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 )
= 10 P(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
Text Solution
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E 1 : Computer is produced by plant T 1
E 2 : Computer is produced by plant T 2
A : Computer is defective
Now, P(A/E 1 ) = 10P(A/E 2 )
Let 
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