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JEE Advanced 2016 Paper 1 Mathematics Question 40

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 the plant \(T_1\)) = 10P(computer turns out to be defective given that it is produced in the 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

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