Answer to Question #145638 in Statistics and Probability for qwerty

Question

6. A certain company has two car assembly plants, A and B. Plant A produces twice as many cars as plant B. Plant A uses engines and transmissions from a subsidiary plant which produces 10% defective engines and 2% defective transmissions. Plant B uses engines and transmissions from another source where 8% of the engines and 4% of the transmissions are defective. Car transmissions and engines at each plant are installed independently.

(a) What is the probability that a car chosen at random will have a good engine?
(b) What is the probability that a car from plant A has a defective engine, or a defective transmission, or both?
(c) What is the probability that a car which has a good transmission and a defective engine was assembled at plant B?

Answer 1

E - Engine is defective.

T - Transmission is defective.

A - Car is produced in plant A; P(A) = 0.667; P(E|A) = 0.1; P(T|A) = 0.02

B - Car is produced in plant B; P(B) = 0.333; P(E|B) = 0.08; P(T|B) = 0.04

(a) The probability that a car chosen at random will have a good engine =

P(A)*(1-P(E|A))+P(B)*(1-P(E|B)) = 0.667*(1-0.1)+0.333*(1-0.08) = 0.907

(b) The probability that a car from plant A has a defective engine, or a defective transmission, or both =

1-((1-P(E|A))*(1-P(T|A))) = 1-((1-0.1)*(1-0.02)) = 0.118

(c) The probability that a car which has a good transmission and a defective engine was assembled at plant B =

[P(B)*(1-P(T|B))*P(E|B)]/{[P(B)*(1-P(T|B))*P(E|B)]+[P(A)*(1-P(T|A))*P(E|A)]} =

[0.333*(1-0.04)*0.08]/{[0.333*(1-0.04)*0.08]+[0.667*(1-0.02)*0.1]} = 0.281