(MASTERY Problem) Section 3.5 & 3.6, Problem 6: An industry uses three methods,
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Question
(MASTERY Problem) Section 3.5 & 3.6, Problem 6: An industry uses three methods, M1, M2, and M3 to manufacture a part. Of all the parts manufactured, 45% are produced by method M1, 32% by method M2, and the rest 23% by method M3. Further it has been noted that 3% of the parts manufactured by method M1 are defective (D), while 296 manufactured by method M2 and 1.5% by method M3 are defective. A randomly selected part is found to be defective. Find the probability that the part was manufactured by (a) method M1, (b) method M2 First, what are the following marginal probabilities P(M1) = P(M2) = PlMal = Second, what are the following conditional probabilities: P(D M1) P(D | M2) = Now, what is the marginal probability of a defective part, P(D) Finally, what are the requested results? (a) The probability that the defective part was manufactured by method M1, P(M1 D) (b) The probability that the defective part was manufactured by method M2, P(M2 | D)=Explanation / Answer
P(M1) = 0.45
P(M2) = 0.32
P(M3) = 0.23
P(D | M1) = 0.03
P(D | M2) = 0.02
P(D | M3) = 0.015
P(D) = P(D | M1)*P(M1) + P(D | M2)*P(M2) + P(D|M3)*P(M3) = 0.03*0.45 + 0.02*0.32 + 0.015*0.23 = 0.02335
a) P(M1 | D) = (0.03*0.45)/0.02335 = 0.57816
b) P(M2 | D) = (0.02*0.32)/0.02335 = 0.2741
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