A computer producing factory has only two plants T<sub>1</sub> and T<sub>2</sub>. Plant T<sub>1</sub> produces 20% and plant T<sub>2</sub> 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 given that it is produced in plant T<sub>1</sub>) = 10 × P(computer turns out to be defective given that it is produced in plant T<sub>2</sub>). 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<sub>2</sub> is:
Let \(P(D|T_2) = p\). Then \(P(D|T_1) = 10p\).
\(0.2 \times 10p + 0.8 \times p = 0.07 \Rightarrow 2.8p = 0.07 \Rightarrow p = 0.025\)
So \(P(\text{not } D|T_1) = 0.75\) and \(P(\text{not } D|T_2) = 0.975\).
\(P(T_2|\text{not } D) = \frac{0.8 \times 0.975}{0.8 \times 0.975 + 0.2 \times 0.75} = \frac{0.78}{0.93} = \frac{78}{93}\)
Answer: (C)
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