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 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

Explanation

Consider the following events :
`E_(1)`=Computer is produced by plant `T_(1)`
`E_(2)`=Computer is produced by plant `T_(2)`
A=Computer produced is defective
It is given that `P(E_(1))=(20)/(100)=(2)/(10), P(E_(2))=(80)/(100)=(8)/(10) " and " P(A)=(7)/(100)`.
It is also given that `P(A//E_(1))=10P(A//E_(2))`.
`therefore P(A)=P(A//E_(1))P(E_(1))+P(A//E_(2))P(E_(2))`
`implies (7)/(100)=10P(A//E_(2))xx(2)/(10)+P(A//E_(2))xx(8)/(10)`
`implies (7)/(100)=(28)/(10)P(A//E_(2)) implies P(A//E_(2))=(1)/(40)`
So, `P(A//E_(1))=10 P(A//E_(2))=(10)/(40)=(1)/(4)`
`therefore P(E_(2)// overline(A))=(P(E_(2))P(overline(A)//E_(2)))/(P(overline(A)))=((8)/(10)xx(1-(1)/(40)))/((1-(7)/(100)))=((8)/(10)xx(39)/(40))/((93)/(100))=(78)/(93)`