A machine that manufactures automobile parts produces defective parts 12% of the time. If 8 parts produced by this machine are randomly selected, what is the probability that more than 1 of the parts are defective?

Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places

This problem can be solved using the binomial distribution. In this scenario, each part can be considered as a Bernoulli trial with two possible outcomes: either the part is defective or it is not.

### Given:
- Probability of a defective part, \( p = 0.12 \)
- Probability of a non-defective part, \( q = 1 - p = 0.88 \)
- Number of parts selected, \( n = 8 \)
- We need to find the probability that more than 1 part is defective, which means \( P(X > 1) \).

### Binomial Probability Formula:
The probability of exactly \( k \) defective parts out of \( n \) is given by:
\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k}
\]
where \( \binom{n}{k} \) is the binomial coefficient:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

### Step 1: Calculate \( P(X = 0) \) and \( P(X = 1) \)
- **For \( X = 0 \):**
\[
P(X = 0) = \binom{8}{0} \cdot (0.12)^0 \cdot (0.88)^8 = 1 \times 1 \times (0.88)^8 \approx 0.4228
\]

- **For \( X = 1 \):**
\[
P(X = 1) = \binom{8}{1} \cdot (0.12)^1 \cdot (0.88)^7 = 8 \times 0.12 \times (0.88)^7 \approx 0.4611
\]

### Step 2: Calculate \( P(X > 1) \)
To find the probability that more than 1 part is defective, we calculate:
\[
P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X = 0) + P(X = 1)]
\]

Substituting the values we found:
\[
P(X > 1) = 1 - [0.4228 + 0.4611] = 1 - 0.8839 = 0.1161
\]

### Final Answer:
The probability that more than 1 of the parts is defective is approximately **0.12** (rounded to two decimal places).

Would you like more details, or do you have any further questions?

---

1. How is the binomial distribution applied in quality control scenarios?
2. What is the significance of the binomial coefficient in the binomial probability formula?
3. Can you explain why we subtract the probabilities of 0 and 1 defective parts from 1?
4. What does the result imply about the machine's production process?
5. How would the result change if the probability of a defective part increased?
6. How does increasing the sample size affect the probability of detecting defective parts?
7. What is the relationship between binomial and normal distributions?
8. How can this probability be interpreted in the context of statistical process control?

**Tip:** When dealing with small probabilities in quality control, binomial distribution is often the first choice for modeling.

Calculate the probability that more than 1 out of 8 randomly selected automobile parts are defective using the binomial distribution. Understand how to apply the binomial probability formula and binomial coefficient in quality control scenarios.

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