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 computers produced in the factory turn out to be defective. It is known that P(computer turns out to bedefective, 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

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)`