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. `78/96`
D. `75/83`

Answer

C

Explanation

`P(T_(1))=1/5`
`P(T_(2))=4/5`
`P(D)=7/100,`
where D is the event that computers produced in the factory turn out to be defective.
Given `P((D)/(T_(1)))=10.P((D)/(T_(2)))`
Let `P((D)/(T_(3)))=x`
`thereforeP(T_(1))xxP((D)/(T_(1)))+P(T_(2)).P((D)/(T_(2)))=P(D)=7/100`
`implies1/5xx10x+4/5xx x=7/100`
`impliesx=1/40`
`thereforeP((T_(2))/(D'))=(P((D')/(T_(2)))P(T_(2)))/(P((D')/T_(1))P(T_(1))+P((D')/(T_(2)))P(T_(2)))`
`=(39/40xx4/5)/(30/40xx1/5+39/40xx4/5)=(4/5xx39/40)/(93/100)=78/93`