1. Define events

Let:

• $T_1$ = computer is produced in plant $T_1$
• $T_2$ = computer is produced in plant $T_2$
• $D$ = computer is defective
• $\overline D$ = computer is not defective

Given: $P(T_1)=0.2=\frac15, \qquad P(T_2)=0.8=\frac45$ $P(D)=0.07=\frac{7}{100}$

Also, $P(D\mid T_1)=10\,P(D\mid T_2)$

2. Assume defect probabilities

Let $P(D\mid T_2)=x$ Then $P(D\mid T_1)=10x$

Using total probability: $P(D)=P(T_1)P(D\mid T_1)+P(T_2)P(D\mid T_2)$ So, $\frac15(10x)+\frac45(x)=\frac{7}{100}$

Simplify: $2x+\frac45x=\frac{7}{100}$ $\frac{14}{5}x=\frac{7}{100}$ $x=\frac{7}{100}\cdot\frac{5}{14}=\frac{1}{40}$

Hence, $P(D\mid T_2)=\frac{1}{40}, \qquad P(D\mid T_1)=\frac{1}{4}$

3. Find non-defective probabilities

$P(\overline D\mid T_2)=1-\frac{1}{40}=\frac{39}{40}$ $P(\overline D\mid T_1)=1-\frac{1}{4}=\frac{3}{4}$

Also, $P(\overline D)=1-P(D)=1-\frac{7}{100}=\frac{93}{100}$

4. Apply Bayes' theorem

We need: $P(T_2\mid \overline D)=\frac{P(T_2)P(\overline D\mid T_2)}{P(\overline D)}$

Substitute values: $P(T_2\mid \overline D)=\frac{\frac45\cdot\frac{39}{40}}{\frac{93}{100}}$

First simplify numerator: $\frac45\cdot\frac{39}{40}=\frac{39}{50}$

So, $P(T_2\mid \overline D)=\frac{39}{50}\cdot\frac{100}{93}$ $=\frac{78}{93}$

This can further reduce to $\frac{26}{31}$, but among the options it appears as: $\boxed{\frac{78}{93}}$

5. Check options

• A: $\frac{36}{73}$ ❌
• B: $\frac{47}{79}$ ❌
• C: $\frac{78}{93}$ ✅
• D: $\frac{75}{83}$ ❌

Therefore, the correct option is C.