TITLE: Average number of couples in dance competition finale
QUESTION [6 upvotes]: 11 pairs of dancers will participate in a dance competition finale. However, on the day of the competition, 7 dancers were tested Covid positive and had to be eliminated.
Given that partners aren’t interchangeable, meaning, if two dancers are left without their pair, they can’t dance together to form another pair, how many pairs on average are going to compete?
Well, I am not sure how to approach this. Best case scenario is when the 7 sick dancers form $3.5$ pairs, that is, $4$ pairs can't participate, so we are left with $7$ pairs.
There are 22 ways to select the first sick dancer. Then for the second, there is $1$ of $21$ left, to choose his pair, so ${22\choose 2}$ for one pair, then ${20\choose 2}$ for a second one and ${18\choose 2}$ for a third.
It is getting very complicated to consider all the remaining cases.
I would appreciate your assistance as it's been years since I worked with combinatorics and probabilities.
Thank you.

REPLY [4 votes]: Define
$$X_i =\begin{cases}
1 \qquad \text{if pair i can dance} \\
0 \qquad \text{otherwise}
\end{cases}$$ for $1 \le i \le 11$.
Then
$$P(X_i = 1) = \frac{15}{22} \cdot \frac{14}{21}$$
The total number of pairs which can dance is $\sum_{i=1}^{11} X_i$, and by linearity of expectation
$$E \left( \sum_{i=1}^{11} X_i \right) = \sum_{i=1}^{11} E(X_i) = 11 \cdot  \frac{15}{22} \cdot \frac{14}{21} = 5$$