A manufacturing firm produces steel pipes in three plants with daily production volumes of 500, 1000, and 2000 units respectively. According to past experience it is known that fractions of defective outputs produced by the three plants are respectively 0.005, 0.008 and 0.010 . if a pipe is selected from day's total production and found to be defective, what is the probability that it came from the first plant ?

Answer

To solve the problem, we will use Bayes' theorem to find the probability that a defective pipe came from the first plant.

### Step-by-Step Solution:

1. **Identify the total production and defective rates:**
- Plant A produces 500 units with a defect rate of 0.005.
- Plant B produces 1000 units with a defect rate of 0.008.
- Plant C produces 2000 units with a defect rate of 0.010.

2. **Calculate the total production:**
\[ \text{Total production} = 500 + 1000 + 2000 = 3500 \text{ units} \]

3. **Calculate the probabilities of selecting a pipe from each plant:**
- Probability of selecting from Plant A:
\[ P(A) = \frac{500}{3500} = \frac{1}{7} \]
- Probability of selecting from Plant B:
\[ P(B) = \frac{1000}{3500} = \frac{2}{7} \]
- Probability of selecting from Plant C:
\[ P(C) = \frac{2000}{3500} = \frac{4}{7} \]

4. **Calculate the probabilities of a pipe being defective given it came from each plant:**
- Probability of a defective pipe from Plant A:
\[ P(E|A) = 0.005 \]
- Probability of a defective pipe from Plant B:
\[ P(E|B) = 0.008 \]
- Probability of a defective pipe from Plant C:
\[ P(E|C) = 0.010 \]

5. **Calculate the total probability of selecting a defective pipe (Law of Total Probability):**
\[ P(E) = P(A) \cdot P(E|A) + P(B) \cdot P(E|B) + P(C) \cdot P(E|C) \]
Substituting the values:
\[ P(E) = \left(\frac{1}{7} \cdot 0.005\right) + \left(\frac{2}{7} \cdot 0.008\right) + \left(\frac{4}{7} \cdot 0.010\right) \]
\[ P(E) = \frac{0.005}{7} + \frac{0.016}{7} + \frac{0.040}{7} \]
\[ P(E) = \frac{0.005 + 0.016 + 0.040}{7} = \frac{0.061}{7} \]

6. **Use Bayes' theorem to find the probability that a defective pipe came from Plant A:**
\[ P(A|E) = \frac{P(E|A) \cdot P(A)}{P(E)} \]
Substituting the values:
\[ P(A|E) = \frac{0.005 \cdot \frac{1}{7}}{\frac{0.061}{7}} \]
Simplifying:
\[ P(A|E) = \frac{0.005}{0.061} \]

7. **Calculate the final probability:**
\[ P(A|E) = \frac{5}{61} \]

### Final Answer:
The probability that a defective pipe came from the first plant is \( \frac{5}{61} \).