In a bolt factory machines A,B,C manufacture respectively 25, 35 and 40 percent of the total. Out of their output 5,4 and 2 percent are defective bolts. A bolt is drawn from the produce and is found defective. What is the probability that it was manufactured by A?

Answer

To solve the problem, we will use Bayes' theorem to find the probability that a defective bolt was manufactured by machine A.

### Step-by-Step Solution:

1. **Identify the probabilities of production by each machine**:
- Let \( P(A) \) be the probability that a bolt is produced by machine A.
- Let \( P(B) \) be the probability that a bolt is produced by machine B.
- Let \( P(C) \) be the probability that a bolt is produced by machine C.

From the problem, we have:
\[ P(A) = 0.25, \quad P(B) = 0.35, \quad P(C) = 0.40 \]

2. **Identify the probabilities of a bolt being defective given the machine**:
- Let \( P(D|A) \) be the probability that a bolt is defective given it was produced by machine A.
- Let \( P(D|B) \) be the probability that a bolt is defective given it was produced by machine B.
- Let \( P(D|C) \) be the probability that a bolt is defective given it was produced by machine C.

From the problem, we have:
\[ P(D|A) = 0.05, \quad P(D|B) = 0.04, \quad P(D|C) = 0.02 \]

3. **Calculate the total probability of drawing a defective bolt, \( P(D) \)**:
Using the law of total probability:
\[ P(D) = P(D|A) \cdot P(A) + P(D|B) \cdot P(B) + P(D|C) \cdot P(C) \]
Substituting the values:
\[ P(D) = (0.05 \times 0.25) + (0.04 \times 0.35) + (0.02 \times 0.40) \]
\[ P(D) = 0.0125 + 0.014 + 0.008 = 0.0345 \]

4. **Calculate the probability that a defective bolt was manufactured by machine A, \( P(A|D) \)**:
Using Bayes' theorem:
\[ P(A|D) = \frac{P(D|A) \cdot P(A)}{P(D)} \]
Substituting the values:
\[ P(A|D) = \frac{0.05 \times 0.25}{0.0345} \]
\[ P(A|D) = \frac{0.0125}{0.0345} \approx 0.3623 \]

5. **Final Result**:
The probability that a defective bolt was manufactured by machine A is approximately \( 0.3623 \) or \( 36.23\% \).