A factory received a shipment of 41 generators, and the vendor who sold the items knows there are 7 generators in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the generators in the sample are defective, he will refuse the shipment.

*For each of the following, give your responses as reduced fractions.*

If a sample of 7 generators is selected, find the probability that all in the sample are defective.

  

If a sample of 7 generators is selected, find the probability that none in the sample are defective.


We are dealing with a scenario involving probability without replacement, which involves hypergeometric probability. 

We have:

- Total generators: 41
- Defective generators: 7
- Non-defective generators: \(41 - 7 = 34\)
- Sample size: 7

We will compute two probabilities: 
1. The probability that all 7 generators in the sample are defective.
2. The probability that none of the 7 generators in the sample are defective.

### 1. Probability that all 7 generators are defective:
We need to select 7 defective generators from the 7 defective ones in the shipment.

The probability is computed as:

\[
P(\text{all defective}) = \frac{\binom{7}{7} \times \binom{34}{0}}{\binom{41}{7}}
\]

Simplifying:

\[
P(\text{all defective}) = \frac{1 \times 1}{\binom{41}{7}} = \frac{1}{\binom{41}{7}}
\]

We now calculate \( \binom{41}{7} \), which is the number of ways to choose 7 generators from 41.

\[
\binom{41}{7} = \frac{41 \times 40 \times 39 \times 38 \times 37 \times 36 \times 35}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 17259390
\]

Thus, the probability is:

\[
P(\text{all defective}) = \frac{1}{17259390}
\]

### 2. Probability that none of the 7 generators are defective:
We need to select 7 non-defective generators from the 34 non-defective ones.

The probability is computed as:

\[
P(\text{none defective}) = \frac{\binom{7}{0} \times \binom{34}{7}}{\binom{41}{7}}
\]

Simplifying:

\[
P(\text{none defective}) = \frac{1 \times \binom{34}{7}}{\binom{41}{7}} = \frac{\binom{34}{7}}{\binom{41}{7}}
\]

We already know that \( \binom{41}{7} = 17259390 \). Now we compute \( \binom{34}{7} \):

\[
\binom{34}{7} = \frac{34 \times 33 \times 32 \times 31 \times 30 \times 29 \times 28}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5379616
\]

Thus, the probability is:

\[
P(\text{none defective}) = \frac{5379616}{17259390}
\]

### Final Answers:

1. The probability that all 7 generators are defective is \( \frac{1}{17259390} \).
2. The probability that none of the 7 generators are defective is \( \frac{5379616}{17259390} \).

Would you like further details or explanations?

Here are five related questions:
1. What is the probability that exactly 1 generator in the sample is defective?
2. How does the hypergeometric distribution apply to this scenario?
3. What is the complement of the probability that none of the generators are defective?
4. How does the sample size affect the probabilities?
5. How would the probabilities change if we were sampling with replacement?

**Tip:** In hypergeometric probability, the number of defective or non-defective items in the sample is key, and choosing without replacement is what distinguishes it from binomial probability.

A factory received a shipment of 41 generators, and the vendor who sold the items knows there are 7 generators in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the generators in the sample are defective, he will refuse the shipment.

If a sample of 7 generators is selected, find the probability that all in the sample are defective.

If a sample of 7 generators is selected, find the probability that none in the sample are defective.

Learn how to calculate the probability that all or none of the selected generators from a shipment are defective. This problem involves the hypergeometric distribution and combinatorics, with a detailed step-by-step solution. Suitable for advanced high school or college-level students.

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