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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 be defective 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 -

1. 73 36
2. 79 47
3. 93 78
4. 83 75

Correct answer: 93 78

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