Question

Twenty individuals consisting of $10$married couples are to be seated at $5$different tables, with $4$people at each table.

(a) If the seating is done“at random,” what is the expected number of married couples that are seated at the same table?

(b) If $2$men and $2$women are randomly chosen to be seated at each table, what is the expected number of married couples that are seated at the same table?

Short answer

According to the information,

a)If the seating is done“at random,” the expected number of married couples that are seated at the same table = $\frac{30}{19}$$\frac{30}{19}$

b) If $2$men and $2$women are randomly chosen to be seated at each table, the expected number of married couples that are seated at the same table is$2$

Step-by-step solution

Given Information (part a)

If the seating is done“at random,” what is the expected number of married couples that are seated at the same table

Explanation (part a)

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

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

Whereby $E_{j,}j=1,2,....,10,$denote the event

$E_j=$"j the 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,

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

Step 3: Explanation (part a)

Consider the next events:

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

${M_j}^i=$" Man from j th married couple is at i th table".

Assume that the seating is done at random. Since there are 5 different tables, with 4 seats at each table, we have:

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

$+\cdots +P\{"j\text{th married couple is at}5\text{th table}"\}=$

$P\left\{W_j^1\right\}P\{M_j^1\mid W_j^1\}+P\left\{W_j^2\right\}P\{M_j^2\mid W_j^2\}+\cdots +P\left\{W_j^5\right\}P\{M_j^5\mid W_j^5\}=$

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

$\text{According to}(\ast )\text{we get:}$

$E\left[X\right]=10\left(\frac{3}{19}\right)=\frac{\mathbf{30}}{\mathbf{19}}$

 Final Answer (part a)

If the seating is done“at random,” the expected number of married couples that are seated at the same table =$\frac{30}{19}$

Given Information (part b)

If $2$ men and $2$ women are randomly chosen to be seated at each table, what is the expected number of married couples that are seated at the same table

Explanation (part b)

Now, assume that $2$men and $2$women are randomly chosen to be seated at each table. We will consider $2$men and $2$women as two different objects. Therefore, with this consideration, instead of $20$spots, we now have $10$spots at $5$different tables, $2$spots at each table. Therefore, since we have $5$'pairs' of $2$men and $5$'pairs' of $2$women, we get:

$P\left\{E_j\right\}=P\{"j\text{th married couple is at 1st table}"\}$$+\cdots +P\{"j\text{th married couple is at}5\text{th table}"\}=$

$P\left\{W_j^1\right\}P\{M_j^1\mid W_j^1\}+P\left\{W_j^2\right\}P\{M_j^2\mid W_j^2\}+\cdots +P\left\{W_j^5\right\}P\{M_j^5\mid W_j^5\}=$

$\frac{2}{10}\left(\frac{1}{5}\right)+\frac{2}{10}\left(\frac{1}{5}\right)+\cdots +\frac{2}{10}\left(\frac{1}{5}\right)=\frac{1}{5}$

$\text{According to}(\ast )\text{weobtain:}$

$E\left[X\right]=10\left(\frac{1}{5}\right)=\mathbf{2}$

Final Answer (part b)

If $2$men and $2$women are randomly chosen to be seated at each table, the expected number of married couples that are seated at the same table is $2$