Question

A batch of 40 parts contains four defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
b. If this experiment is repeated, with replacement, what is the probability that both parts are defective?

Answer 1

a) the probability of picking one part with a defect is 4/40 = 1/10
So, the probability of picking 2 defective parts without replacement is:
4/40 * 3/39 = 1/10 * 1/13 = 1/130

the 2nd fraction is 3/39 since you pick one that is defective in the first pick, so there are 3 defective ones left and thus 39 parts in total.

b) same logic as in a), but now you put back the part with initially picked. Thus, you have the same amount of defective parts in the 1st pick and 2nd pick:

so the probability is 4/40 * 4/40 = 1/10 * 1/10 = 1/100