Assume that the seating is done at random. Then,

 $P\left\{E_j\right\}=P\{"j\mathrm{th}$ married couple is at $1$ table " } $+\cdots +P\{"j$ married couple is at $7$  table "} .

Step 4: Explanation

Function,

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

$P\left\{W_j^4\right\}P\left\{M_j^4\mid W_j^4\right\}+P\left\{W_j^5\right\}P\left\{M_j^5\mid W_j^5\right\}+P\left\{W_j^6\right\}P\left\{M_j^6\mid W_j^6\right\}+P\left\{W_j^7\right\}P\left\{M_j^7\mid W_j^7\right\}=$

Stepping,

$\frac{4}{20}\left(\frac{3}{19}\right)+\frac{4}{20}\left(\frac{3}{19}\right)+\frac{4}{20}\left(\frac{3}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)+\frac{2}{20}\left(\frac{1}{19}\right)=\frac{11}{95}$

and therefore according to $(*)$ we get:

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

Step 5: Final answer

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