To solve these problems, we can use combinatorics to calculate the probabilities. The key concept here is that we are dealing with a hypergeometric distribution because we are sampling without replacement.

Part 1: Probability that all 4 lightbulbs in the sample are defective

1. Total number of lightbulbs = 26
2. Number of defective lightbulbs = 4
3. Sample size = 4

We want the probability that all 4 sampled lightbulbs are defective. The number of ways to choose 4 defective lightbulbs out of 4 available defective lightbulbs is given by:

$\text{Number of ways to choose 4 defective bulbs} = \binom{4}{4} = 1$

The total number of ways to choose 4 lightbulbs out of 26 is:

$\text{Total number of ways to choose 4 bulbs} = \binom{26}{4} = \frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} = 14950$

So, the probability that all 4 lightbulbs in the sample are defective is:

$\text{Probability} = \frac{\binom{4}{4}}{\binom{26}{4}} = \frac{1}{14950}$

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

Now, we want the probability that none of the lightbulbs in the sample are defective.

1. Total number of non-defective lightbulbs = 26 - 4 = 22
2. Sample size = 4

The number of ways to choose 4 non-defective lightbulbs out of 22 available non-defective lightbulbs is given by:

$\text{Number of ways to choose 4 non-defective bulbs} = \binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315$

The probability that none of the lightbulbs in the sample are defective is:

$\text{Probability} = \frac{\binom{22}{4}}{\binom{26}{4}} = \frac{7315}{14950} = \frac{7315 \div 7315}{14950 \div 7315} = \frac{1}{2}$

Final Answers:

1. Probability that all 4 lightbulbs in the sample are defective: $\frac{1}{14950}$
2. Probability that none of the lightbulbs in the sample are defective: $\frac{1}{2}$

Would you like any further details or have any questions?

Here are five related questions:

1. What is the probability that exactly 2 out of the 4 sampled lightbulbs are defective?
2. How would the probability change if the sample size was increased to 5?
3. What is the expected number of defective lightbulbs in a random sample of 4?
4. How would the calculation differ if the sampling was done with replacement?
5. What is the probability that at least one lightbulb in the sample is defective?

Tip: When dealing with hypergeometric distributions, always remember to consider whether the sampling is with or without replacement. This affects the calculations significantly.