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?

  • $\begingroup$ 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? $\endgroup$
    – Ken Duna
    Jun 23, 2016 at 20:14

3 Answers 3


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 $\frac{45}{49}$


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