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To solve this, let’s analyze each part carefully.

### Given Information:
- Sample size, \( n = 1200 \) U.S. adults.
- Out of 1200 adults, 205 dine out more than once per week.

The probability of selecting an adult who dines out more than once per week, \( P(\text{dines out more than once}) \), is:
\[
\frac{205}{1200} = 0.1708
\]

Let:
- \( p = 0.1708 \), the probability that a randomly selected adult dines out more than once per week.
- \( q = 1 - p = 0.8292 \), the probability that a randomly selected adult does **not** dine out more than once per week.

Now, let’s work through each part.

---

### Part (a)
**The probability that neither adult dines out more than once per week.**

Since two adults are selected without replacement:
1. The probability the first adult does not dine out more than once per week is \( q = 0.8292 \).
2. The probability the second adult also does not dine out more than once per week, given the first does not, is slightly adjusted for the reduction in the sample size. It will be \( \frac{994}{1199} \).

Thus:
\[
P(\text{neither dines out more than once}) = q \times \frac{994}{1199}
= 0.8292 \times \frac{994}{1199} \approx 0.688
\]

### Part (c)
**The probability that at least one of the two adults dines out more than once per week.**

This probability is the complement of both adults not dining out more than once per week.

\[
P(\text{at least one dines out more than once}) = 1 - P(\text{neither dines out more than once})
\]

Substitute the value from Part (a):
\[
= 1 - 0.688 = 0.312
\]

### Summary of Answers:
- **(a)** The probability that neither adult dines out more than once per week is approximately **0.688**.
- **(c)** The probability that at least one of the two adults dines out more than once per week is approximately **0.312**.

Would you like a further breakdown of the calculations? 

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#### Related Questions:
1. What if the probability of dining out more than once a week was different? How would that change the results?
2. How would these probabilities change if more than two adults were selected?
3. What is the impact of sampling without replacement versus with replacement in probability calculations?
4. How can these probability techniques be applied to other real-life scenarios?
5. How do we calculate probabilities for more complex scenarios, like selecting multiple groups?

---

#### Tip:
When calculating probabilities without replacement, remember to adjust the sample size after each selection to maintain accuracy.

In a sample of 1200 U.S. adults, 205 dine out at a restaurant more than once per week. Two U.S. adults are selected at random from the population of all U.S. adults without replacement. Assuming the sample is representative of all U.S. adults, complete parts (a) through (d).

(a) The probability that neither adult dines out more than once per week.
(c) The probability that at least one of the two adults dines out more than once per week.

Calculate the probability that two randomly selected U.S. adults dine out more than once per week. This probability problem involves concepts like sampling without replacement and the complement rule, making it ideal for high school students.

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