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To solve this problem, let's work through the probability calculation step by step.

### Given Data
- Total sample size = 1200 U.S. adults
- Adults who dine out more than once per week = 205
- Two adults are selected without replacement.

### (a) Finding the Probability That Both Adults Dine Out More Than Once Per Week

1. **Calculate the probability that the first selected adult dines out more than once per week**:
   \[
   P(\text{first adult}) = \frac{205}{1200}
   \]

2. **Calculate the probability that the second selected adult also dines out more than once per week, given that the first adult already dines out more than once per week**:
   Since we are selecting without replacement, after the first selection, there are now 204 adults who dine out more than once per week out of a total of 1199 adults left.
   \[
   P(\text{second adult} | \text{first adult}) = \frac{204}{1199}
   \]

3. **Multiply these probabilities** to get the combined probability that both selected adults dine out more than once per week:
   \[
   P(\text{both adults}) = \frac{205}{1200} \times \frac{204}{1199}
   \]

### Calculation

\[
P(\text{both adults}) = \frac{205 \times 204}{1200 \times 1199} 
\]

Let's calculate this result:

\[
P(\text{both adults}) \approx 0.0291
\]

### Answer
The probability that both adults dine out more than once per week is approximately **0.029** (rounded to three decimal places).

Would you like further details, or do you have any questions about this solution?

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#### Additional Practice Questions

1. If three adults were selected instead, what would the probability be that all three dine out more than once per week?
2. What is the probability that at least one of the two selected adults dines out more than once per week?
3. How would the probability change if the sample size was 2400 with the same proportion dining out more than once per week?
4. If the selection were with replacement, how would the probability calculation change?
5. What is the probability that neither of the two selected adults dines out more than once per week?

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#### Tip
When dealing with probability questions involving "without replacement," remember that each successive selection slightly changes the probabilities.

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, find the probability that both adults dine out more than once per week.

Learn how to calculate the probability of selecting two U.S. adults who dine out more than once per week from a sample of 1200, with 205 adults who dine out frequently. This solution covers key probability concepts, including conditional probability and sampling without replacement, suitable for Grades 9-12.

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