A factory received a shipment of 16 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.

The problem deals with sampling without replacement from a shipment of 16 generators, of which 7 are defective. We need to find two probabilities: (1) the probability that all 7 generators sampled are defective, and (2) the probability that none of the 7 generators sampled are defective.

### Part 1: Probability that all 7 generators in the sample are defective

To calculate this, we need to determine how many ways we can choose 7 defective generators from the 7 defective generators in the shipment, divided by how many ways we can choose any 7 generators from the 16 generators in the shipment.

- **Number of ways to choose 7 defective generators from the 7 defective generators**:  
  \[
  \binom{7}{7} = 1
  \]
- **Total number of ways to choose 7 generators from the 16 generators**:  
  \[
  \binom{16}{7} = \frac{16!}{7!(16 - 7)!} = \frac{16!}{7!9!}
  \]

Thus, the probability that all 7 generators in the sample are defective is:
\[
P(\text{all defective}) = \frac{\binom{7}{7}}{\binom{16}{7}} = \frac{1}{\binom{16}{7}}
\]
Now, we compute \(\binom{16}{7}\):
\[
\binom{16}{7} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 11440
\]
So, the probability is:
\[
P(\text{all defective}) = \frac{1}{11440}
\]

### Part 2: Probability that none of the generators in the sample are defective

For this case, we need to calculate the number of ways to choose 7 non-defective generators from the 9 non-defective ones, divided by the total number of ways to choose 7 generators from the 16 generators.

- **Number of ways to choose 7 non-defective generators from the 9 non-defective generators**:  
  \[
  \binom{9}{7} = \frac{9!}{7!(9 - 7)!} = \frac{9 \times 8}{2 \times 1} = 36
  \]
- **Total number of ways to choose 7 generators from the 16 generators** is the same as in Part 1:  
  \[
  \binom{16}{7} = 11440
  \]

Thus, the probability that none of the generators in the sample are defective is:
\[
P(\text{none defective}) = \frac{\binom{9}{7}}{\binom{16}{7}} = \frac{36}{11440}
\]
Simplifying this fraction:
\[
P(\text{none defective}) = \frac{9}{2860}
\]

### Final Results:
1. The probability that all 7 generators in the sample are defective is:  
   \[
   \frac{1}{11440}
   \]
2. The probability that none of the generators in the sample are defective is:  
   \[
   \frac{9}{2860}
   \]

Would you like further details or have any questions?  

Here are 5 related questions to explore:
1. What is the probability that exactly 1 generator in the sample is defective?
2. What is the probability that at least one generator in the sample is defective?
3. How does the probability change if only 5 generators are sampled?
4. What is the expected number of defective generators in a sample of 7?
5. How would the calculation change if the sampling was with replacement?

**Tip:** When calculating probabilities from large combinations, simplifying fractions helps avoid numerical errors and improves clarity.

A factory received a shipment of 16 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.

Explore the probability problem of selecting defective generators from a shipment of 16. Learn how to calculate the probability that all 7 selected generators are defective or that none are defective, using combinatorics and probability theory. Ideal for Grades 10-12.

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