How many ways can a couple be among $15$ people chosen from $60$? Suppose $30$ couples attend a gathering. The $60$ people all participate in a raffle and $15$ winners are chosen at random. On average how many couples will there be such that both members of the couple are among the winners?
I used the method of indicators but I can't find the probability that both members of the i-th couple are among the winners. I know the denominator is $\binom{60}{15}$ but I can't get the number of ways that the i-th couple can be among the $15$ winners chosen.
 A: We have to choose $13$ people from the remaining $58$. So the numerator is $\binom{58}{13}$.
Then the method of indicator random variables shows that the mean is $30\cdot\dfrac{\binom{58}{13}}{\binom{60}{15}}$.
A: For a given number $k$ of couples who are entirely in the $15$, you have
$${30\choose k}{30-k\choose 15-2k}2^{15-2k}$$
possible ways to select the winners. The first "choose" is which couples won together. The second is which couples only had one winner. The doubling comes from choosing the person in each couple with only one winner.
From there, since you asked two different questions, you can find the probability of any number of couples from $0$ to $7$ by substituting the value into the expression above and dividing by the total number of selections without consideration of the winning couples: $60\choose15$
The average number of couples winning together is then found as the expected value. That is, multiply the probability by the number of couples for that probability, then sum them.
For the probabilities:
$$\begin{array}{lr}k&P(k)\\0&\frac{98304}{102873}\\1&\frac{46080}{146969}\\2&\frac{898560}{2498473}\\3&\frac{457600}{2498473}\\4&\frac{2059200}{47470987}\\5&\frac{216216}{47470987}\\6&\frac{8580}{47470987}\\7&\frac{585}{332296909}\end{array}$$
And the expected value is $\frac{105}{59}\approx1.7797$
