Question

A container has 3 defective and 4 non defective items. if a sample of 2 items are drawn one [05] after another without replacement what is the probability that, the sample will at most one defective

Answer 1

P(at most one defective) = 1 - P(2 defective)

P(2 defective) = P(1st defective)*P(2nd defective) = 3/7*2/6 = 1/7

P(at most one defective) = 1-1/7 = 6/7

Answer: 6/7