Question

Suppose that chips for an integrated circuit are tested and that the probability that they are detected if they are defective is .95, and the probability that they are declared sound if in fact they are sound is .97. If .5% of the chips are faulty, what is the probability that a chip that is declared faulty is sound?

Easy and full explanation thanks

Answer 1

We are given here for each chip that:
P( detected | defective ) = 0.95, therefore P( sound | defective ) = 1 - 0.95 = 0.05
P( sound | sound ) = 0.97, therefore P( detected | sound ) = 1 - 0.97 = 0.03

We also given here that: P( defective) = 0.005, therefore P(sound) = 1 - 0.005 = 0.995

Using law of total probability, we have here:
P( detected ) = P( detected | defective ) P(defective) + P( detected | sound ) P(sound)

P(detected) = 0.95*0.005 + 0.03*0.995 = 0.0346

Using Bayes theorem, we get here:
P( sound | detected ) = P( detected | sound ) P(sound) / P(detected ) = 0.03*0.995 / 0.0346 = 0.8627

Therefore 0.8627 is the required probability here.