Let x=P (computer turns out to be defective, given that it is produced in plant T2 ) ⇒x=P(T2D)...(ii) where, D= Defective computer∴P (computer turns out to be defective given that is produced in plant T1 ) =10xi.e.,P(T1D)=10x...(ii) Also, P(T1)=10020andP(T2)=10080Given, P(,)=1007 defective computer P(D)=1007i.e.,P(D)=P(T1)⋅P(T1D)+P(T2)⋅(T2D)Using law of total probability, ∴1007=(10020)⋅10x+(10080)⋅x⇒7=(280)x⇒x=401∴P(T2D)=401....(iii) and P(T1D)=4010⇒P(T2D)=1−401=4039P(T1D)=4030 and P(DT2)=P(T1)⋅P(T1D)+P(T2)⋅P(T2D)P(T2)⋅P(T2D)...(iv) Using Baye's theorem,=10020⋅4030+10080⋅403910080⋅4039=9378.7947