A car factory has three manufacturing lines in its facility. The first line prod
ID: 3252058 • Letter: A
Question
A car factory has three manufacturing lines in its facility. The first line produces 50% of the vehicles, the second produces 30% and the third produces 20%. The defect rates of the three lines .06, .03, .09, respectively.
a) What is the probability of a randomly selected car from the factory having a defect?
b) What is the probability that a random car came from the third line, if I know it is defective?
c) If the factory produced 12 cars today and 2 of the were defective, how many ways could a quality control inspector select three and not pick one of the defective vehicles?
Explanation / Answer
first line produces 50% of the vehicles, the second produces 30% and the third produces 20%
defect rates of the three lines .06, .03, .09, respectively
a)
probability of a randomly selected car from the factory having a defect = P(selecting car from first line)*P(car from first line being defective)+P(selecting car from second line)*P(car from second line being defective)+P(selecting car from third line)*P(car from thrid line being defective)
= 0.5*0.06+0.3*0.03+0.2*0.09 = 0.057
b)
probability that a random car came from the third line and it is defective = P(selecting car from third line)*P(car from thrid line being defective) = 0.2*0.09 = 0.018
probability that a random car came from the third line, if I know it is defective = probability that a random car came from the third line and it is defective/probability of a randomly selected car from the factory having a defect = 0.018/0.057 = 0.3158
c)
factory produced 12 cars today and 2 of the were defective
no of defective cars = 2
no of non defective cars = 12-2 = 10
no of ways of selecting three cars from 12 = 12C3 = 220
no fo ways of selecting three and not pick one of the defective vehicles = no of ways of selecting three cars from 12 - no fo ways of selecting three and pick both of the defective vehicles
no fo ways of selecting three and pick both of the defective vehicles= 10*1 = 10
no fo ways of selecting three and not pick one of the defective vehicles = 220-10 = 210
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