A factory has three machines that all manufacture the same engine part. From long historical records the company has determined that the first machine produces 4% defectives, the second machine produces 6% defectives, and the third machine produces 1% defectives. Suppose also that the first machine produces 30%, the second machine 20%, and the third machine 50% of the engine parts. An engine part is selected at random. If the part is defective, what is the probability that was produced by the first machine?

Question

Helen Mathews

Mathematics

5 July, 14:48

A factory has three machines that all manufacture the same engine part. From long historical records the company has determined that the first machine produces 4% defectives, the second machine produces 6% defectives, and the third machine produces 1% defectives. Suppose also that the first machine produces 30%, the second machine 20%, and the third machine 50% of the engine parts. An engine part is selected at random. If the part is defective, what is the probability that was produced by the first machine?

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Jacqueline · Answer 1

Step-by-step explanation:

P (D/M₁) =.04 (probability of defect from a 1 st machine)

P (D/M₂) =.06

P (D/M₃) =.01

P (M₁) =.3 (probability of manufacture from 1 st machine)

P (M₂) =.2

P (M₃) =.5

P (M₁/D) (Probability of manufacture from 1 st machine, given it is defective)

= P (M₁) xP (D/M₁) / [P (M₁) xP (D/M₁) + P (M₂) xP (D/M₂) + P (M₃) xP (D/M₃) ]

=.3 x. 06 /.3 x. 06 +.2 x. 06 +.5 x. 01

=.018 /.035

= 18/35