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


We can solve this problem using **Bayes' Theorem**, which helps us find the probability of an event given some prior information.

In this case, we want to find the probability that the defective part came from supplier \( S1 \) given that it is defective. This can be expressed as:

\[
P(S1 \mid D)
\]

Where:
- \( S1 \) is the event that the part came from supplier S1.
- \( D \) is the event that the part is defective.

Bayes' Theorem formula is:

\[
P(S1 \mid D) = \frac{P(D \mid S1) P(S1)}{P(D)}
\]

Where:
- \( P(D \mid S1) \) is the probability that a part is defective given that it came from S1.
- \( P(S1) \) is the prior probability that a part came from S1.
- \( P(D) \) is the total probability that a part is defective, regardless of the supplier.

### Step 1: Assign values to the known probabilities.

- \( P(S1) = 0.50 \) (since 50% of the parts come from S1)
- \( P(S2) = 0.20 \) (since 20% of the parts come from S2)
- \( P(S3) = 0.30 \) (since 30% of the parts come from S3)

- \( P(D \mid S1) = 0.05 \) (5% of the parts from S1 are defective)
- \( P(D \mid S2) = 0.03 \) (3% of the parts from S2 are defective)
- \( P(D \mid S3) = 0.06 \) (6% of the parts from S3 are defective)

### Step 2: Find \( P(D) \) (the total probability that a part is defective).

Using the law of total probability, we calculate \( P(D) \) as follows:

\[
P(D) = P(D \mid S1)P(S1) + P(D \mid S2)P(S2) + P(D \mid S3)P(S3)
\]

Substituting the known values:

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

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

### Step 3: Use Bayes' Theorem to find \( P(S1 \mid D) \).

Now, apply the values to Bayes' Theorem:

\[
P(S1 \mid D) = \frac{(0.05)(0.50)}{0.049}
\]

\[
P(S1 \mid D) = \frac{0.025}{0.049} \approx 0.5102
\]

### Conclusion:

The probability that the defective part was supplied by S1 is approximately **0.5102**, or **51.02%**.

---

Would you like any further details or have any questions? Here are 5 related questions to consider:

1. How would the result change if the defect rates for the suppliers were different?
2. Can you apply Bayes' Theorem in other contexts outside defect rates, like medical tests?
3. What if the supplier S3 had a 10% defect rate? How would that impact the outcome?
4. How do we interpret the overall defect rate (P(D)) in a business context?
5. Can Bayes'

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. 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 using Bayes' Theorem. A computer maker sources parts from three suppliers (S1, S2, S3) with differing defect rates. Learn how to calculate the probability that a defective part came from supplier S1, with step-by-step explanations of the law of total probability and Bayesian inference. Suitable for high school students (Grades 10-12).

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