From a group of 3 men, 2 women, and 4 children (9 people in total), four persons are chosen at random. What is the probability that exactly 2 of the selected persons are children?

Introduction / Context:
This is a combinatorial probability problem involving selection from different categories: men, women, and children. We are asked to find the probability that among four selected people, exactly two are children. This tests counting with combinations and careful handling of category constraints.

Given Data / Assumptions:

• There are 3 men, 2 women, and 4 children, for a total of 9 people.
• Four persons are selected at random without replacement.
• All groups of four are equally likely.
• We require that exactly 2 of the selected persons are children; the remaining 2 must be adults (men or women).

Concept / Approach:
We first compute the total number of ways to choose any 4 persons out of 9 using combinations. Then we compute the number of favourable selections that contain exactly 2 children. For the favourable selections, we choose 2 children from 4 and 2 adults from the 5 adults (3 men + 2 women). The probability is the ratio of favourable selections to total selections.

Step-by-Step Solution:
Total people = 3 men + 2 women + 4 children = 9. Total number of ways to choose 4 people from 9 = C(9,4). Compute C(9,4) = 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126. Number of children = 4, so ways to choose exactly 2 children = C(4,2) = 6. Number of adults (men + women) = 3 + 2 = 5. Ways to choose 2 adults from 5 = C(5,2) = 10. Total favourable groups with exactly 2 children = 6 * 10 = 60. Required probability = favourable / total = 60 / 126. Simplify 60 / 126 by dividing numerator and denominator by 6 to get 10 / 21.

Verification / Alternative check:
We can verify that we have not missed any case because the requirement is very clear: exactly 2 children and exactly 2 adults. There is no overlap between the count of children and adults. Double counting does not occur because we select from disjoint categories. The fraction 10/21 is in simplest form, and numerical evaluation gives approximately 0.476, which is a reasonable probability for this situation.

Why Other Options Are Wrong:
9/29: This fraction does not match the ratio 60/126 and simplifies to a different decimal value.
12/21: This simplifies to 4/7, which would correspond to 72 favourable selections, not 60.
14/19: This is greater than 0.7, which is too large for such a specific composition constraint.
None of these: This is incorrect because 10/21 is one of the given choices and matches the correct computation.

Common Pitfalls:
A typical mistake is to include cases with three or four children by accident or to forget that the remaining members after choosing children must be adults. Some learners also mistakenly treat combinations as permutations and overcount by considering order. Always remember that combinations C(n,r) count unordered selections and are appropriate for group selection problems like this one.

Final Answer:
Therefore, the probability that exactly 2 of the selected persons are children is 10/21.