Chapter 10: Problem 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 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) \(\frac{36}{73}\) (B) \(\frac{47}{79}\) (C) \(\frac{78}{93}\) (D) \(\frac{75}{83}\)
Short Answer
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Key Concepts
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