JEE Math Practice Question

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 $T_{1}$ ) $=10 P\left(\right.$ computer turns out to be defective given that it is produced in plant $\left.T_{2}\right)$, 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

1. A.$\frac{36}{73}$
2. B.$\frac{47}{79}$
3. C.$\frac{78}{93}$
4. D.$\frac{75}{83}$