Question

In a shipment of 10 computers, 3 are defective. Three computers are randomly selected and tested. What is the probability that all three are defective if the first and second ones are not replaced after being tested?

Answer 1

First one: probability=3/20

Second one: p= 2/19

third one: p= 1/18

So, the probability that all three are defective if the first and second ones are not replaced after being tested is 3/20*2/19*1/18 = 1/(3*20*19) =1/1140