An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects.
1) What is the probability that exactly 10 of these are from the second section?
My assumptions for a hypergeometric rv X are as follows: You have a population of N objects. Each object can be characterized as S or F. There are M successes in the population. A sample of n objects is selected without replacement in such a way that each subset of size n is equally likely to be chosen. If x is the number of S's in a completely random sample of size n drawn from a population consisting of M S's and (N-M) F's, then the probability distribution of X, called the hypergeometric distribution, is given by
$$ P(X = x) = h(x; n, M, N) = \frac{{M \choose x} {N-M \choose n-x}}{N \choose n} $$
Analyzing this problem, it seemed to me that the assumptions led to this correspondence: Population of 50 students (N = 50). Class 1 represented as F and Class 2 represented as S. M would be 30, since M is the number of S in the population. A sample of n objects is selected (15 papers chosen), and the question asks what the probability is that 10 of these is a S (10 are in class 2).
So my understanding of the problem was that the answer would be h(10; 15, 30, 50), which is 0.2069.
However, when I was checking my work against the solutions, I found that the correct answer was h(x; 10, 15, 50) and I have difficulty understanding why this is the case... The sample is 15, and then you want to find the probability that 10 of them are S, so how is it that the correct sample is 10? I would deeply appreciate someone explaining this to me.
Thank you.