# Expected number of married couples chosen out of 50 different people

I've encountered this problem, and would like to know if my approach is right.

We select 10 people out of a group of 25 married couples, what is the expected number of married couples chosen?

Well, I've defined the following indicator:

$$X_i = \begin{cases} \text{1, if couple i was chosen } \\ \text{0, otherwise } \end{cases}$$

Then, I computed the probability of couple $i$ to be selected, by looking at the husband $i$:

The sample space is obviously $\binom{49}{9}$, the options to choose wife $i$ is $\binom{1}{1}$, and the combinations for the rest of the 8 people is $\binom{48}{8}$.

All together:

$$\mathbb{P}(X_i=1) = \frac{\displaystyle\binom{1}{1} \cdot \binom{48}{8}}{\displaystyle\binom{49}{9}} = \frac{9}{49}$$

Then, we want to sum the probabilities for the 5 chosen "couples":

$$\sum_{i=1}^{5} X_i = X_1+X_2+X_3+X_4+X_5 = 5 \cdot \frac{9}{49} = \frac{45}{49}$$

Am I on the right path or totally lost it?

• Here are we assuming that each couple is a hetero couple? If so, are you choosing $5$ random men and $5$ random women? Or are you choosing $10$ random people? Jun 23, 2016 at 20:14

Yet another way of getting the same answer:

Let $X_i$ for $1\leq i\leq 10$ be the indicator random variable that the $i$'th person is male and his wife is among the other nine chosen people. (We specifically want the male so that we do not accidentally overcount the couples.)

$Pr(X_i=1)=\frac{1}{2}\cdot \frac{9}{49}$

The total number of married couples present is $X=X_1+X_2+\dots+X_{10}$

The expected number is then $E[X]=\sum\limits_{i=1}^{10}E[X_i]=10\cdot \frac{1}{2}\cdot \frac{9}{49}=\frac{45}{49}$

Yes, an indicator random variable argument will do the job. Here is one version.

Let $Y_i=1$ if Couple $i$ is chosen, and $0$ otherwise. If $W$ is the total number of couples chosen, then $W=Y_1+\cdots +Y_{25}$, so by the linearity of expectation we have $E(W)=25E(Y_1)=25\Pr(Y_1=1)$.

So we need to find the probability Couple $1$ is chosen. There are $\binom{50}{10}$ ways to choose $10$ people. We assume that these are equally likely. There are $\binom{48}{8}$ ways to choose $10$ people including Couple $1$. Thus $\Pr(Y_1=1)=\frac{\binom{48}{8}}{\binom{50}{10}}$, and we now multiply by $25$.

There is substantial simplification, and we end up with $\frac{45}{49}$.

If I choose any two people at random, what is the chance that I choose a married couple.

$\frac 1{49}$

If we choose $10$ people, that makes for $45$ ways to match any pair of the $10$ chosen. Each pair has a $\frac 1{49}$ of being a couple.

The expected number of couples is $\frac{45}{49}$