Does the hypergeometric distribution follow a Bernoulli process? There is a comment in an online resource that I need help understanding:

In a Bernoulli process, given that there are M successes among N trials, the number X
of successes among the first n trials has a Hypergeometric(N; M; n) distribution. source:  https://mathcs.clarku.edu/~djoyce/ma217/distributions2.pdf

Problem:  I am confused because my understanding is that the Bernoulli process requires repeated independent Bernoulli trials with FIXED probability p; however, sampling from a hypergeometric distribution are not independent, since a given sampling alters the probability of subsequent samplings.  I must be missing something or reading it incorrectly.
Question:  How or why does Bernoulli process come into play for hypergeometric distributions?
 A: The steps in the comment that prompted the question could be described a little more clearly as follows:

*

*There was, in the past, a Bernoulli process that generated $M$
successes in $N$ trials.

*Now, we take a subset of the $N$ trials, in this case, the first $n$ of them, and see how many of them are successes ($X$).

*$X$ has a Hypergeometric distribution with parameters $(N,M,n)$.

The reason the Bernoulli nature of the process is important is that it guarantees the locations of the $M$ successes among the $N$ trials are completely random; all possible permutations of the individual successes and failures are equally likely. Without this, you can't get to the Hypergeometric for the number of successes in the first $n$ trials. If you randomly sampled $n$ of the results you would end up with a Hypergeometric variate, even without the initial process being Bernoulli, but that's not what's happening here. Imagine a generating process that first produced $M$ successes and then produced $N−M$ failures; the number of successes in the first $n$ trials would not be Hypergeometric.
