91Ó°ÊÓ

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$=?$

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

$=\frac{10×10}{190}$

$=\frac{10}{19}âˆ\P Äi=1,2,…,10$

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

And $\mathrm{E}\left[{\mathrm{X}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{j}}\right]=1·\mathrm{E}\left[{\mathrm{X}}_{\mathrm{j}}=1âˆ\pounds {\mathrm{X}}_{\mathrm{i}}=1\right]·\mathrm{P}\left[{\mathrm{X}}_{\mathrm{i}}=1\right]âˆ\P Äi≠j=1,2,….,10$

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

$=\frac{9}{17}×\frac{10}{19}$

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

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

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

$=0.001629$

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

$∴E\left[X\right]=10·\frac{10}{19}$

$=\frac{100}{19}$

Calculate the total number of possible pairs,

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

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

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

$=\frac{1530}{6137}$

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

$=\frac{16200}{6137}$

Now suppose that 20 people consisted of 10 married couples.

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

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

$∴E\left(X\right)=\frac{1}{19}×10$

$=\frac{10}{19}$

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

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

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

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

$=\frac{19-17}{19×19×17}$

$=\frac{2}{6137}$

$Var\left(X_i\right)=E\left[X_i^2\right]-{\left(E\left[X_i\right]\right)}^2âˆ\P Ä\mathrm{i}=1,2,….,10$

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

$=\frac{19-1}{19×19}$

$=\frac{18}{19×19}$

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

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

$=\frac{3240}{6137}$

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$\frac{3240}{6137}.$