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

Mathematics

Creedence

27 February, 11:17

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?

+4

Answers (1)

Know the Answer?

Maya Shaw · 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