Question

Three machines turn out all the products in a factory, with the first machine producing 15% of the products, the second machine 25%, and the third machine 60%. The first machine produces defective products 20% of the time, the second machine 3% of the time and the third machine 5% of the time. What is the probability that a non-defective product came from the second machine? (Round your answer to four decimal places.)

Answer 1

Let F, S and T denote the events that the product is produced by first machine, second machine and third machine respectively

Let D denote the event that a defective product is produced

P(F) = 0.15, P(S) = 0.25, P(T) = 0.60

P(D | F) = 0.20

P(D | S) = 0.03

P(D | T) = 0.05

Thus, P(D) = 0.20*0.15 + 0.03*0.25 + 0.05*0.60 = 0.0675

P(D') = 1 - P(D) = 0.9325

Probability that a non defective product came from the second machine = P(S | D')

= P(D' | S)*P(S)/P(D')

= 0.97*0.25/0.9325

= 0.2601