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Question

(A) Let $E_1$ be the event that the motorcycle is manufactured at plant $A$,and $E_2$ be the event that it is manufactured at plant $B$. Let $S$ be th

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(A) Let $E_1$ be the event that the motorcycle is manufactured at plant $A$,and $E_2$ be the event that it is manufactured at plant $B$. Let $S$ be the event that the motorcycle is of standard quality.Given:$P(E_1) = 0.60$$P(E_2) = 0.40$$P(S|E_1) = 0.80$$P(S|E_2) = 0.90$Using Bayes' Theorem,the probability $p$ that the motorcycle was manufactured at plant $B$ given it is of standard quality is:$p = P(E_2|S) = \frac{P(S|E_2)P(E_2)}{P(S|E_1)P(E_1) + P(S|E_2)P(E_2)}$Substituting the values:$p = \frac{0.90 	imes 0.40}{(0.80 	imes 0.60) + (0.90 	imes 0.40)}$$p = \frac{0.36}{0.48 + 0.36} = \frac{0.36}{0.84} = \frac{36}{84} = \frac{3}{7}$We need to find $126p$:$126p = 126 	imes \frac{3}{7} = 18 	imes 3 = 54$Thus,the value is $54$.

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