A computer producing factory has only two plants $T_1$ and $T_2$ 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 $\mathrm{T}_1$ )
$=10 \mathrm{P}$ (computer turns out to be defective given that it is produced in plant $\mathrm{T}_2$ ), 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