Assume there are three machines A, B, and C in a semiconductor manufacturing fac
ID: 3061700 • Letter: A
Question
Assume there are three machines A, B, and C in a semiconductor manufacturing facility that make chips. They manufacture, respectively, 25, 35, and 40 percent of the total semiconductor chips there. Of their outputs, respectively, 6, 4, and 2 percent of the chips are defective. A chip is drawn randomly from the combined output of the three machines and is found defective. What is the probability that this defective chip was manufactured by machine A? by machine B? by machine C? Answer only when you are sure your answer is correct, please. I will be thankful to you.
Explanation / Answer
p(A)=0.25,p(B)=0.35,p(C)=0.4
p(Item made by matchine A is defective)=p(D/A)=0.06
p(D/B)=0.04 & p(D/C)=0.02
What is the probability that this defective chip was manufactured by machine A?
i.e p(A/D)=p(D/A)*p(A)/p(D)
where
P(D) = P(D|A) P(A)+P(D|B) P(B)+P(D|C) P(C)
=0.06*0.25+0.04*0.35+0.02*0.4
=0.037
Hence p(A/D)=0.06*0.25/0.037
probability that this defective chip was manufactured by machine A=0.40
similarlly
probability that this defective chip was manufactured by machine B=p(B/D)
=0.04*0.35/0.037
probability that this defective chip was manufactured by machine B=0.378378
probability that this defective chip was manufactured by machine C=p(C/D)
=0.02*0.4/0.037
probability that this defective chip was manufactured by machine C=0.21