Question

In a manufacturing plant, three machines, A, B, and C, produce 37%, 36%, and 27%, respectively, of the total production. The company's quality-control department has determined that 1.5% of the items produced by Machine A, 2% of the items produced by Machine B, and 2.5% of the items produced by Machine C are defective. If an item is selected at random and found to be defective, what is the probability that it was produced by Machine B? (Round your answer to three decimal places.)

Answer 1

It will be similar to the previous problem.
http://www.jiskha.com/display.cgi?id=1498184495

The steps are to use law of total probability and then Bayes theorem.

I'll start, and let you finish.

Events:
A=part was produced by machine A
B=part was produced by machine B
C=part was produced by machine C
D=part was defective

We're given:
P(A)=0.37
P(B)=0.36
P(C)=0.27
P(D|A)=0.015 (defective given produced by A)
P(D|B)=0.02
P(D|C)=0.025

Also, by definition of conditional probability:
P(X|Y)=P(X∩Y)P(Y)

Then by the law of total probability:
P(D)=P(D∩A)+P(D∩B)+P(D∩C)
=P(D|A)P(A)+P(D|B)P(B)+P(D|C)*P(C)
= ? (all necessary probabilities are known)

Using Bayes Theorem,
P(B|D) (produced by B given defective)
=P(B∩D)P(D)
=P(D∩B)P(D)
=[P(D|B)/P(B)]*P(D) [from P(D|B)=P(D∩B)P(B)]
=? (all necessary probabilities are known or calculated before)