JEE Advanced 2016 Paper 1 Mathematics Q4

Mathematics

Problem

A computer producing factory has only two plants T1T_1 and T2T_2. Plant T1T_1 produces 20% and plant T2T_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 T1)=10P(computer turns out to be defective given that it is produced in plant T2)P(\text{computer turns out to be defective given that it is produced in plant }T_1)=10P(\text{computer turns out to be defective given that it is produced in plant }T_2), where P(E)P(E) denotes the probability of an event EE. 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 T2T_2 is

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