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The problem on the screen involves calculating probabilities for defective capacitors in a shipment.

We are given:
- There are 30 capacitors in total.
- 4 of these capacitors are defective.

### 1. First Question: 
**What is the probability that all 4 capacitors selected are defective?**

To calculate this, we use combinations and conditional probability. The total number of ways to choose 4 capacitors from the 30 is:

\[
\text{Total number of ways} = \binom{30}{4}
\]

The number of ways to choose 4 defective capacitors out of the 4 defective capacitors is:

\[
\text{Defective combinations} = \binom{4}{4}
\]

Thus, the probability of all 4 being defective is the ratio of these two:

\[
P(\text{all defective}) = \frac{\binom{4}{4}}{\binom{30}{4}} = \frac{1}{\binom{30}{4}}
\]

Now let's calculate the value of \(\binom{30}{4}\):

\[
\binom{30}{4} = \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} = 27,405
\]

So, the probability of selecting 4 defective capacitors is:

\[
P(\text{all defective}) = \frac{1}{27,405}
\]

### 2. Second Question:
**What is the probability that none of the capacitors in the sample are defective?**

In this case, we are choosing all 4 capacitors from the 26 non-defective ones. The number of ways to choose 4 non-defective capacitors from 26 is:

\[
\binom{26}{4}
\]

Thus, the probability that none are defective is:

\[
P(\text{none defective}) = \frac{\binom{26}{4}}{\binom{30}{4}} = \frac{\binom{26}{4}}{27,405}
\]

Let's calculate \(\binom{26}{4}\):

\[
\binom{26}{4} = \frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} = 14,950
\]

Therefore, the probability of selecting no defective capacitors is:

\[
P(\text{none defective}) = \frac{14,950}{27,405}
\]

Would you like to see more details, or do you have any other questions?

### 5 Follow-Up Questions:
1. How would the probability change if there were 5 defective capacitors in the shipment?
2. What is the probability of selecting exactly 1 defective capacitor in the sample of 4?
3. How would the problem change if you sampled 5 capacitors instead of 4?
4. Can this situation be modeled using a binomial distribution? Why or why not?
5. How does the hypergeometric distribution relate to this problem?

### Tip:
Remember, when sampling without replacement (as in this case), it's important to use the **hypergeometric distribution**, which considers the finite population size.

A factory received a shipment of 30 capacitors, and the vendor who sold the items knows there are 4 capacitors in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the capacitors in the sample are defective, he will refuse the shipment. If a sample of 4 capacitors is selected, find the probability that all in the sample are defective. If a sample of 4 capacitors is selected, find the probability that none in the sample are defective.

Calculate the probability of selecting defective capacitors from a sample in a shipment of 30 capacitors, 4 of which are defective. This problem uses combinatorics and hypergeometric probability concepts, suitable for high school students in Grades 11-12.

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