In a bolt factory, machines A, B and C manufacture 60%, 25% and 15% respectively. Of the total of their outputs 1%, 2%, 1% are defectively from these machines respectively. A bold is drawn at random from the total production and found to be defective. From which machine, the defective bolt is most expected to have been manufactured?

Options

A. Machine A
B. Machine B
C. Machine C
D. None of these

Answer

To solve the problem, we need to find out from which machine (A, B, or C) the defective bolt is most likely to have originated. We will use Bayes' theorem to find the conditional probabilities.

### Step-by-Step Solution:

1. **Define the Events:**
- Let \( D \) be the event that a bolt is defective.
- Let \( A \), \( B \), and \( C \) be the events that the bolt is produced by machines A, B, and C respectively.

2. **Given Data:**
- Probability of selecting a bolt from machine A: \( P(A) = 0.6 \)
- Probability of selecting a bolt from machine B: \( P(B) = 0.25 \)
- Probability of selecting a bolt from machine C: \( P(C) = 0.15 \)
- Probability of a bolt being defective given it is from machine A: \( P(D|A) = 0.01 \)
- Probability of a bolt being defective given it is from machine B: \( P(D|B) = 0.02 \)
- Probability of a bolt being defective given it is from machine C: \( P(D|C) = 0.01 \)

3. **Calculate Total Probability of Defectiveness:**
We need to calculate \( P(D) \), the total probability of a bolt being defective:
\[ P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \]
Substituting the values:
\[ P(D) = (0.01 \times 0.6) + (0.02 \times 0.25) + (0.01 \times 0.15) \]
\[ P(D) = 0.006 + 0.005 + 0.0015 = 0.0125 \]

4. **Calculate Conditional Probabilities:**
Now we will calculate the conditional probabilities \( P(A|D) \), \( P(B|D) \), and \( P(C|D) \) using Bayes' theorem:
\[ P(A|D) = \frac{P(D|A)P(A)}{P(D)} \]
\[ P(B|D) = \frac{P(D|B)P(B)}{P(D)} \]
\[ P(C|D) = \frac{P(D|C)P(C)}{P(D)} \]

- For machine A:
\[ P(A|D) = \frac{0.01 \times 0.6}{0.0125} = \frac{0.006}{0.0125} = 0.48 \]

- For machine B:
\[ P(B|D) = \frac{0.02 \times 0.25}{0.0125} = \frac{0.005}{0.0125} = 0.4 \]

- For machine C:
\[ P(C|D) = \frac{0.01 \times 0.15}{0.0125} = \frac{0.0015}{0.0125} = 0.12 \]

5. **Conclusion:**
The probabilities are:
- \( P(A|D) = 0.48 \)
- \( P(B|D) = 0.4 \)
- \( P(C|D) = 0.12 \)

Since \( P(A|D) > P(B|D) > P(C|D) \), the defective bolt is most likely to have been manufactured by machine A.

### Final Answer:
The defective bolt is most expected to have been manufactured by **Machine A**.