Defects in the articles produced by a manufacturing company can be the result of

Question

Defects in the articles produced by a manufacturing company can be the result of one of the following independent causes:
1. Malfunction of the machinery, which occurs 5% of the production time
2. Carelessness of workers, which occurs 8% of the production time
Defective articles are produced only by these two causes. Also, if only the machinery malfunctions, the probability of a defective article is 0.10, whereas whenever there is only carelessness of workers, the probability of a defective article is 0.20. However, when both causes occur, the probability of a defective article is 0.80.
(a) What is the probability of producing a defective article in the company?
(b) If a defective article is discovered, what is the probability that it was caused by carelessness of workers?

Explanation / Answer

Call P(mm) the probability a machine is malfunctioning, P(mm) = 0.05 Call P(cw) the probability careless workers, P(cw) = 0.08 When P(mm) occurs, the risk of producing a defective article is 0.10. So we will call this probability defect due to machinery P(dm) = 0.10 When P(cw) occurs, the risk of producing a defective article is 0.20. So we will call this probability defect due to workers P(dc) = 0.20 Finally when both occursthe risk of producing a defective article rises to 0.80. So we will call this probability defect due to both P(db) = 0.80 To have a defective article one of these things must happen: (1) Machinery Malfunction (P(mm)), NO worker carelessness (1-P(cw)), defective production due to mm (Pdm): That is case 1: P(c1) = P(mm).(1-P(cw)).P(dm) P(c1) = (0.05) . (1 - 0.08). (0.10) P(c1) = (0.05) . (0.92). (0.10) P(c1) = 0.0046 (2) Worker carelessness (P(cw)), NO machine malfunction (1-P(mm)), defective production due to carelessness (P(dc)): That is case 2: P(c2) = P(cw).(1-P(mm)).P(dc) P(c2) = (0.08).(1-0.05).(0.20) P(c2) = (0.08).(0.95).(0.20) P(c2) = (0.08).(0.19) P(c2) = 0.0152 (3) Finally Machinery Malfunction (P(mm)), AND worker carelessness(P(cw)), defective production then is P(db): That is case 3: P(3) =( 0.05).(0.08).(0.80) P(3) = (0.05).(0.64) P(3) = 0.032 So: (a) Probability to produce a defective item P(di): P(di) = P(1) + P(2) + P(3) P(di) = (0.0046) + (0.0152) + (0.032) P(di) = 0.0518 Now: (b) When a defective is produced, the probability it is caused by worker carelessness is a conditional probability P(cc) (conditional carelessness). To have P(cc) of these things must happen: (1) Worker carelessness (P(cw)), NO machine malfunction (1-P(mm)), defective production due to carelessness (P(dc)): That is P(cw).(1-P(mm)).P(dc) = P(2) again (2) OR Machinery Malfunction (P(mm)), AND worker carelessness(P(cw)), defective production then is P(db): That is P(mm).P(cw).P(db) = P(3) again And as it is conditional on P(di) we must divide this total by (P(di) to get P(cc): P(cc) = [P(2) + P(3) ] / P(di) P(cc) = [(0.0152) + (0.032)] / (0.0518) P(cc) = 0.911969111969111969111….. = 911969/999999 (that is a little known math rule about infinitely periodically repeating decimal numbers are fractions of that type)

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