2. Inspecting incoming car parts from suppliers is a crucial step in automobile industry quality
control procedures. Let's say they received a shipment of 50 car spoiler parts. Each part has
a probability of 0.1 being defective. After thoroughly examining them, they'll return the
whole shipment if 20% or more defectives were found.
a. What is the probability of a whole shipment to get returned?
b. What is the probability of finding at most 5 defects per shipment?
c. Determine the average number of defects found in a shipment

We can solve this problem by modeling it as a binomial distribution. Each car spoiler part can either be defective or non-defective, with a given probability. The probability of finding a defective part is 0.1, and there are 50 parts in the shipment.

Let’s break down the three parts of the problem:

---

### a. **Probability of the whole shipment being returned**:
The shipment will be returned if 20% or more of the parts are defective. Since there are 50 parts, 20% of 50 is:

\[
20\% \times 50 = 10 \text{ defective parts}.
\]

We need to calculate the probability of finding 10 or more defective parts in the shipment. This can be modeled using the binomial distribution formula:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k},
\]
where:
- \( n = 50 \) (total number of parts),
- \( p = 0.1 \) (probability of a defective part),
- \( k \) is the number of defective parts.

We want \( P(X \geq 10) \), i.e., the probability that 10 or more parts are defective. This is:

\[
P(X \geq 10) = 1 - P(X \leq 9).
\]

To solve this, we need the cumulative binomial probability for \( X \leq 9 \). I will calculate this probability.

---

### b. **Probability of finding at most 5 defects per shipment**:
This means we need to calculate \( P(X \leq 5) \), which is the probability of finding 5 or fewer defective parts. This is a cumulative binomial probability:

\[
P(X \leq 5) = \sum_{k=0}^{5} \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}.
\]

I will calculate this value as well.

---

### c. **Average number of defects in a shipment**:
The expected value \( E(X) \) of a binomial distribution is given by:

\[
E(X) = n \cdot p,
\]
where \( n = 50 \) and \( p = 0.1 \).

Thus, the average number of defects is:

\[
E(X) = 50 \times 0.1 = 5.
\]

---

Let me now compute the exact probabilities for parts (a) and (b).Here are the results:

### a. **Probability of the whole shipment being returned**:
The probability of finding 10 or more defective parts (and hence returning the shipment) is approximately:

\[
P(X \geq 10) = 0.0245 \quad \text{(or 2.45%)}.
\]

### b. **Probability of finding at most 5 defects per shipment**:
The probability of finding 5 or fewer defective parts is approximately:

\[
P(X \leq 5) = 0.6161 \quad \text{(or 61.61%)}.
\]

### c. **Average number of defects in a shipment**:
The expected (average) number of defective parts in a shipment is:

\[
E(X) = 5 \quad \text{defective parts}.
\]

---

Would you like more details on any of these solutions, or do you have any other questions?

### Related questions:
1. What is the probability of finding exactly 7 defective parts?
2. What is the probability of finding fewer than 4 defective parts?
3. How would the probabilities change if the defective rate increased to 15%?
4. What is the probability of returning the shipment if the threshold is 15% defects?
5. Can you calculate the variance and standard deviation of the number of defects?

### Tip:
The expected number of defects is proportional to the total number of items and the defect probability, making it easy to scale across different shipments.

Inspecting incoming car parts from suppliers is a crucial step in automobile industry quality control procedures. Let's say they received a shipment of 50 car spoiler parts. Each part has a probability of 0.1 being defective. After thoroughly examining them, they'll return the whole shipment if 20% or more defectives were found. a. What is the probability of a whole shipment to get returned? b. What is the probability of finding at most 5 defects per shipment? c. Determine the average number of defects found in a shipment.

Learn how to calculate the probability of returning a shipment of car parts based on defect rates. This problem involves the binomial distribution, expected values, and cumulative probability, with solutions for finding 20% or more defective parts, as well as the probability of finding at most 5 defects. Suitable for Grades 11-12.

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