Example 54. A factory produces a certain type of output by three types of machines. The

respective daily production figures are Machine I: 3,000 units, Machine II: 2,500 units and Machine III: 4,500 units. = 800

Past experience shows that 1% of the output produced by Machine I is defective. The corresponding fraction of defectives for the other two machines are 1.2% and 2% respectively. An item is drawn at random from the day's production run and is found to be defective. What is the probability that it comes from the output of (a) Machine 1, (b) Machine II and (c) Machine III?

This problem can be solved using **Bayes' Theorem**, which helps in determining conditional probabilities. The event we are interested in is that an item is found defective, and we need to find the probability that this defective item came from Machine I, Machine II, or Machine III.

Let’s define the events:

- \( A_1 \): The item is produced by Machine I.
- \( A_2 \): The item is produced by Machine II.
- \( A_3 \): The item is produced by Machine III.
- \( D \): The item is defective.

We are asked to find:
1. \( P(A_1 | D) \) — the probability that the defective item came from Machine I.
2. \( P(A_2 | D) \) — the probability that the defective item came from Machine II.
3. \( P(A_3 | D) \) — the probability that the defective item came from Machine III.

### Step 1: Find the prior probabilities

The daily production by each machine gives us the prior probabilities \( P(A_1) \), \( P(A_2) \), and \( P(A_3) \).

- Total production = \( 3000 + 2500 + 4500 = 10000 \) units.
- \( P(A_1) = \frac{3000}{10000} = 0.3 \)
- \( P(A_2) = \frac{2500}{10000} = 0.25 \)
- \( P(A_3) = \frac{4500}{10000} = 0.45 \)

### Step 2: Find the likelihoods

The likelihoods represent the probability of drawing a defective item given that it came from a specific machine. These are provided in the problem:

- \( P(D | A_1) = 0.01 \) (1% defective from Machine I)
- \( P(D | A_2) = 0.012 \) (1.2% defective from Machine II)
- \( P(D | A_3) = 0.02 \) (2% defective from Machine III)

### Step 3: Apply the Law of Total Probability

We first need to find the overall probability of drawing a defective item, \( P(D) \), by using the law of total probability:

\[
P(D) = P(D | A_1)P(A_1) + P(D | A_2)P(A_2) + P(D | A_3)P(A_3)
\]

Substituting the known values:

\[
P(D) = (0.01)(0.3) + (0.012)(0.25) + (0.02)(0.45)
\]

\[
P(D) = 0.003 + 0.003 + 0.009 = 0.015
\]

### Step 4: Use Bayes' Theorem

Now we can apply Bayes' Theorem to calculate the conditional probabilities:

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

#### (a) Probability that the defective item came from Machine I:

\[
P(A_1 | D) = \frac{(0.01)(0.3)}{0.015} = \frac{0.003}{0.015} = 0.2
\]

#### (b) Probability that the defective item came from Machine II:

\[
P(A_2 | D) = \frac{(0.012)(0.25)}{0.015} = \frac{0.003}{0.015} = 0.2
\]

#### (c) Probability that the defective item came from Machine III:

\[
P(A_3 | D) = \frac{(0.02)(0.45)}{0.015} = \frac{0.009}{0.015} = 0.6
\]

### Final Answer:

- \( P(A_1 | D) = 0.2 \)
- \( P(A_2 | D) = 0.2 \)
- \( P(A_3 | D) = 0.6 \)

Thus, if the item is found to be defective:
- There is a **20% chance** it came from Machine I,
- A **20% chance** it came from Machine II, and
- A **60% chance** it came from Machine III.

---

Would you like further clarification on any of the steps or calculations? Here are some related questions to explore:

1. How does Bayes' Theorem apply to real-life quality control processes?
2. How would the result change if Machine III produced 5% defective items instead of 2%?
3. What happens if all machines produced the same number of units but had different defect rates?
4. How can we extend this method to more than three machines with similar conditions?
5. What would be the effect of increasing the overall production while keeping defect percentages constant?

**Tip:** Bayes' Theorem is particularly useful when we need to update our belief about a cause (which machine produced a defective item) based on observed evidence (the defect).

Example 54. A factory produces a certain type of output by three types of machines. The respective daily production figures are Machine I: 3,000 units, Machine II: 2,500 units, and Machine III: 4,500 units. Past experience shows that 1% of the output produced by Machine I is defective. The corresponding fraction of defectives for the other two machines are 1.2% and 2% respectively. An item is drawn at random from the day's production run and is found to be defective. What is the probability that it comes from the output of (a) Machine I, (b) Machine II, and (c) Machine III?

Solve a probability problem using Bayes' Theorem to determine the likelihood that a defective item comes from one of three machines. Explore daily production outputs and defect percentages for Machine I, II, and III. Learn about conditional probability and how the law of total probability is applied.

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