CUET UG 2025 — General Test Quantitative Reasoning

The group contains:

Men = 3

Women = 2

Children = 4

Total people = 3 + 2 + 4 = 9

---

The total number of ways to choose 4 people from 9:

$$C(9,4) = \frac{9!}{4! \times 5!}$$C(9,4)=4!×5!9!​

$$= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$$=4×3×2×19×8×7×6​

$$= \frac{3024}{24}$$=243024​

$$= 126$$=126

---

For exactly 2 children, the other 2 must be non-children (men or women).

Non-children = 3 + 2 = 5 people

Choosing 2 children from 4:

$$C(4,2) = \frac{4 \times 3}{2 \times 1} = 6$$C(4,2)=2×14×3​=6

Choosing 2 non-children from 5:

$$C(5,2) = \frac{5 \times 4}{2 \times 1} = 10$$C(5,2)=2×15×4​=10

Total favorable outcomes = $6 \times 10 = 60$6×10=60

---

The probability is:

$$P = \frac{60}{126}$$P=12660​

$$= \frac{10}{21}$$=2110​

Therefore, the probability that exactly 2 of them are children is $\frac{10}{21}$2110​.