A computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20\%$ and plant $T_2$ produces $80\%$ of the total computers produced. $7\%$ of computers produced in the factory turn out to be defective. It is known that $P$(computer turns out to be defective given that it is produced in plant $T_1$) $= 10P$(computer turns out to be defective given that it is produced in plant $T_2$), where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is (A) $\frac{36}{73}$ (B) $\frac{47}{79}$ (C) $\frac{78}{93}$ (D) $\frac{75}{83}$
A computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20\%$ and plant $T_2$ produces $80\%$ of the total computers produced. $7\%$ of computers produced in the factory turn out to be defective. It is known that\\
$P$(computer turns out to be defective given that it is produced in plant $T_1$)\\
$= 10P$(computer turns out to be defective given that it is produced in plant $T_2$),\\
where $P(E)$ denotes the probability of an event $E$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is\\
(A) $\frac{36}{73}$\\
(B) $\frac{47}{79}$\\
(C) $\frac{78}{93}$\\
(D) $\frac{75}{83}$