At a local clinic there are 8 men, 5 women, and 3 children in the waiting room. If three patients are randomly selected, find the probability that there is at least 1 child among them.

To solve the problem of finding the probability that there is at least 1 child among the 3 randomly selected patients, we can use the complement rule. The complement rule says that the probability of an event happening is 1 minus the probability that the event does not happen.

### Steps:

1. **Total number of people in the waiting room**:
   \[
   8 \text{ men} + 5 \text{ women} + 3 \text{ children} = 16 \text{ people total}
   \]

2. **Total number of ways to select 3 people from 16**:
   The number of ways to choose 3 patients from the total of 16 people is given by the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \(n\) is the total number of people and \(r\) is the number of people being selected:
   \[
   \binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560
   \]

3. **Number of ways to select 3 people with no children (only from men and women)**:
   There are \(8 + 5 = 13\) men and women combined. The number of ways to select 3 patients from these 13 people (excluding children) is:
   \[
   \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286
   \]

4. **Number of ways to select 3 people with at least 1 child**:
   The number of ways to select 3 people that includes at least 1 child is the complement of selecting 3 people with no children. So:
   \[
   \text{Number of ways with at least 1 child} = \binom{16}{3} - \binom{13}{3} = 560 - 286 = 274
   \]

5. **Probability of selecting at least 1 child**:
   The probability of selecting at least 1 child is the ratio of the favorable outcomes (selecting at least 1 child) to the total outcomes (selecting any 3 people):
   \[
   P(\text{at least 1 child}) = \frac{274}{560} \approx 0.4893
   \]

Thus, the probability that there is at least 1 child among the 3 randomly selected patients is approximately **0.4893**, or **48.93%**.

Would you like more details or have any questions?

### Related Questions:
1. How would the probability change if there were 4 children instead of 3?
2. What is the probability of selecting exactly 2 children from the 3 randomly selected patients?
3. If we instead select 2 patients, what is the probability that both are children?
4. How would the probability change if the number of men increased to 10?
5. What is the probability of selecting no men from the 3 patients?

### Tip:
Always consider using the complement rule when calculating probabilities involving "at least" or "at most" conditions—it simplifies complex problems!

Learn how to calculate the probability of selecting at least 1 child from a group of 8 men, 5 women, and 3 children in a waiting room. This detailed solution involves combinatorics and the complement rule, making it suitable for students in Grades 9-12.

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