Two machines turn out all the products in a factory, with the first machine producing 45%...
Two machines turn out all the products in a factory, with the first machine producing 45% of the product and the second 55%. The first machine produces defective products 3% of the time and the second machine 8% of the time. (a) What is the probability that a defective part is produced at this factory given that it was made on the first machine? (b) What is the probability that a defective part is produced at this factory?
Homework Answers
P(1'st) = 0.45 , P(2nd ) = 0.55
P(defective | 1st ) = 0.03 , P(defective | second) = 0.08
a)
Using Bayes' rule
P(1st | defective) = P(1st) * P(defective | 1st ) / [ P(1st) * P(defective | 1st ) + P(2nd) * P(defective | 2nd ) ]
= (0.45 * 0.03) / ( 0.45 * 0.03 + 0.55 * 0.08)
= 0.2348
b)
P(defective) = P(1st) * P(defective | 1st ) + P(2nd) * P(defective | 2nd )
= 0.45 * 0.03 + 0.55 * 0.08
= 0.0575
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