A company has three plants. Plant A produces 23% of the company’s widgets, plant

Question

A company has three plants.  Plant A produces 23% of the company’s widgets, plant B produces 36% of the company’s widgets, and plant C produces the rest of the company’s widgets.  From past records, 5% of the widgets produced from plant A are good, 3% of the widgets produced from plant B are good, and 9% of the widgets produced from plant C are good.  If one widget from the company is selected at random and found to be defective, what is the probability that the widget was produced by plant B? Round to 4 places after the decimal point.

Explanation / Answer

Let A,B,C be the events that the item come from plants A, B, Crespectively. Let D be the event the item is defective. P(A) = .23, P(B) = .36, P(C) = 1 - .23 - .36 = .41 P(D|A) = 1 - .05 = .95 P(D|B) = 1 - .03 = .97 P(D|C) = 1 - .09 = .91 We want to know: P(B|D) = ? P(B|D) = P(BD)/P(D) (Baye's Formula) P(BD) = P(D|B)P(B) = (.97)(.36) = .3492 P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) P(D) = (.95)(.23) + (.97)(.36) + (.91)(.41) = .9408 So P(D|B) = .3492/.9408 = .3712

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