Finding conditional probability

Question

A manufacturer knows that 12 percent of its 200 computer monitors are defective. If 3 monitors are randomly selected, how do I find the probability of the following outcomes?

P(1st monitor is defective)
P(2nd is defective | 1st is defective)
P(2nd is not defective | 1st is defective)
P(2nd is defective | 1st is not defective)

and
P(3rd is defective | first 2 are defective)

 

I am having a lot of difficulty with this problem as the textbook does not explain much, help is very appreciated.

Answer 1

12% of 200 computers are defective ⇒ Number of defective computers out of 200 is 24.
P(1st monitor is defective)

= 0.12 (=24/200)

If the first selection is defective, there are 23 defective computers out of the remaining 199 computers, so
P(2nd is defective | 1st is defective)

= 23/199

P(2nd is not defective | 1st is defective)

= 1 − P(2nd is defective | 1st is defective)
= 1 − 23/199

Using the same arguments

P(2nd is defective | 1st is not defective) = 24/199
P(3rd is defective | first 2 are defective) = 22/198