A manufacturing company of bulbs has 3 units | vedclass

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(A) Let $E_1, E_2, E_3$ be the events that the bulb is produced by units $A, B, C$ respectively. Let $D$ be the event that the bulb is defective.Given

Vedclass · Accepted Answer

$\frac{28}{69}$

Vedclass · Answer

(A) Let $E_1, E_2, E_3$ be the events that the bulb is produced by units $A, B, C$ respectively. Let $D$ be the event that the bulb is defective.Given probabilities are:$P(E_1) = 0.25, P(E_2) = 0.35, P(E_3) = 0.40$$P(D|E_1) = 0.05, P(D|E_2) = 0.04, P(D|E_3) = 0.02$Using the Law of Total Probability,the probability that a bulb is defective is:$P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)$$P(D) = (0.25 	imes 0.05) + (0.35 	imes 0.04) + (0.40 	imes 0.02)$$P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345$Using Bayes' Theorem,the probability that the defective bulb was produced by unit $B$ is:$P(E_2|D) = \frac{P(E_2)P(D|E_2)}{P(D)}$$P(E_2|D) = \frac{0.35 	imes 0.04}{0.0345} = \frac{0.0140}{0.0345} = \frac{140}{345}$Dividing numerator and denominator by $5$,we get:$P(E_2|D) = \frac{28}{69}$

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