Question

A factory produces products of two different models: 60% of model 1 and 40% of model 2.

4% of the products of model 1are faulty. 3% of the products of model 2 are faulty.

What is the chance that a faulty product is model 1?

Answer 1

Let A be the event product is Model 1, B be the event product is Model 2, C be the event product is faulty.

Given,

P(A product is model 1) = P(A) = 0.6

P(A product is model 2) = P(B) = 0.4

P(Product is faulty / Product is model 1) = P(C/A) = 0.04

P(Product is faulty / Product is model 2) = P(C/B) = 0.03

Now, using Baye's theorem

P(Product is Model 1 / Product is faulty)

= P(A/C)

= P(C/A) * P(A) / (P(C/A) * P(A) + P(C/B) * P(B))

= 0.04 * 0.6 / (0.04 * 0.6 + 0.03 * 0.4)

= 0.667