91Ó°ÊÓ

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

The indicator variables, whereby,

$E_j,j=1,2,…,10$denotes the events,

$E_j="j$ married couple is at the same table.

Then, $X=\underset{j=1}{\overset{10}{∑}}{I}_j$

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

$E\left[X\right]=E\left[\underset{j=1}{\overset{10}{∑}}{I}_j\right]=\underset{j=1}{\overset{10}{∑}}E\left[I_j\right]=\underset{j=1}{\overset{10}{∑}}P\left\{E_j\right\}·(*)$

$W_j^i="$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 " } $+⋯+P\{"j$married couple is at $7$ table "} .

Function,

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

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

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:

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

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