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

Let x=P (computer turns out to be defective, given that it is produced in plant `T_(2)`)
`implies" "x=P((D)/(T_(2)))" "...(i)`
where, D = Defective computer
`:.P("computer turns out to be defective given that is produced in plant" T_(2))=10x`
i.e. `" "P((D)/(T_(1)))=10x" "...(ii)`
Also, `" "P(T_(1))=(20)/(100)" and "P(T_(2))=(80)/(100)`
Given, P (defective computer)`=(7)/(100)`
i.e. `" "P(D)=(7)/(100)`
Using law of total probability,
`P(D)=9(T_(1)).P((D)/(T_(1)))+P(T_(2)).P((D)/(T_(2)))`
`:." "(7)/(100)=((20)/(100)).10x+((80)/(100)).x`
`implies" "7=(280)ximpliesx=(1)/(40)" "...(iii)`
`:." "P((D)/(T_(2)))=(1)/(40)" and "P((D)/(T_(1)))=(10)/(40)`
`impliesP((bar(D))/(T_(2)))=1-(1)/(40)=(39)/(40)" and "P((bar(D))/(T_(1)))=1-(10)/(40)=(30)/(40)" "...(iv)`
Using Beye's theorem,
`P((T_(2))/(D))=(P(T_(2)nnbar(D)))/(P(T_(1)nnbar(D))+P(T_(2)nnbar(D)))`
`=(P(T_(2)).P((bar(D))/(T_(2))))/(P(T_(1)).P((bar(D))/(T_(1)))+P(T_(2)).P((bar(D))/(T_(2))))`
`=((80)/(100).(39)/(93))/((20)/(100).(30)/(40)+(80)/(100).(39)/(40))=(78)/(93)`