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Question

# A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant 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

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B

4779

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C

7893

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D

7583

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Solution

## The correct option is C 7893 Let x = P (computer turns out to be defective, given that it is produced in plant T2) ⇒ x=P(D|T2) ...(i) Where, D = Defective computer ∴P (computer turns out to be defective given that it is produced in plant T1) = 10x i.e P(D|T1)=10x ...(ii) Also, P(T1)=20100 and P(T2)=80100 Given, P (defective computer)=7100 i.e. P(D)=7100 Using law of total probability, P(D)=P(T1).P(D|T1)+P(T2).P(D|T2)∴7100=(20100).10x+(80100).x⇒7=(280)x⇒x=140 ...(iii) ∴P(D|T2)=140 and P(D|T1)=1040⇒P(¯D|T2)=1−140=3940 and P(¯D|T1)=1−1040=3040 ...(iv) Using Bayes' theorem, P(T2|¯D)=P(T2∩¯D)P(T1∩¯D)+P(T2∩¯D)=P(T2).P(¯D|T2)P(T1).P(¯D|T1)+P(T2).P(¯D|T2)80100.394020100.3040+80100.3940=7893

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