24. A factory produces items from two machines: Machine A and Machine B. Machine

A produces 60% of the total items, while Machine B produces 40%. The probability

that an item produced by Machine A is defective is P(D|A) = 0.03. The probability

that an item produced by Machine B is defective is P(D|B) = 0.05.

(a) What is the probability that a randomly selected product be defective, P(D)?

(b) If a randomly selected item from the production line is defective, calculate the

probability that it was produced by Machine A, P(A|D).

Let's break down the problem step by step.

### Given Information:
- **Machine A** produces 60% of the items → \( P(A) = 0.60 \)
- **Machine B** produces 40% of the items → \( P(B) = 0.40 \)
- The probability that an item produced by **Machine A** is defective → \( P(D|A) = 0.03 \)
- The probability that an item produced by **Machine B** is defective → \( P(D|B) = 0.05 \)

We are asked to calculate the following:

### Part (a): Probability that a randomly selected product is defective, \( P(D) \)

We will use **Total Probability Theorem** for this. The probability that a randomly selected item is defective can be found by considering both machines:

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

Substituting the given values:

\[
P(D) = (0.03 \times 0.60) + (0.05 \times 0.40)
\]
\[
P(D) = 0.018 + 0.02 = 0.038
\]

So, the probability that a randomly selected product is defective is \( \boxed{0.038} \), or 3.8%.

---

### Part (b): Probability that an item was produced by Machine A, given that it is defective, \( P(A|D) \)

We will use **Bayes' Theorem** to find this conditional probability. Bayes' Theorem states:

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

We already know the values:
- \( P(D|A) = 0.03 \)
- \( P(A) = 0.60 \)
- \( P(D) = 0.038 \)

Substituting these values into Bayes' Theorem:

\[
P(A|D) = \frac{(0.03 \times 0.60)}{0.038}
\]
\[
P(A|D) = \frac{0.018}{0.038} \approx 0.4737
\]

So, the probability that a defective item was produced by **Machine A** is approximately \( \boxed{0.4737} \), or about 47.37%.

---

### Summary:
- (a) The probability that a randomly selected product is defective is 0.038.
- (b) The probability that a defective product came from Machine A is approximately 0.4737.

---

Would you like more details or clarification on any part? Here are some related questions you can explore:

1. What happens if the probability of defects changes for either machine?
2. How does increasing Machine A's defect rate affect \( P(A|D) \)?
3. What if the distribution between the machines changes (e.g., 50% from Machine A and 50% from Machine B)?
4. How do you calculate the expected number of defective items if 1000 products are produced?
5. How would this problem be approached if more than two machines were involved?

**Tip:** When using Bayes' Theorem, always check if you have the full set of conditional probabilities, priors, and total probability before calculating.

A factory produces items from two machines: Machine A and Machine B. Machine A produces 60% of the total items, while Machine B produces 40%. The probability that an item produced by Machine A is defective is P(D|A) = 0.03. The probability that an item produced by Machine B is defective is P(D|B) = 0.05. (a) What is the probability that a randomly selected product be defective, P(D)? (b) If a randomly selected item from the production line is defective, calculate the probability that it was produced by Machine A, P(A|D).

Explore how to calculate the probability of a defective item and the probability that it was produced by Machine A using Bayes' Theorem. This problem involves key concepts in probability, such as conditional probability and the Total Probability Theorem, and is ideal for students in Grades 10-12.

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