A foreman for an injectionmolding firm admits that on 46 of

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 4% to 23%. The plant manager randomly selects a molding from the early morning run and discovers it is defective. What is the probability that the foreman forgot to shut off the machine the previous night?

Solution

Define events
1. F = event of forgetting to shut off machine.
2. D = event of product being defective.

P(F and D) = 0.46*0.23 = 0.1058
P(F and ~D) = 0.46*(1-0.23) = 0.3542
P(~F and D) = (1-0.46)*0.04 = 0.0216
P(~F and ~D) = (1-0.46)*(1-0.04) = 0.5184

By definition of conditional probability
P(F|D) = P(F and D)/P(D)
          = P(F and D)/[P(F and D)+P(~F and D)]
          = 0.1058/(0.1058+0.0216)
          = 0.8305 (approx.)

Answer:
The probability that the foreman forgot to shut the machine is 0.8305, given that the product was defective.