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, $P\left(E\right)$ 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

  $\frac{47}{79}$
• Correct answer
  C

  $\frac{78}{93}$
• D

  $\frac{75}{83}$

Hence, the correct answer is option: 'C'.