In a tape recorder factory three machines A, B and C produced 50%, 30% and 20% of total production. The percentage of the defective output of these machines are 3%, 4% and 5% respectively. A tape recorder is selected randomly and found to be defective, find the probability that it is produced by machine A.

Answer

To solve the problem, we will use Bayes' theorem. We need to find the probability that a defective tape recorder was produced by machine A, given that it is defective.

### Step-by-Step Solution:

1. **Define the Events:**
- Let \( A \) be the event that a tape recorder is produced by machine A.
- Let \( B \) be the event that a tape recorder is produced by machine B.
- Let \( C \) be the event that a tape recorder is produced by machine C.
- Let \( D \) be the event that a tape recorder is defective.

2. **Given Information:**
- Probability of production by machine A: \( P(A) = 0.50 \)
- Probability of production by machine B: \( P(B) = 0.30 \)
- Probability of production by machine C: \( P(C) = 0.20 \)
- Probability of a defective tape recorder from machine A: \( P(D|A) = 0.03 \)
- Probability of a defective tape recorder from machine B: \( P(D|B) = 0.04 \)
- Probability of a defective tape recorder from machine C: \( P(D|C) = 0.05 \)

3. **Calculate Total Probability of Defectiveness \( P(D) \):**
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:
\[ P(D) = (0.03 \cdot 0.50) + (0.04 \cdot 0.30) + (0.05 \cdot 0.20) \]
\[ P(D) = 0.015 + 0.012 + 0.01 = 0.037 \]

4. **Apply Bayes' Theorem:**
We want to find \( P(A|D) \):
\[ P(A|D) = \frac{P(D|A) \cdot P(A)}{P(D)} \]
Substituting the known values:
\[ P(A|D) = \frac{0.03 \cdot 0.50}{0.037} \]
\[ P(A|D) = \frac{0.015}{0.037} \]

5. **Calculate the Final Probability:**
\[ P(A|D) \approx 0.4054 \]
Converting to a fraction:
\[ P(A|D) = \frac{15}{37} \]

### Final Answer:
The probability that the defective tape recorder was produced by machine A is \( \frac{15}{37} \).