$4n$ children at a party are paired at random, with each pair being equally likely. If there are $n$ boys and $3n$ girls, find the expected number of boy-girl pairs. (Ordering does not matter within boy-girl pairs or between pairs)
So far I've attempted:
Let $x$ be the number of pairs consisting of a boy and a girl.
Possible values of $x$ are from $0$ to $n$.
$E(x)=∑_{i,j=0}^{n} P(x_{i,j})$
where $x_{i,j}$ is an indicator random variable that is equal to $1$ if boy $i$ is paired with girl $j$, and 0 otherwise.
However, I'm not sure how to calculate $P(x_{i,j})$
My guess is that it would be $\frac{\binom{n}{1}\cdot\binom{3n}{1}}{\binom{4n}{2}}$, but I'm not sure if this is overcounting.
Also, after we find $P(x_{i,j})$, do we sum $P(x_{i,j})$ over $n$ possible pairs to find $E(x)$?
Best Answer
You ask specifically about your probability $P(X_{i,j})$ the probability that boy $i$ is paired with girl $j$.
The specific boy, boy $i$, will be in one of the pairs. It matters not which to us. The partner paired with boy $i$ is equally likely to be any of the other $4n-1$ children, exactly one of which is the specific girl, girl $j$. The probability then you ask about is simply $$P(X_{i,j})=\dfrac{1}{4n-1}$$
"After we find $P(X_{i,j})$ do we sum $P(X_{i,j})$ over $n$ possible pairs to find $E[X]$?" No, we are summing over all possible boy-girl pairings. There are $n\times (3n)$ possible pairings, namely boy 1 with girl 1, boy 1 with girl 2, boy 1 with girl 3, ... boy n with girl 3n-1, boy n with girl 3n.
We can see that $X = \sum\limits_{i=1}^n\sum\limits_{j=1}^{3n} X_{i,j}$ and continue from there with linearity of expectation.
An alternate approach would have been instead of looking at each of the possible pairings (of which there are $3n^2$) to look at each of the pairs (of which there are only $2n$).
(Note, with $4n$ children, there are $2n$ pairs made... not just $n$)
Letting $Y_i$ be the indicator random variable which equals $1$ if the $i$'th pair has one boy and one girl and $0$ otherwise, there are $\binom{4n}{2}$ equally likely pairs of children which could be in this $i$'th pair, $n\times 3n$ of which are a boy-girl pairing. This gives $\Pr(Y_i) = \dfrac{n\times 3n}{\binom{4n}{2}}$.
We can then recognize that $X = \sum\limits_{i=1}^{2n} Y_i$ and continue from there with linearity of expectation.
We could have broken up yet another way... letting $Z_i$ be the indicator random variable corresponding to whether or not boy $i$ was partnered with a girl. You'd have $X = \sum\limits_{i=1}^n Z_i$ and you'd have $P(Z_i)=\frac{3n}{4n-1}$. Similarly you could have done this from the girls' perspectives.