A foreman for an injection-molding firm admits that on 46% of his shifts, he forgets to shut off the injection machine on his line. This causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 5% to 19%. If a molding is randomly selected from the early morning run of a random day, what is the probability that it is defective?

Question

I got:
P(F and D) = 0.46*0.19 = 0.0874
P(~F and D) = 0.54*0.05 = 0.027
P(F|D) = P(F and D) / [P(F and D) + P(~F and D)] = 0.0874 / 0.1144 = 0.76398601398

But my online homework's saying the answers incorrect

Mike · Accepted Answer

P(FD) = .46 P(notFD) = . 54 If he forgets, then increase is P(D) = .19 If doesn’t forget, remain P(D) .05 Therefore, P(FD)*P(D) +P(NotFD)*P(D) (.46*.19) +(.54*.05) =.1144