Three machines make parts in a factory. The following information about the production line is available.
Machine 1 makes 50% of the parts
Machine 2 makes 25% of the parts
Machine 3 makes 25% of the parts
5% of the parts by Machine 1 are defective
10% of the parts by Machine 2 are defective
12% of the parts by Machine 3 are defective
If a part is randomly selected, what is the probability that it is defective?
The solution goes to a fundamental rule of probability that relates total probability to marginal and conditional probabilities.
P(A) = P(A ∩ B1) + P(A ∩ B2) + … + P(A ∩ Bk)
P(A) = P(B1)P(A|B1) + P(B2)P(A|B2) + … + P(Bk)P(A|Bk)
Using this equation, we get
P(Defective) = P(Machine1)P(Defective|Machine1) + P(Machine2)P(Defective|Machine2) + P(Machine3)P(Defective|Machine3)
= 0.5×0.05 + 0.25×0.1 + 0.25×0.12 = 0.08 or 8% chance.
