Is it right way to calculate pdf

Question

Which of these is the right way to calculate pmf. P(X=0) = (5/6)^4 or P(X=0) = (number of ways of choosing 4 out of 25)/ (number of ways of choosing 4 from 30)

Answer 1

We are choosing without replacement. There are $$\dbinom{30}{4}\tag{1}$$ equally likely ways to choose $4$ stoves. Let $0\le k\le 4$. The number of ways to choose $k$ defectives and $4-k$ non-defectives is $$\dbinom{5}{k}\dbinom{25}{4-k}.\tag{2}$$ This is because there are $\binom{5}{k}$ ways to choose $k$ defectives, and for each of these ways there are $\binom{25}{4-k}$ ways to select the non-defectives that will join them.

For the probability that $X=k$, divide (2) by (1).

Remark: The random variable $X$ of the problem has hypergeometric distribution.

If the manager were choosing from an enormous supply, with probability $\frac{5}{30}$ that an individual stove is defective, we could safely use a "with replacement" model. Then the probability of exactly $k$ defective would have for all practical purposes binomial distribution, with $$\binom{4}{k}\left(\frac{5}{30}\right)^k \left(\frac{25}{30}\right)^{4-k}.$$