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1)

A computer producing factory has only two plants   T1   and T2. Plant  T1 produces 20% and T2 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 be defective. given that it is produced in plant T1)= 10P  (Computer turns out to be defective, given that it is produced in plant T2, 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 T2, is


A) 3673

B) 4779

C) 7893

D) 7583

Answer:

Option C

Explanation:

Let x=P   (computer turns out to be defective, given that it is produced in plant T2)

        x=P(DT2)           .............(i)

where  , D= Defective computer

P (computer  turns out to be defective given that is produced  in plant  T1)=10x

 i.e,     P(DT1)=10x            .......(ii)

Also,   P(T1)=20100  and  P(T2)=80100

Given, P  (defective computer )=  7100

Using law of total probability,

 P(D)=9(T1).P(DT1)+P(T2).P(DT2)

       7100=(20100).10x+(80100).x

     7= (280)x

     x=140          ..........(iii)

    P(DT2)=140  and   P(DT1)=1040

    P(¯DT2)=1140=3940

and    P(¯DT1)=11040=3040                       .........(iv)

 Using Baye's theorem,

 P(T2D)=P(T2¯D)P(T1¯D)+P(T2¯D)

 =P(T2).P(¯DT2)P(T1).P(¯DT1)+P(T2).P(¯DT2)

 =80100.394020100.3040+80100.3940=7893