Four persons are chosen at random from a group of 3 men, 2 women, and 4 children. What is the probability that the selected group contains exactly 1 man, 1 woman, and 2 children?

Introduction / Context:
This problem is a classic example of probability with combinations, where we are selecting people from different categories. The group has men, women, and children, and we want the probability that the chosen group of four contains exactly one man, one woman, and two children. This tests understanding of counting methods, combinations, and probability based on equally likely combinations.

Given Data / Assumptions:

• Total people = 3 men + 2 women + 4 children = 9 persons.
• We must select exactly 4 persons.
• Required composition: 1 man, 1 woman, and 2 children.
• All combinations of 4 persons from the 9 are equally likely.
• Sampling is without replacement.

Concept / Approach:
We use combinations nCr to count the number of ways to choose people from each category. The probability is computed as (number of favourable groups) divided by (total number of possible groups of 4). We count favourable outcomes by choosing 1 man from 3, 1 woman from 2, and 2 children from 4, and we multiply these because the choices are independent categorywise. Then we divide by the total number of ways to choose 4 people from 9.

Step-by-Step Solution:
Total number of ways to choose 4 persons from 9 = C(9,4). Compute C(9,4) = 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126. Number of ways to choose 1 man from 3 = C(3,1) = 3. Number of ways to choose 1 woman from 2 = C(2,1) = 2. Number of ways to choose 2 children from 4 = C(4,2) = 6. Total favourable groups = 3 * 2 * 6 = 36. Required probability = favourable / total = 36 / 126. Simplify 36 / 126 by dividing numerator and denominator by 18: 36 / 126 = 2 / 7.

Verification / Alternative check:
An alternative way is to verify that there is no other composition counted by mistake. Since the requirement is exactly 1 man, 1 woman, and 2 children, there is no overlap with any other composition such as 2 men or 3 children. Therefore, every counted group matches the requirement. The simplified fraction 2/7 is already in lowest terms, so the probability is correct.

Why Other Options Are Wrong:
3/7: This would require 54 favourable groups, which is larger than the actual count of 36.
4/7: This is too large and would imply 72 favourable groups, which is impossible since the total number of groups is only 126.
5/21: This fraction is approximately 0.238, which does not match the exact ratio 36/126 = 0.2857.
None of these: This is incorrect because 2/7 is a valid and correct option in the list.

Common Pitfalls:
A common error is to forget that we are dealing with combinations, not permutations, and to multiply in the wrong way. Another pitfall is to misinterpret the phrase "exactly 1 man, 1 woman, and 2 children" and accidentally include groups with more children or more adults. Some learners also forget to compute the total number of groups correctly and instead use an incorrect denominator, which changes the probability value completely.

Final Answer:
Hence, the probability of selecting exactly 1 man, 1 woman, and 2 children is 2/7.