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 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