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 bedefective, 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

Options

A. `36/73`
B. `47/79`
C. `78/96`
D. `75/83`

Explanation

To solve the problem step by step, we will use the information provided and apply the concepts of probability.

### Step 1: Define the probabilities for each plant
Let:
- \( P(T_1) = 0.2 \) (Probability that a computer is produced in plant \( T_1 \))
- \( P(T_2) = 0.8 \) (Probability that a computer is produced in plant \( T_2 \))

### Step 2: Define the probability of a computer being defective
Let:
- \( P(D) = 0.07 \) (Probability that a computer is defective)

### Step 3: Define the conditional probabilities of defectiveness
Let:
- \( P(D | T_1) = 10x \) (Probability that a computer is defective given it is produced in plant \( T_1 \))
- \( P(D | T_2) = x \) (Probability that a computer is defective given it is produced in plant \( T_2 \))

From the problem, we know that:
\[ P(D | T_1) = 10 P(D | T_2) \]
Thus, we can express \( P(D | T_1) \) in terms of \( x \):
\[ P(D | T_1) = 10x \]
\[ P(D | T_2) = x \]

### Step 4: Calculate the total probability of defectiveness
Using the law of total probability:
\[ P(D) = P(T_1) \cdot P(D | T_1) + P(T_2) \cdot P(D | T_2) \]
Substituting the values:
\[ 0.07 = 0.2 \cdot (10x) + 0.8 \cdot x \]
\[ 0.07 = 2x + 0.8x \]
\[ 0.07 = 2.8x \]

### Step 5: Solve for \( x \)
\[ x = \frac{0.07}{2.8} = \frac{7}{280} = \frac{1}{40} \]

### Step 6: Calculate \( P(D | T_1) \) and \( P(D | T_2) \)
Now we can find:
\[ P(D | T_1) = 10x = 10 \cdot \frac{1}{40} = \frac{10}{40} = \frac{1}{4} \]
\[ P(D | T_2) = x = \frac{1}{40} \]

### Step 7: Calculate the probability of not being defective
Let \( P(D') \) be the probability that a computer is not defective:
\[ P(D') = 1 - P(D) = 1 - 0.07 = 0.93 \]

### Step 8: Calculate \( P(D' | T_1) \) and \( P(D' | T_2) \)
\[ P(D' | T_1) = 1 - P(D | T_1) = 1 - \frac{1}{4} = \frac{3}{4} \]
\[ P(D' | T_2) = 1 - P(D | T_2) = 1 - \frac{1}{40} = \frac{39}{40} \]

### Step 9: Use Bayes' theorem to find \( P(T_2 | D') \)
We want to find \( P(T_2 | D') \):
\[ P(T_2 | D') = \frac{P(T_2) \cdot P(D' | T_2)}{P(T_2) \cdot P(D' | T_2) + P(T_1) \cdot P(D' | T_1)} \]
Substituting the values:
\[ P(T_2 | D') = \frac{0.8 \cdot \frac{39}{40}}{0.8 \cdot \frac{39}{40} + 0.2 \cdot \frac{3}{4}} \]
Calculating the denominator:
\[ = \frac{0.8 \cdot \frac{39}{40} + 0.2 \cdot \frac{3}{4}}{0.8 \cdot \frac{39}{40} + 0.15} \]
Calculating:
\[ = \frac{0.8 \cdot \frac{39}{40} + 0.15}{0.8 \cdot \frac{39}{40} + 0.15} \]
After simplifying, we find:
\[ P(T_2 | D') = \frac{78}{93} \]

### Final Answer
Thus, the probability that a randomly selected computer, which is not defective, was produced in plant \( T_2 \) is:
\[ \frac{78}{93} \]