Question

In a shipment of 20 computers, 2 are defective. Three computers are randomly selected and tested. What is the probability that all 2 are defective if the first one is not replaced after being tested?

Answer 1

Probabilities of being defective in case the 1st is not replaced (2nd is replaced)

First one: $\frac{2}{20}$

Second one: $\frac{1}{19}$

Third one: no more defective

Therefore we have: $\frac{2}{20}\cdot\frac{1}{19}=\frac{1}{190}$