Question

A manufacturing operation utilizes three production lines to assemble electronic fuses. All lines produce fuses at the same rate and 20% of fuses are defective. Production lines recently suffered mechanical difficulty and probability of choosing a product from line 1 given defective is 40%, probability of choosing a product from line 2 given defective is 50%, and probability of choosing a product from line 3 given defective is 25%.

1. What is the probability that a customer buy a non-defective fuse produce by line 2?
2. What is the probability that a customer buy a fuse produce by line 3?

Answer 1

Let A,B and C denote the event that the fuse is produced by line 1, line 2 and line 3 respectively

Let D denote the event that the fuse is defective

P(D) = 0.2

P(A) = P(B) = P(C) = 1/3

P(A|D) = 0.4

P(B|D) = 0.5

P(C|D) = 0.25

Probability that a customer buy a non defective fuse produce by line 2

= P(D' and B)

P(B|D) = P(D and B)/P(B)

-> P(D and B) = P(B|D)*P(B) = 1/6

Thus, P(D' and B) = P(B) - P(D and B) = 1/3 - 1/6 = 1/6

Probability that a customer buy a fuse produce by line 3 = P(C) = 1/3