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Exercise 9.6 · Q2

Q.In a factory which manufactures bulbs, units A, B and C manufacture respectively 25%, 35% and 40% of the bulbs. Of their outputs, 5, 4 and 2 percent are respectively defective bulbs. A bulb is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the unit B?

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Bayes' theorem across the three production units gives P(bulb from unit B∣defective)=2869≈0.406P(\text{bulb from unit B}\mid\text{defective})=\dfrac{28}{69}\approx0.406.

Bayes' theorem with partition {A,B,C}\{A,B,C\} = manufacturing units and event D=D= "bulb is defective":

P(B∣D)=P(B) P(D∣B)P(A)P(D∣A)+P(B)P(D∣B)+P(C)P(D∣C)P(B\mid D)=\frac{P(B)\,P(D\mid B)}{P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)}

  1. Given production shares (priors).

P(A)=0.25,P(B)=0.35,P(C)=0.40P(A)=0.25,\qquad P(B)=0.35,\qquad P(C)=0.40

  1. Given defect rates.

P(D∣A)=0.05,P(D∣B)=0.04,P(D∣C)=0.02P(D\mid A)=0.05,\qquad P(D\mid B)=0.04,\qquad P(D\mid C)=0.02

  1. Compute each joint term.

P(A)P(D∣A)=0.25×0.05=0.0125P(A)P(D\mid A)=0.25\times0.05=0.0125

P(B)P(D∣B)=0.35×0.04=0.0140P(B)P(D\mid B)=0.35\times0.04=0.0140

P(C)P(D∣C)=0.40×0.02=0.0080P(C)P(D\mid C)=0.40\times0.02=0.0080

  1. Total probability a randomly drawn bulb is defective.

P(D)=0.0125+0.0140+0.0080=0.0345P(D)=0.0125+0.0140+0.0080=0.0345

  1. Apply Bayes' theorem for unit B. …

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