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A computer producing factory has only tw...

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 bedefective, 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

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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) produced 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) denptes 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

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 the computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective givent 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

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 in the factory turn out to be defective. It is known that P (computer turns out to be defective givent 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

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 in the factory turn out to be defective. It is known that P (computer turns out to be defective givent 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

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