In a bolt factory, machines X. Y. Z manufacture bolt that are respectively 25%. 35% and 40% of the factor's total output. The machines X. Y, Z respectively produce 2%, 4% and 5% defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine X?

Options

A. `(5)/(39)`
B. `(11)/(39)`
C. `(20)/(39)`
D. `(34)/(39)`

Answer

To solve the problem, we will use Bayes' theorem. We want to find the probability that a defective bolt was produced by machine X, given that we know the bolt is defective.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Probability that a bolt is produced by machine X, \( P(X) = 0.25 \)
- Probability that a bolt is produced by machine Y, \( P(Y) = 0.35 \)
- Probability that a bolt is produced by machine Z, \( P(Z) = 0.40 \)
- Probability that a bolt from machine X is defective, \( P(D|X) = 0.02 \)
- Probability that a bolt from machine Y is defective, \( P(D|Y) = 0.04 \)
- Probability that a bolt from machine Z is defective, \( P(D|Z) = 0.05 \)

2. **Calculate the Total Probability of Picking a Defective Bolt, \( P(D) \):**
\[ P(D) = P(X) \cdot P(D|X) + P(Y) \cdot P(D|Y) + P(Z) \cdot P(D|Z) \]
\[ P(D) = (0.25 \cdot 0.02) + (0.35 \cdot 0.04) + (0.40 \cdot 0.05) \]
\[ P(D) = 0.005 + 0.014 + 0.02 = 0.039 \]

3. **Apply Bayes' Theorem to Find \( P(X|D) \):**
\[ P(X|D) = \frac{P(D|X) \cdot P(X)}{P(D)} \]
\[ P(X|D) = \frac{0.02 \cdot 0.25}{0.039} \]
\[ P(X|D) = \frac{0.005}{0.039} \]

4. **Calculate the Final Probability:**
\[ P(X|D) = \frac{5}{39} \]

### Conclusion:
The probability that the defective bolt was manufactured by machine X is \( \frac{5}{39} \).