An electric instrument consists of two units. | vedclass

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(D) Let $A$ be the event that the first unit functions,so $P(A) = 0.9$ and $P(A^c) = 0.1$.Let $B$ be the event that the second unit functions,so $P(B)

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) Let $A$ be the event that the first unit functions,so $P(A) = 0.9$ and $P(A^c) = 0.1$.Let $B$ be the event that the second unit functions,so $P(B) = 0.8$ and $P(B^c) = 0.2$.The instrument operates only if both units function. The probability that the instrument operates is $P(A \cap B) = 0.9 	imes 0.8 = 0.72$.The probability that the instrument fails to operate is $P(F) = 1 - 0.72 = 0.28$.The failure occurs in three mutually exclusive cases: $1$. First unit fails,second unit functions: $P(A^c \cap B) = 0.1 	imes 0.8 = 0.08$.$2$. First unit functions,second unit fails: $P(A \cap B^c) = 0.9 	imes 0.2 = 0.18$.$3$. Both units fail: $P(A^c \cap B^c) = 0.1 	imes 0.2 = 0.02$.Note that $0.08 + 0.18 + 0.02 = 0.28$,which matches $P(F)$.We are given that the instrument failed. We need the conditional probability $p$ that only the first unit failed (i.e.,$A^c \cap B$ occurred) given that the instrument failed $(F)$.$p = P(A^c \cap B | F) = \frac{P(A^c \cap B)}{P(F)} = \frac{0.08}{0.28} = \frac{8}{28} = \frac{2}{7}$.Therefore,$98p = 98 	imes \frac{2}{7} = 14 	imes 2 = 28$.

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