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(B) Let $E_1$, $E_2$, and $E_3$ be the events that the battery is produced by machines $P$, $Q$, and $R$ respectively. Let $A$ be the event that the b

Vedclass · Accepted Answer

$\frac{33}{2000}$

Vedclass · Answer

(B) Let $E_1$, $E_2$, and $E_3$ be the events that the battery is produced by machines $P$, $Q$, and $R$ respectively. Let $A$ be the event that the battery is defective. Given probabilities are: $P(E_1) = 0.20$, $P(E_2) = 0.30$, $P(E_3) = 0.50$. Conditional probabilities of defective batteries are: $P(A|E_1) = 0.01$, $P(A|E_2) = 0.015$, $P(A|E_3) = 0.02$. Using the Law of Total Probability: $P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)$ $P(A) = (0.20 	imes 0.01) + (0.30 	imes 0.015) + (0.50 	imes 0.02)$ $P(A) = 0.002 + 0.0045 + 0.010 = 0.0165$ Converting to fraction: $P(A) = \frac{165}{10000} = \frac{33}{2000}$.

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