In a factory which manufactures bolts,machine | vedclass

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Question

(A) Let events $B_1, B_2, B_3$ be defined as follows:$B_1$: The bolt is manufactured by machine $A$.$B_2$: The bolt is manufactured by machine $B$.$B_

Vedclass · Accepted Answer

$\frac{28}{69}$

Vedclass · Answer

(A) Let events $B_1, B_2, B_3$ be defined as follows:$B_1$: The bolt is manufactured by machine $A$.$B_2$: The bolt is manufactured by machine $B$.$B_3$: The bolt is manufactured by machine $C$.Clearly,$B_1, B_2, B_3$ are mutually exclusive and exhaustive events.Let $E$ be the event that the bolt is defective.Given probabilities:$P(B_1) = 0.25, P(B_2) = 0.35, P(B_3) = 0.40$.Conditional probabilities of defective bolts:$P(E|B_1) = 0.05, P(E|B_2) = 0.04, P(E|B_3) = 0.02$.By Bayes' Theorem,the probability that the bolt is manufactured by machine $B$ given it is defective is:$P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(B_1)P(E|B_1) + P(B_2)P(E|B_2) + P(B_3)P(E|B_3)}$$P(B_2|E) = \frac{0.35 	imes 0.04}{(0.25 	imes 0.05) + (0.35 	imes 0.04) + (0.40 	imes 0.02)}$$P(B_2|E) = \frac{0.0140}{0.0125 + 0.0140 + 0.0080} = \frac{0.0140}{0.0345}$$P(B_2|E) = \frac{140}{345} = \frac{28}{69}$.

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Let $H_1, H_2, \ldots, H_{n}$ be mutually exclusive and exhaustive events with $P(H_i) > 0, i = 1, 2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
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$STATEMENT-2$: $\sum_{i=1}^{n} P(H_i) = 1$.

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