In a factory, machine A produces 40% of the output and machine B produces 60%. On the average, 9 items in 1000 produced by A are defective and 1 item in 250 produced by B is defective. An item drawn at random from a day’s output is defective.  What is the probability that it was produced by A or B.


This problem can be solved using **Bayes' Theorem**.

Let:
- \( A \) represent the event that the item was produced by machine A.
- \( B \) represent the event that the item was produced by machine B.
- \( D \) represent the event that the item is defective.

### Given Information:
- \( P(A) = 0.40 \) (40% of the output is produced by machine A)
- \( P(B) = 0.60 \) (60% of the output is produced by machine B)
- Defective rates:
  - \( P(D | A) = \frac{9}{1000} = 0.009 \) (the probability of an item being defective if it is produced by machine A)
  - \( P(D | B) = \frac{1}{250} = 0.004 \) (the probability of an item being defective if it is produced by machine B)

We are tasked with finding:
- \( P(A | D) \): the probability that the item was produced by machine A given that it is defective.
- \( P(B | D) \): the probability that the item was produced by machine B given that it is defective.

### Step 1: Find the total probability that the item is defective, \( P(D) \), using the Law of Total Probability:

\[
P(D) = P(D | A) P(A) + P(D | B) P(B)
\]

Substitute the given values:

\[
P(D) = (0.009 \times 0.40) + (0.004 \times 0.60)
\]
\[
P(D) = 0.0036 + 0.0024 = 0.006
\]

### Step 2: Use **Bayes' Theorem** to find the conditional probabilities.

#### (a) Probability that the defective item was produced by machine A:

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

Substitute the known values:

\[
P(A | D) = \frac{0.009 \times 0.40}{0.006} = \frac{0.0036}{0.006} = 0.6
\]

#### (b) Probability that the defective item was produced by machine B:

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

Substitute the known values:

\[
P(B | D) = \frac{0.004 \times 0.60}{0.006} = \frac{0.0024}{0.006} = 0.4
\]

### Final Answer:
- The probability that the defective item was produced by machine A is \( P(A | D) = 0.6 \).
- The probability that the defective item was produced by machine B is \( P(B | D) = 0.4 \).

Would you like further details on Bayes' Theorem or the Law of Total Probability?

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### 5 Relative Questions:
1. How does Bayes' Theorem apply in real-world problems?
2. What if machine C was added with a different defective rate, how would that affect the calculation?
3. Can you calculate the overall defect rate if the defect rate for machine A changes?
4. How can the Law of Total Probability be extended to more machines?
5. What is the importance of conditional probability in decision making?

### Tip:
Always ensure to correctly identify the "prior probabilities" (like \( P(A) \) and \( P(B) \)) and the "likelihoods" (like \( P(D | A) \) and \( P(D | B) \)) when applying Bayes' Theorem.

In a factory, machine A produces 40% of the output and machine B produces 60%. On the average, 9 items in 1000 produced by A are defective and 1 item in 250 produced by B is defective. An item drawn at random from a day’s output is defective. What is the probability that it was produced by A or B?

Explore the probability of a defective item being produced by machine A or B using Bayes' Theorem. Learn how to apply the Law of Total Probability to calculate conditional probabilities based on defect rates. This problem is ideal for students in Grades 10-12, covering core concepts of probability and conditional probability.

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