Question

Two machines turn out all the products in a factory, with the first machine producing 60% of the product and the second 40%. The first machine produces defective products 2% of the time and the second machine 9% of the time. Given a defective product, what is the probability it was produced on the first machine? (Round your answer to four decimal places.)

Answer 1

Answer:

Let machine 2 be represented as M2 and machine 1 be represented as M1. Then we are given that:

P(M1) = 0.60 and P(M2) = 0.40

Let D represent defective products and ND represent non defective products . Then we are given that:

P(D | M1) = 0.02 and P(D | M2) = 0.09

Therefore, P(D) = P(D | M1)P(M1) + P(D | M2)P(M2) = 0.02*0.60 + 0.09*0.40 = 0.012 + 0.036= 0.048

a) Given that the product is made on first machine 1, Probability that a defective part is produced:

P( D | M1) = 0.02( already given )

P(defective part is produced given that it was made on first machine) = 2% = 0.02

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