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 $P(\text{computer turns out to be defective given that it is produced in plant } T_1) = 10P(\text{computer turns out to be defective given that it is produced in plant } T_2)$, 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

$\text{Given: } P(T_1) = \dfrac{20}{100}, \quad P(T_2) = \dfrac{80}{100}$

$P(D) = \dfrac{7}{100}, \quad P(D\mid T_1) = 10\,P(D\mid T_2)$

$\text{To find: } P(T_2 \mid D^{c})$

$P(T_2 \mid D^{c}) = \dfrac{P(T_2)\,P(D^{c} \mid T_2)}{P(D^{c})}$ $\text{Let } P(D\mid T_2) = x \Rightarrow P(D\mid T_1) = 10x$ $P(D) = P(T_1)P(D\mid T_1) + P(T_2)P(D\mid T_2)$ $0.2(10x) + 0.8x = 0.07$ $2.8x = 0.07 \Rightarrow x = \dfrac{0.07}{2.8} = 0.025 = \dfrac{1}{40}$ $P(D\mid T_2) = \dfrac{1}{40}, \quad P(D\mid T_1) = 10x = \dfrac{1}{4}$ $P(D) = 0.2 \times \dfrac{1}{4} + 0.8 \times \dfrac{1}{40}$ $= 0.05 + 0.02 = 0.07\ \checkmark$ $P(D^{c}) = 1 - P(D) = 1 - 0.07 = 0.93$ $P(T_2 \cap D^{c}) = P(T_2)\,P(D^{c}\mid T_2) = 0.8 \times 0.975 = 0.78$ $P(T_2 \mid D^{c}) = \dfrac{0.78}{0.93} = \dfrac{78}{93} = \dfrac{26}{31}$ $\text{Answer: } \dfrac{78}{93} = \dfrac{26}{31} \Rightarrow \text{option (3)}$