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 the computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective givent that it is produced in plant `T_(1))` = 10 P ( 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

Options

A. `(36)/(73)`
B. `(47)/(79)`
C. `(78)/(93)`
D. `(75)/(83)`

Answer

C