Let's solve the problem using the binomial distribution.

Given:

• Probability of a part being defective ($p$) = 0.13
• Number of parts selected ($n$) = 9
• We want to find the probability that more than 1 part is defective, i.e., $P(X > 1)$.

Step 1: Calculate $P(X = 0)$ and $P(X = 1)$

The binomial probability formula is:

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

a) Probability that no parts are defective ($P(X = 0)$):

$P(X = 0) = \binom{9}{0} \cdot (0.13)^0 \cdot (0.87)^9$ $P(X = 0) = 1 \cdot 1 \cdot 0.87^9 \approx 1 \times 0.4870 = 0.4870$

b) Probability that exactly 1 part is defective ($P(X = 1)$):

$P(X = 1) = \binom{9}{1} \cdot (0.13)^1 \cdot (0.87)^8$ $P(X = 1) = 9 \cdot 0.13 \cdot 0.87^8 \approx 9 \times 0.13 \times 0.5679 = 0.6642$

Step 2: Calculate $P(X \leq 1)$:

$P(X \leq 1) = P(X = 0) + P(X = 1)$ $P(X \leq 1) = 0.4870 + 0.4322 = 0.9192$

Step 3: Calculate $P(X > 1)$:

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9192 = 0.0808$

Final Answer:

The probability that more than 1 of the parts are defective is 0.08 (rounded to two decimal places).

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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 15%?
2. What is the probability that exactly 3 of the 9 parts are defective?
3. How is the binomial distribution used in quality control processes?
4. What is the cumulative distribution function (CDF) for this scenario?
5. How would you interpret the result in the context of manufacturing quality?
6. What is the expected number of defective parts in a batch of 9?
7. How does increasing the sample size affect the probability of detecting defects?
8. Can you calculate the variance and standard deviation of defective parts in this case?

Tip: When working with probabilities, always ensure the sum of all possible outcomes equals 1 to check your work.