In a bolt factory,machines A, B and C manufac | vedclass

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(C) Let $E_1, E_2, E_3$ be the events that the bolt is manufactured by machines $A, B$ and $C$ respectively,and $D$ be the event that the bolt is defe

Vedclass · Accepted Answer

$\frac{5}{14}$

Vedclass · Answer

(C) Let $E_1, E_2, E_3$ be the events that the bolt is manufactured by machines $A, B$ and $C$ respectively,and $D$ be the event that the bolt is defective.Given probabilities:$P(E_1) = 0.20 = \frac{20}{100}$,$P(E_2) = 0.30 = \frac{30}{100}$,$P(E_3) = 0.50 = \frac{50}{100}$Conditional probabilities of defective bolts:$P(D|E_1) = \frac{3}{100}$,$P(D|E_2) = \frac{4}{100}$,$P(D|E_3) = \frac{2}{100}$Using Bayes' Theorem,the probability that the defective bolt is manufactured by machine $C$ is:$P(E_3|D) = \frac{P(E_3) 	imes P(D|E_3)}{P(E_1) 	imes P(D|E_1) + P(E_2) 	imes P(D|E_2) + P(E_3) 	imes P(D|E_3)}$$P(E_3|D) = \frac{\frac{50}{100} 	imes \frac{2}{100}}{\frac{20}{100} 	imes \frac{3}{100} + \frac{30}{100} 	imes \frac{4}{100} + \frac{50}{100} 	imes \frac{2}{100}}$$P(E_3|D) = \frac{100}{60 + 120 + 100} = \frac{100}{280} = \frac{10}{28} = \frac{5}{14}$

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