Q.7.42 - A First Course in Probability

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.

Step-by-Step Solution

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 1: 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$=?$

Step 2: 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 $E\left(X_i\right)=1.P\left[X_i=1\right]$

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

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

Step 3: 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.

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

$=\frac{100}{19}$

Step 4: 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 $=\left(\begin{array}{l}20\\ 2\end{array}\right)=190$

Step 5: 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.

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

$=\frac{10}{19}$

Step 6: Explanation

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

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

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

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

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

$=\frac{2}{6137}$

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

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

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

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

Step 7: Explanation

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

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

$=\frac{3240}{6137}$

Step 8: 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 $\frac{3240}{6137}.$