A computer maker receives parts from three suppliers, S1, S2, and S3. Fifty percent come 
from S1, twenty percent from S2, and thirty percent from S3. Among all the parts supplied by 
S1, 5% are defective. For S2 and S3, the portion of defective parts is 3% and 6%, 
respectively. What portion of all the parts is defective? A customer complains that a certain 
part in her recently purchased computer is defective. What is the probability that it was 
supplied by S1?

Let's break this problem into two parts:

### Part 1: What portion of all the parts is defective?

We can solve this part using the **law of total probability**.

Define the following events:
- \( D \): The event that a part is defective.
- \( S_1 \): The event that a part comes from supplier \( S_1 \).
- \( S_2 \): The event that a part comes from supplier \( S_2 \).
- \( S_3 \): The event that a part comes from supplier \( S_3 \).

The problem provides the following information:
- \( P(S_1) = 0.50 \) (50% of parts come from \( S_1 \))
- \( P(S_2) = 0.20 \) (20% of parts come from \( S_2 \))
- \( P(S_3) = 0.30 \) (30% of parts come from \( S_3 \))

Also, the probability that a part is defective given that it came from each supplier:
- \( P(D | S_1) = 0.05 \) (5% of \( S_1 \)'s parts are defective)
- \( P(D | S_2) = 0.03 \) (3% of \( S_2 \)'s parts are defective)
- \( P(D | S_3) = 0.06 \) (6% of \( S_3 \)'s parts are defective)

The total probability that a part is defective can be found using the law of total probability:

\[
P(D) = P(D | S_1) \cdot P(S_1) + P(D | S_2) \cdot P(S_2) + P(D | S_3) \cdot P(S_3)
\]

Now, substitute the values:

\[
P(D) = (0.05 \times 0.50) + (0.03 \times 0.20) + (0.06 \times 0.30)
\]

\[
P(D) = 0.025 + 0.006 + 0.018 = 0.049
\]

Thus, the total portion of defective parts is **4.9%**.

### Part 2: What is the probability that the defective part was supplied by \( S_1 \)?

For this part, we can use **Bayes' Theorem**. We want to find \( P(S_1 | D) \), the probability that a defective part came from \( S_1 \), given that the part is defective.

Bayes' Theorem states:

\[
P(S_1 | D) = \frac{P(D | S_1) \cdot P(S_1)}{P(D)}
\]

We already know the following:
- \( P(D | S_1) = 0.05 \)
- \( P(S_1) = 0.50 \)
- \( P(D) = 0.049 \)

Now substitute the values:

\[
P(S_1 | D) = \frac{0.05 \times 0.50}{0.049}
\]

\[
P(S_1 | D) = \frac{0.025}{0.049} \approx 0.5102
\]

Thus, the probability that a defective part was supplied by \( S_1 \) is approximately **51.02%**.

### Summary:
1. **The portion of all parts that are

A computer maker receives parts from three suppliers, S1, S2, and S3. Fifty percent come from S1, twenty percent from S2, and thirty percent from S3. Among all the parts supplied by S1, 5% are defective. For S2 and S3, the portion of defective parts is 3% and 6%, respectively. What portion of all the parts is defective? A customer complains that a certain part in her recently purchased computer is defective. What is the probability that it was supplied by S1?

Solve a probability problem involving defective parts from three suppliers using the law of total probability and Bayes' Theorem. Learn how to calculate the overall defective rate and the probability that a defective part came from a specific supplier. Suitable for advanced high school or college-level students.

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