A computer program has two modules X and Y an | vedclass

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(D) Let $E_1$ be the event that module $X$ has an error and $E_2$ be the event that module $Y$ has an error. Given $P(E_1) = 0.1$ and $P(E_2) = 0.3$.S

Vedclass · Accepted Answer

$\frac{31}{125}$

Vedclass · Answer

(D) Let $E_1$ be the event that module $X$ has an error and $E_2$ be the event that module $Y$ has an error. Given $P(E_1) = 0.1$ and $P(E_2) = 0.3$.Since the events are independent, $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2) = 0.1 	imes 0.3 = 0.03$.We define the mutually exclusive scenarios for errors:$1$. Error in $X$ only: $P(E_1 \cap E_2^c) = P(E_1) - P(E_1 \cap E_2) = 0.1 - 0.03 = 0.07$.$2$. Error in $Y$ only: $P(E_1^c \cap E_2) = P(E_2) - P(E_1 \cap E_2) = 0.3 - 0.03 = 0.27$.$3$. Error in both $X$ and $Y$: $P(E_1 \cap E_2) = 0.03$.Let $C$ be the event that the program crashes. The conditional probabilities are given as $P(C|X 	ext{ only}) = 0.5$, $P(C|Y 	ext{ only}) = 0.7$, and $P(C|X \cap Y) = 0.8$.Using the law of total probability:$P(C) = P(C|X 	ext{ only})P(X 	ext{ only}) + P(C|Y 	ext{ only})P(Y 	ext{ only}) + P(C|X \cap Y)P(X \cap Y)$$P(C) = (0.5 	imes 0.07) + (0.7 	imes 0.27) + (0.8 	imes 0.03)$$P(C) = 0.035 + 0.189 + 0.024 = 0.248$$P(C) = \frac{248}{1000} = \frac{31}{125}$.

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