How to find the probability that not all of the infected students end up in the same van The question states that there are $15$ students (N), and $3$ of them have an asymptomatic disease (r). They will be randomly divided into two busses, $8$ going to one van and $7$ going to the other. The question is asking to find the probability that the three infected students do not end up in the same van.
I used the hypergeometric distribution. I assumed that $P(\text{all not in the same van}) = 1 - P(\text{all in the same van})$? I got my answer to be $$1 -\frac{C(3,3)C(12,4)}{C(15,7)}$$ Wondering if this thought process is correct.
 A: It is true that the probability that not all the students are in the same van by subtracting the probability that they are all in the same van from $1$.  However, you did not calculate the probability that all the infected students are in the same van correctly.
There are two possibilities.  They are all in the van with seven students or they are all in the van with eight students.  You calculated the probability that all three infected students are in the van with seven students.  The probability that they are not all in the van with eight students is
$$\frac{\dbinom{3}{3}\dbinom{12}{5}}{\dbinom{15}{8}}$$
since if all three infected students are in that van, five of the other $12$ students must be in the van with them.
Thus, the probability that the three infected students are not all in the same van is
$$\Pr(\text{not all infected students in the same van}) = 1 - \frac{\dbinom{3}{3}\dbinom{12}{4}}{\dbinom{15}{7}} - \frac{\dbinom{3}{3}\dbinom{12}{5}}{\dbinom{15}{8}}$$
We could also calculate the probability that not all students are in the same van directly.  If not all the infected students are in the same van, then there are either one infected student and six non-infected students or two infected and five non-infected students in the van with seven students.  Hence,
$$\Pr(\text{not all infected students in the same van}) =  \frac{\dbinom{3}{1}\dbinom{12}{6} + \dbinom{3}{2}\dbinom{12}{5}}{\dbinom{15}{7}}$$
