Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is

### Concept & Logic To find the probability of exactly 2 children being chosen out of 4 people, we must select 2 children from the available children and 2 other people (adults) from the remaining group. $$ P = \frac{\text{Ways to choose 2 children} \times \text{Ways to choose 2 adults}}{\text{Total ways to choose 4 people}} $$ ### Step-by-Step Solution * **Group composition:** 3 men + 2 women + 4 children = 9 persons total. * **Adults (men + women):** $3 + 2 = 5$. * **Total ways to choose 4 persons from 9:** ${}^{9}C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$. * **Ways to choose exactly 2 children (out of 4):** ${}^{4}C_2 = \frac{4 \times 3}{2} = 6$. * **Ways to choose the other 2 persons from the 5 adults:** ${}^{5}C_2 = \frac{5 \times 4}{2} = 10$. * **Total favorable ways:** $6 \times 10 = 60$. * **Probability:** $\frac{60}{126}$. * Divide numerator and denominator by 6 to simplify: $\frac{10}{21}$. ### Exam Strategy & Shortcut Recognize that "exactly 2 children" implies the other 2 chosen must NOT be children. Grouping the men and women together as a single "non-children" category simplifies the problem, avoiding the need to calculate separate scenarios (e.g., 2 children + 1 man + 1 woman). ### Common Pitfall Calculating the probability of choosing 2 children but forgetting to account for the remaining 2 people needed to complete the group of 4. You must select the full group to satisfy the denominator's condition. ### Final Answer Therefore, the correct answer is **$\frac{10}{21}$**.