A company receives large shipments of parts from two sources. 70% of the shipments come from supplier A whose shipments typically contain 10% defectives while the remainders are from supplier B whose shipments typically contain 20% defectives. A manager receives
the shipment but doesn’t know the source. Randomly selected 20 items from this shipment are
tested and one of the parts is found to be defective. Find probability that this shipment came
from the more reliable supplier.

We can solve this problem using **Bayes' Theorem**, which is ideal for determining conditional probabilities. Here's the breakdown:

### Given:
- 70% of the shipments come from **supplier A** (reliable), i.e., \( P(A) = 0.7 \).
- 30% of the shipments come from **supplier B**, i.e., \( P(B) = 0.3 \).
- Supplier A’s shipments typically contain 10% defectives, i.e., \( P(D|A) = 0.1 \).
- Supplier B’s shipments typically contain 20% defectives, i.e., \( P(D|B) = 0.2 \).
- Out of 20 parts selected, 1 part is found to be defective.

We need to calculate the probability that the shipment came from supplier A, given that 1 defective part was found. Mathematically, this is \( P(A|D) \), where \( D \) indicates that 1 defective part is found out of the 20.

### Step 1: Applying Bayes' Theorem

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

Where:
- \( P(D|A) \): Probability of finding 1 defective part in 20 from supplier A.
- \( P(D|B) \): Probability of finding 1 defective part in 20 from supplier B.
- \( P(D) \): Total probability of finding 1 defective part in 20 from either supplier (A or B).

### Step 2: Calculating \( P(D|A) \) and \( P(D|B) \)

The number of defectives in a sample of 20 follows a **binomial distribution**. We are interested in the probability of exactly 1 defective part being found, i.e., \( k = 1 \) for a binomial distribution.

For supplier A:
\[
P(D|A) = \binom{20}{1} (0.1)^1 (0.9)^{19} = 20 \times 0.1 \times (0.9)^{19}
\]

For supplier B:
\[
P(D|B) = \binom{20}{1} (0.2)^1 (0.8)^{19} = 20 \times 0.2 \times (0.8)^{19}
\]

### Step 3: Total probability \( P(D) \)

The total probability \( P(D) \) is a weighted sum of the probabilities for supplier A and supplier B:
\[
P(D) = P(D|A) \cdot P(A) + P(D|B) \cdot P(B)
\]

### Step 4: Substituting into Bayes' Theorem

Finally, we substitute the values back into Bayes' Theorem:
\[
P(A|D) = \frac{P(D|A) \cdot P(A)}{P(D)}
\]

Let's compute the numerical values.The probability that the shipment came from the more reliable supplier (Supplier A) given that one defective part was found is approximately **91.62%**.

Would you like more details or have any questions?

Here are some related questions to deepen your understanding:
1. What would happen to the probability if two defective parts were found instead of one?
2. How does the size of the sample affect the probability outcome?
3. What is Bayes' Theorem, and how is it applied in real-world problems like this?
4. How would the result change if the defective rates for suppliers A and B were closer?
5. How can we generalize this approach to shipments with more than two suppliers?

**Tip:** Bayes' Theorem is particularly powerful for updating probabilities when new evidence is available. It's commonly used in medical diagnosis, spam filters, and reliability testing.

A company receives large shipments of parts from two sources. 70% of the shipments come from supplier A whose shipments typically contain 10% defectives while the remainders are from supplier B whose shipments typically contain 20% defectives. A manager receives the shipment but doesn’t know the source. Randomly selected 20 items from this shipment are tested and one of the parts is found to be defective. Find the probability that this shipment came from the more reliable supplier.

Learn how to calculate the probability that a shipment came from the more reliable supplier using Bayes' Theorem. This problem involves concepts like probability, binomial distribution, and a step-by-step application of Bayes' Theorem. Suitable for advanced high school or college-level students.

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