Let's tackle each part of this problem step by step.

We know:

• Total products = 100
• Defective products = 12
• Non-defective products = 100 - 12 = 88
• Owen samples 4 products without replacement.

Part (b): Probability that none of the products in the sample are defective.

To calculate this, we need the probability of selecting only non-defective products.

The number of ways to choose 4 non-defective products from 88 is given by the combination formula: $\binom{88}{4} = \frac{88!}{4!(88-4)!}$ The total number of ways to choose 4 products from 100 is: $\binom{100}{4} = \frac{100!}{4!(100-4)!}$ Thus, the probability that none of the products in the sample are defective is: $P(\text{none defective}) = \frac{\binom{88}{4}}{\binom{100}{4}}$

Now let's compute this.

Part (c): Probability that at least one product is defective.

The complement of "none of the products are defective" is "at least one product is defective." Therefore, the probability of at least one defective product is: $P(\text{at least one defective}) = 1 - P(\text{none defective})$

Let’s calculate these probabilities.

I'll compute the values now.### Results:

• (b) The probability that none of the products in the sample are defective is approximately 0.59.
• (c) The probability that at least one product is defective (and the whole batch is sent back) is approximately 0.41.

Do you need more details or have any further questions?

Here are 5 related questions to explore:

1. What is the probability of selecting exactly 2 defective products in the sample?
2. How would the probability change if the sample size increased to 5?
3. What is the probability of selecting 3 non-defective and 1 defective product?
4. How does the probability change if the number of defective products is reduced to 10?
5. How would the analysis differ if Owen replaced the products after inspecting them?

Tip: The concept of "without replacement" greatly affects probabilities by reducing the sample space after each selection.