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 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 \(T_1\)) = 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
Let \(P(D|T_2) = p\), so \(P(D|T_1) = 10p\).
\(0.2(10p) + 0.8p = 0.07 \Rightarrow p = 0.025\)
\(P(D'|T_1) = 0.75\) and \(P(D'|T_2) = 0.975\).
\(P(T_2|D') = \frac{0.8 \times 0.975}{0.8 \times 0.975 + 0.2 \times 0.75} = \frac{0.78}{0.93} = \frac{78}{93}\)
Answer: (C)
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