Question

Companies​ A, B, and C produce 10​%, 20​%, and 70​%, respectively, of the major appliances sold in a certain area. In that​ area, 3​% of the company A​ appliances, 1 ¹/₂ % of the company B​ appliances, and 1​% of the company C appliances need service within the first year. Suppose a defective appliance is chosen at​ random; find the probability that it was manufactured by Company B.

Answer 1

We are given here that:
P(A) = 0.1,
P(B) = 0.2,
P(C) = 0.7

Also, we are given here that:
P(servicing | A) = 0.03,
P(servicing | B) = 0.015
P(servicing | C) = 0.01

Using law of total probability, we get here:
P( servicing ) = P(servicing | A)P(A) + P(servicing | B)P(B) + P(servicing | C)P(C)

P(servicing) = 0.03*0.1 + 0.015*0.2 + 0.01*0.7 = 0.013

Now given that a defective appliance is chosen at​ random,  the probability that it was manufactured by Company B is computed using Bayes theorem here as:

P(B | servicing ) = P(servicing | B)P(B) / P( servicing )

= 0.015*0.2 / 0.013

= 3/13

Therefore 3/13 = 0.2308 is the required probability here.