Question

A group of 20 people consisting of 10 men and 10 women is randomly arranged into 10 pairs of 2 each. Compute the expectation and variance of the number of pairs that consist of a man and a woman. Now suppose the 20 people consist of 10 married couples. Compute the mean and variance of the number of married couples that are paired together.

Short answer

The expectation and variance of the number of pairs that consist of a man and a woman is $\frac{10}{19}$

The mean of the number of married couples that are paired together is $E\left[\sum _{i=1}^{10} X_i\right]=\frac{10}{19}$

The variance of the number of married couples that are paired together is$\frac{3240}{6137}$

Step-by-step solution

Given Information

Group of people$=$10 men and 10 women are randomly arranged into 10 pairs of 2 each.

Compute the expectation and variance of the number of pairs that consist of a man and a woman$=?$

Number of married couples in 20 people $=10$

Compute the mean and variance of the number of married couples that are paired together$=?$

Explanation

If a group of 20 people consisting of 10 Men and 10 women is randomly arranged into 10 pairs of 2 each then the total number of possible pairs are,

$\left(\begin{array}{l}20\\ 2\end{array}\right)=190$

Let$X_i=\left\{\begin{array}{l}1\text{If}{i}^{\text{th}}\text{pair consist of a man and a woman}\\ 0\text{other wise}\end{array}\right.$

Let$E\left(X_i\right)=1.P\left[X_i=1\right]$

$=\frac{10\times 10}{190}$

$=\frac{10}{19}\forall i=1,2,\dots ,10$

Explanation

Calculate the value of $\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{j}}\right],$

And $\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{j}}\right]=1·\mathrm{E}\left[{\mathrm{X}}_{\mathrm{j}}=1\mid {\mathrm{X}}_{\mathrm{i}}=1\right]·\mathrm{P}\left[{\mathrm{X}}_{\mathrm{i}}=1\right]\forall i\ne j=1,2,\dots .,10$

$=\frac{9\times 9}{\left(\begin{array}{c}18\\ 2\end{array}\right)}·\frac{10}{19}$

$=\frac{9}{17}\times \frac{10}{19}$

Calculate the value of $Cov\left(X_i,X_j\right)$,

$\therefore Cov\left(X_i,X_j\right)=E\left[X_i{X}_j\right]-E\left[X_i\right]·E\left[X_j\right]$

$=\frac{90}{19\times 17}-\frac{100}{(19{)}^2}$

$=0.001629$

Now $\mathrm{X}={\mathrm{X}}_1+{\mathrm{X}}_2+\dots +{\mathrm{X}}_{10}=$Total number of pairs that consist of a man and a woman.

$\therefore E\left[X\right]=10·\frac{10}{19}$

$=\frac{100}{19}$

Explanation

Calculate the total number of possible pairs,

$Var\left(X\right)=10·Var\left(X_i\right)+10·9·Cov\left(X_i,X_j\right)$

$Var\left(X_i\right)=E\left[X_i^2\right]-{\left(E\left(X_i\right)\right)}^2$

$=\left(\frac{10}{19}\right)-{\left(\frac{10}{19}\right)}^2$

$=\frac{1530}{6137}$

$\Rightarrow Var\left(X\right)=10·\frac{1530}{6137}+\left(9\right)\frac{10.10}{6137}$

$=\frac{16200}{6137}$

Now suppose that 20 people consisted of 10 married couples.

Total number of possible pairs$=\left(\begin{array}{l}20\\ 2\end{array}\right)=190$

Explanation

Let ${\mathrm{X}}_{\mathrm{i}}=\left\{\begin{array}{l}1{\text{If}}^{\text{ith}}\text{pair consist of married couple}\\ 0\text{other wise}\end{array}\right.$

Let $\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}}\right]=1.\mathrm{P}\left[{\mathrm{X}}_{\mathrm{i}}=1\right]$

$=1·\frac{\left(\begin{array}{l}10\\ 1\end{array}\right)}{\left(\begin{array}{l}20\\ 2\end{array}\right)}$

$=\frac{1}{19}$

Let $X=X_1+X_2+\dots \dots +X_{10}=$Total of number of married couple that are paired together.

$\therefore E\left(X\right)=\frac{1}{19}\times 10$

$=\frac{10}{19}$

Explanation

Now $Var\left(\mathrm{X}\right)=10·Var\left({\mathrm{X}}_{\mathrm{i}}\right)+\left(10\right)\left(9\right)·Cov\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)$

$Cov\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)=\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right]-\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}}\right]·\mathrm{E}\left[{\mathrm{X}}_{\mathrm{j}}\right];i\ne j=1,2,\dots .,10$

$=\frac{1}{19}·\frac{9}{\left(\begin{array}{l}18\\ 2\end{array}\right)}-\left(\frac{1}{19}\right)\left(\frac{1}{19}\right)$

$=\frac{1}{19}·\frac{1}{17}-\frac{1}{19\times 19}$

$=\frac{19-17}{19\times 19\times 17}$

$=\frac{2}{6137}$

$Var\left(X_i\right)=E\left[X_i^2\right]-{\left(E\left[X_i\right]\right)}^2\forall \mathrm{i}=1,2,\dots .,10$

$=\frac{1}{19}-\frac{1}{(19{)}^2}$

$=\frac{19-1}{19\times 19}$

$=\frac{18}{19\times 19}$

Explanation

Calculate the value of $Var\left(X\right),$

$\therefore Var\left(\mathrm{X}\right)=10\times \frac{18}{19\times 19}+90\times \frac{2}{6137}$

$=\frac{3240}{6137}$

Final Answer

Therefore, the The expectation and variance of the number of pairs that consist of a man and a woman is $\frac{10}{19}$.

The mean and variance of the number of married couples that are paired together is$\frac{3240}{6137}.$