Three machines I, II, and III manufacture 30%, 30%, and 40%, respectively, of the total output of 
certain items. Of them, 4%, 3%, and 2%, respectively are defective. One item is drawn at random 
from the total output and is tested.
(i) What is the probability that the item is defective?
(ii) If it is found to be defective, what is the probability the item was produced by machine I?
(iii) Same question as in part (ii) for each one of the machines II and III

Given the problem, we have three machines (I, II, and III) producing a certain percentage of items, with each machine having a certain defect rate. We need to solve for the probability that an item is defective and the probability that a defective item was produced by each specific machine.

### Let's define the problem:

- **Machine I:**
  - Produces 30% of the total items.
  - Defective rate: 4%.
  
- **Machine II:**
  - Produces 30% of the total items.
  - Defective rate: 3%.
  
- **Machine III:**
  - Produces 40% of the total items.
  - Defective rate: 2%.

### (i) Probability that an item is defective:

This can be calculated using the law of total probability. Let \( D \) be the event that the item is defective.

\[
P(D) = P(D | I) \cdot P(I) + P(D | II) \cdot P(II) + P(D | III) \cdot P(III)
\]

Where:
- \( P(D | I) = 0.04 \)
- \( P(I) = 0.30 \)
- \( P(D | II) = 0.03 \)
- \( P(II) = 0.30 \)
- \( P(D | III) = 0.02 \)
- \( P(III) = 0.40 \)

Substituting the values:

\[
P(D) = (0.04 \times 0.30) + (0.03 \times 0.30) + (0.02 \times 0.40)
\]
\[
P(D) = 0.012 + 0.009 + 0.008 = 0.029
\]

So, the probability that the item is defective is **0.029** or **2.9%**.

### (ii) Probability that the defective item was produced by Machine I:

This is a conditional probability problem. We need to find \( P(I | D) \), the probability that the defective item was produced by Machine I given that the item is defective. Using Bayes' theorem:

\[
P(I | D) = \frac{P(D | I) \cdot P(I)}{P(D)}
\]

Substituting the values:

\[
P(I | D) = \frac{0.04 \times 0.30}{0.029} = \frac{0.012}{0.029} \approx 0.414
\]

So, the probability that the defective item was produced by Machine I is approximately **0.414** or **41.4%**.

### (iii) Probability that the defective item was produced by Machine II and Machine III:

Similarly, we can find \( P(II | D) \) and \( P(III | D) \).

For Machine II:

\[
P(II | D) = \frac{P(D | II) \cdot P(II)}{P(D)}
\]

Substituting the values:

\[
P(II | D) = \frac{0.03 \times 0.30}{0.029} = \frac{0.009}{0.029} \approx 0.310
\]

So, the probability that the defective item was produced by Machine II is approximately **0.310** or **31.0%**.

For Machine III:

\[
P(III | D) = \frac{P(D | III) \cdot P(III)}{P(D)}
\]

Substituting the values:

\[
P(III | D) = \frac{0.02 \times 0.40}{0.029} = \frac{0.008}{0.029} \approx 0.276
\]

So, the probability that the defective item was produced by Machine III is approximately **0.276** or **27.6%**.

### Summary:

1. Probability that the item is defective: **2.9%**
2. Probability that the defective item was produced by Machine I: **41.4%**
3. Probability that the defective item was produced by Machine II: **31.0%**
4. Probability that the defective item was produced by Machine III: **27.6%**

Would you like more details or have any other questions?

Here are 5 related questions to deepen your understanding:
1. How would the probabilities change if Machine I had a higher defect rate?
2. What if Machine III produced 50% of the total items instead of 40%?
3. How would you calculate the probability if there were four machines instead of three?
4. What is the impact on the overall defect probability if all machines had the same defect rate?
5. How can you use these probabilities to improve quality control in manufacturing?

**Tip:** Bayes' theorem is a powerful tool for updating probabilities when new information is available, particularly in scenarios involving conditional probabilities like this one.

Explore the probability of a defective item being produced by machines I, II, and III using Bayes' theorem and the law of total probability. Understand how to calculate conditional probabilities in manufacturing scenarios. Suitable for advanced high school students and beyond.

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