Question

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

• A. \( \frac{36}{73} \)
• B. \( \frac{47}{79} \)
• C. \( \frac{78}{93} \)
• D. \( \frac{75}{83} \)

Hint:

To solve problems involving conditional probability, use Bayes' Theorem to calculate the desired probability based on known conditions.

Answer 1

The Correct Option is C

Let the total number of computers produced be 100. Then:
- \( P(T_1) = 0.20 \) (probability that a computer is produced in plant \(T_1\))
- \( P(T_2) = 0.80 \) (probability that a computer is produced in plant \(T_2\))
Let \( P(D|T_1) \) represent the probability that a computer is defective given it is produced in \(T_1\), and \( P(D|T_2) \) represent the probability that a computer is defective given it is produced in \(T_2\). We are given that:
\[ P(D|T_1) = 10P(D|T_2) \] Using the law of total probability, we know that the total probability of selecting a defective computer is:
\[ P(D) = P(D|T_1)P(T_1) + P(D|T_2)P(T_2) \] Substitute the values into the equation:
\[ P(D) = 10P(D|T_2) \times 0.20 + P(D|T_2) \times 0.80 \] Simplifying:
\[ P(D) = P(D|T_2) \times (2 + 0.80) = P(D|T_2) \times 2.8 \] We are also told that 7% of the computers are defective, so \( P(D) = 0.07 \). Therefore:
\[ 0.07 = 2.8P(D|T_2) \] \[ P(D|T_2) = \frac{0.07}{2.8} = 0.025 \] Now we can calculate \( P(D|T_1) \):
\[ P(D|T_1) = 10P(D|T_2) = 10 \times 0.025 = 0.25 \] Next, we want to find the probability that the computer was produced in \(T_2\) given that it is not defective. Using Bayes' Theorem:
\[ P(T_2| \neg D) = \frac{P(\neg D|T_2)P(T_2)}{P(\neg D)} \] Where:
\[ P(\neg D|T_2) = 1 - P(D|T_2) = 1 - 0.025 = 0.975 \] \[ P(\neg D) = 1 - P(D) = 1 - 0.07 = 0.93 \] Substitute the values into Bayes' Theorem:
\[ P(T_2| \neg D) = \frac{0.975 \times 0.80}{0.93} = \frac{0.78}{0.93} = \frac{78}{93} \] Thus, the probability that the computer was produced in plant \( T_2 \) given that it is not defective is \( \boxed{\frac{78}{93}} \).