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(C) Let $E_1, E_2, E_3$ be the events that the refrigerator is from companies $C_1, C_2, C_3$ respectively. Let $D$ be the event that the refrigerator

Vedclass · Accepted Answer

$\frac{14}{37}$

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(C) Let $E_1, E_2, E_3$ be the events that the refrigerator is from companies $C_1, C_2, C_3$ respectively. Let $D$ be the event that the refrigerator is defective.Given probabilities are:$P(E_1) = 0.25, P(E_2) = 0.35, P(E_3) = 0.40$$P(D|E_1) = 0.03, P(D|E_2) = 0.02, P(D|E_3) = 0.01$Using Bayes' Theorem, the probability that the defective refrigerator is from $C_2$ is:$P(E_2|D) = \frac{P(E_2) \cdot P(D|E_2)}{P(E_1) \cdot P(D|E_1) + P(E_2) \cdot P(D|E_2) + P(E_3) \cdot P(D|E_3)}$$P(E_2|D) = \frac{0.35 	imes 0.02}{(0.25 	imes 0.03) + (0.35 	imes 0.02) + (0.40 	imes 0.01)}$$P(E_2|D) = \frac{0.0070}{0.0075 + 0.0070 + 0.0040} = \frac{0.0070}{0.0185}$$P(E_2|D) = \frac{70}{185} = \frac{14}{37}$

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