Question

suppose you receive a shipment of ten televisions. Three of the televisions are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not work?

Answer 1

P(both work) = ((10-3)/10)*((10-3-1)/9) = (7/10)*(6/9) = 42/90

P(at least one does not work) = 1 - P(both work) = 1 - 42/90 = 48/90 = 24/45.