Problem 1. In a semiconductor manufacturing facility there are 3 machines, M1, M
ID: 3351787 • Letter: P
Question
Problem 1. In a semiconductor manufacturing facility there are 3 machines, M1, M2, and M3 that make chips. M1 makes 25% of the total production, M2 makes 35%, and M3 makes 40%. Of the chips made by M, five percent are defective, of those made by M2 four percent are defective, and of those made by M3 two percent are defective. A chip is drawn at random from the combined output of the three machines (a) What is the probability that the selected chip is defective!? (b) The selected chip was tested and found to be defective. What is the probability that this defective chip was manufactured by machine Mi?Explanation / Answer
Given: Pr (M1) = 0.25
Pr (M2) = 0.35
Pr (M3) = 0.40
Let D be the event of obtaining a defective.
Then,
Pr (D/M1) = 0.05
Pr (D/M2) = 0.04
Pr (D/M3) = 0.02
(a) To find the probability that selected chip is defective, we have the formula,
Let us define the conditional probabilities:
Pr (D/M1) = Probability of drawing a defective chip from M1
Similarly, Pr (D/M2) and Pr (D/M3) be that drawn from M2 and M3
By theorem of total Probability,
Pr (D) = Pr (M1)* Pr (D/M1) + Pr (M2)* Pr (D/M2) + Pr (M3)* Pr(D/M3)
= 0.25 *0.05 + 0.35 *0.04 + 0.4 *0.02
Pr (D) = 0.0345
(b) To find the probability that the selected chip which was found to be defective was manufactured by M1
i.e. to find Pr(M1/D)
By Bayes’ theorem,
Pr (M1/D) = [ Pr(M1) * Pr(D/M1)] / [Pr (M1) * Pr (D/M1) + Pr (M2) * Pr (D/M2) + Pr (M3) * Pr (D/M3)]
= [0.25 * 0.05] / [0.25 * 0.05 + 0.35 *0.04 + 0.4 *0.02]
Pr (M1/D) = 0.3623
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