A machine that manufactures automobile parts produces defective parts 13% of the time. If 9 parts produced by this machine are randomly selected, what is the probability that more than 1 of the parts are defective?
arry your intermediate computations to at least four decimal places, and round your answer to two decimal places

To solve this problem, we'll use the binomial distribution. The probability of a specific number of defective parts \( k \) in \( n \) trials (or parts selected) is given by the binomial probability formula:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}
\]

where:
- \( n = 9 \) (the number of parts selected),
- \( p = 0.13 \) (the probability that a part is defective),
- \( k \) is the number of defective parts.

We need to find the probability that more than 1 part is defective, which means \( P(X > 1) \). This can be calculated as:

\[
P(X > 1) = 1 - P(X \leq 1)
\]

\( P(X \leq 1) \) is the sum of the probabilities that 0 or 1 part is defective:

\[
P(X \leq 1) = P(X = 0) + P(X = 1)
\]

Now, let's calculate these probabilities step by step:

### 1. Probability that no parts are defective (\( P(X = 0) \)):

\[
P(X = 0) = \binom{9}{0} \cdot (0.13)^0 \cdot (0.87)^9 = 1 \cdot 1 \cdot 0.87^9
\]

\[
P(X = 0) \approx 0.3987
\]

### 2. Probability that exactly 1 part is defective (\( P(X = 1) \)):

\[
P(X = 1) = \binom{9}{1} \cdot (0.13)^1 \cdot (0.87)^8 = 9 \cdot 0.13 \cdot 0.87^8
\]

\[
P(X = 1) \approx 0.5342
\]

### 3. Sum of the probabilities (\( P(X \leq 1) \)):

\[
P(X \leq 1) = 0.3987 + 0.5342 = 0.9329
\]

### 4. Probability that more than 1 part is defective (\( P(X > 1) \)):

\[
P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9329 = 0.0671
\]

### Final Answer:

The probability that more than 1 of the parts are defective is **0.07** (rounded to two decimal places).

---

Would you like any further details or have any questions?

Here are some related questions you might find interesting:
1. How would the probability change if the defect rate was 20%?
2. What is the probability that exactly 2 of the 9 parts are defective?
3. How does the binomial distribution apply to quality control in manufacturing?
4. Can you explain the meaning of the binomial coefficient in this context?
5. What is the expected number of defective parts in a sample of 9?
6. How would you calculate the variance of the number of defective parts?
7. What if 15 parts were selected instead of 9? How would the probability change?
8. How is this problem related to real-world quality assurance practices?

**Tip:** In problems involving the binomial distribution, recognizing when to use cumulative probabilities (like \( P(X \leq 1) \)) is key to simplifying calculations.

Explore the probability that more than 1 out of 9 automobile parts are defective using the binomial distribution. This problem involves calculating probabilities of defective parts with a given defect rate. Suitable for high school students.

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