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(C) Let $T_1$ and $T_2$ be the events that the computer is produced in plant $T_1$ and $T_2$ respectively. Let $D$ be the event that the computer is d

Vedclass · Accepted Answer

$\frac{78}{93}$

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(C) Let $T_1$ and $T_2$ be the events that the computer is produced in plant $T_1$ and $T_2$ respectively. Let $D$ be the event that the computer is defective.Given $P(T_1) = 0.2$,$P(T_2) = 0.8$,and $P(D) = 0.07$.We are given $P(D | T_1) = 10 P(D | T_2)$. Let $P(D | T_2) = p$,then $P(D | T_1) = 10p$.Using the law of total probability: $P(D) = P(D | T_1)P(T_1) + P(D | T_2)P(T_2)$.$0.07 = (10p)(0.2) + (p)(0.8) = 2p + 0.8p = 2.8p$.$p = \frac{0.07}{2.8} = \frac{7}{280} = \frac{1}{40}$.So,$P(D | T_2) = \frac{1}{40}$ and $P(D | T_1) = 10 	imes \frac{1}{40} = \frac{1}{4}$.The probability that a computer is not defective is $P(\bar{D}) = 1 - P(D) = 1 - 0.07 = 0.93$.We need to find $P(T_2 | \bar{D}) = \frac{P(\bar{D} | T_2)P(T_2)}{P(\bar{D})}$.$P(\bar{D} | T_2) = 1 - P(D | T_2) = 1 - \frac{1}{40} = \frac{39}{40}$.$P(T_2 | \bar{D}) = \frac{(\frac{39}{40}) 	imes 0.8}{0.93} = \frac{0.78}{0.93} = \frac{78}{93}$.

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