In an international school, 60 students, of whom 15 are Korean, 20 are French, eight are Greek, and the rest, 17, are Chinese, are divided randomly into four classes of 15 each. If there are a total of eight French and six Korean students in classes A and B, what is the probability that class C has four of the remaining 12 French and three of the remaining nine Korean students?
This is a problem of P(12C4 and 9C3 | A or B has 20C8 and 15C6), I think. I think I keep adding too much information to my solution.
There are $15$ places available in Class $C$.
There are $12$ French, $9$ Korean and $30-12-9=9$ others available to fill those places (the rest are already in Classes $A$ and $B$).
This is a Multivariate Hypergeometric Distribution with $N=20$, $K_{French}=12$, $K_{Korean}=9$, $K_{Other}=9$ and we are selecting $15$. So:
$$\begin{align}P(French=4,Korean=3,Others=8)&={{12\choose4}{9\choose3}{9\choose8}\over{30\choose15}}\\ &={495\times84\times9\over155,117,520}\\ &={374,220\over155,117,520}\\ &\approx0.0024\end{align}$$