Probability of more than one, but no more than three?

Question

I am having trouble wrapping my head around the following problem... Hope someone can enlighten me.

On the motherboard of a device there are 20 different processors of which 5 of them are faulty. We know that there is 1 or more faults on the motherboard. In an attempt to find the error, we select 4 random processors for testing.

What is the probability that more than 1 of the processors is faulty, but no more than 3?

Solution tldr: Given the suggestion per BruceET, using the hypergeometric probability distribution, I figured the probability as follows.

$\frac{\binom 5 2 \binom {15}2}{\binom {20}4} + \frac{\binom 53 \binom {15}1}{\binom {20}4} = 0.2477\ \ \ $which checks with the problem's reference values.

Answer 1

I think sampling of processors is without replacement, so this would be a hypergeometric distribution instead of a binomial. Thus one of the two terms is as follows:

$$P(X = 2) = \frac{{5 \choose 2}{15 \choose 2}}{{20 \choose 4}}.$$

In R statistical software, the answer (total of two terms as in the comment of @N.F.Taussig) is computed as:

sum(dhyper(2:3,  5, 15,  4))
## 0.247678