A factory received a shipment of 49 generators, and the vendor who sold the items knows there are 6 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 6 generators is selected, find the probability that all in the sample are defective.

  

If a sample of 6 generators is selected, find the probability that none in the sample are defective.

  

The given scenario involves selecting a sample of generators from a larger shipment, with a certain number known to be defective. We will calculate the probabilities using combinations.

### Key information:
- Total number of generators: \( 49 \)
- Defective generators: \( 6 \)
- Non-defective (good) generators: \( 49 - 6 = 43 \)
- Sample size: \( 6 \)

### 1. **Probability that all 6 in the sample are defective:**

We need to calculate the probability that all 6 generators selected in the sample are defective. This can be found by dividing the number of ways to choose 6 defective generators by the total number of ways to choose 6 generators from the shipment.

- The number of ways to select 6 defective generators from 6 defective ones:
\[
\binom{6}{6} = 1
\]

- The total number of ways to select 6 generators from 49 generators:
\[
\binom{49}{6} = \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 13983816
\]

Thus, the probability that all generators are defective is:
\[
P(\text{all defective}) = \frac{1}{13983816}
\]

### 2. **Probability that none in the sample are defective:**

Now, we calculate the probability that none of the 6 selected generators are defective, meaning all are from the 43 non-defective generators.

- The number of ways to select 6 non-defective generators from 43 non-defective ones:
\[
\binom{43}{6} = \frac{43 \times 42 \times 41 \times 40 \times 39 \times 38}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 6096454
\]

Thus, the probability that none of the selected generators are defective is:
\[
P(\text{none defective}) = \frac{6096454}{13983816}
\]

This fraction is already in its reduced form.

### Summary of results:
- Probability that all generators are defective: \( \frac{1}{13983816} \)
- Probability that none of the generators are defective: \( \frac{6096454}{13983816} \)

Would you like any details on how these combinations were computed, or any other part of the solution? Here are 5 related questions to expand your understanding:

1. What is the probability that exactly one generator in the sample is defective?
2. How would the probabilities change if the sample size was 5 instead of 6?
3. What is the probability that at least one generator in the sample is defective?
4. Can you explain how combinations differ from permutations in these calculations?
5. How would you calculate the expected number of defective generators in the sample?

**Tip:** When calculating probabilities from combinations, it's helpful to remember that \( \binom{n}{k} \) represents the number of ways to choose \( k \) items from a set of \( n \) without regard to order.

A factory received a shipment of 49 generators, and the vendor who sold the items knows there are 6 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 6 generators is selected, find the probability that all in the sample are defective. If a sample of 6 generators is selected, find the probability that none in the sample are defective.

Find out how to calculate the probability of selecting defective generators from a shipment using combinatorics and probability principles. This problem covers selecting a sample of 6 generators from a total of 49, with 6 defective. Learn step-by-step how to compute the probability that all or none of the selected generators are defective. Suitable for high school students in Grades 10-12.

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