A group consists of 3 men, 2 women and 4 children.\nFour persons are chosen at random from the group.\nWhat is the probability that exactly two of the selected persons are women?

Introduction / Context:
This problem involves combinations and basic probability. From a mixed group of men, women and children, we choose four persons and want the probability that exactly two of the chosen persons are women. Because there are only two women in the entire group, this means both women must be selected.

Given Data / Assumptions:

• Number of men = 3.
• Number of women = 2.
• Number of children = 4.
• Total persons in the group = 3 + 2 + 4 = 9.
• We select 4 persons at random, all selections equally likely.
• The required event is: exactly two women are selected.

Concept / Approach:
Because there are only two women in the entire group, having exactly two women in the selected four means both women must be included. The remaining two positions must be filled from the seven non-women. We will use combinations: Number of ways to choose r items from n = nCr. Probability is then the ratio of favourable combinations to total combinations.

Step-by-Step Solution:
Total persons = 9. Total ways to choose any 4 persons = 9C4. Compute 9C4 = 9! / (4! * 5!) = 126. To have exactly two women, we must select both women: 2C2 = 1 way. We then choose 2 more persons from the remaining 7 (3 men + 4 children): 7C2. Compute 7C2 = 7! / (2! * 5!) = 21. Favourable ways = 1 * 21 = 21. Probability = favourable / total = 21 / 126 = 1 / 6. So the required probability is 1/6.

Verification / Alternative check:
We can verify numerically. If we expand 9C4 as 126 and observe that there is no way to get exactly one woman or more than two women because only two exist, then 21 favourable combinations is reasonable. Dividing 21 by 126 gives 0.1666..., which is exactly 1/6.

Why Other Options Are Wrong:
Option 1/5 corresponds to 25.2 favourable selections, which is not an integer. Option 1/7 is about 0.1429, which would require 18 favourable selections. Option 1/9 is about 0.1111 and does not match the computed value of 1/6.

Common Pitfalls:
Many learners mistakenly use permutations instead of combinations, or they forget that the two women must both be chosen, not just one. Another common error is to choose four persons and then adjust by conditional reasoning, which is more complex and prone to mistakes. It is much simpler to count combinations carefully as done here.

Final Answer:
The probability that exactly two of the four selected persons are women is 1/6.