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

$\frac{10}{21}$

The correct answer is Option (1) → $\frac{10}{21}$

Step 1: Total group

• 3 men + 2 women + 4 children = 9 people
• Total ways to choose 4 people out of 9:

$\begin{pmatrix}9\\4\end{pmatrix} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = 126$

Step 2: Ways to choose exactly 2 children

• Number of ways to choose 2 children out of 4:

$\begin{pmatrix}4\\2\end{pmatrix} = 6$

• Remaining 2 people must be chosen from 3 men + 2 women = 5 people:

$\begin{pmatrix}5\\2\end{pmatrix} = 10$

• Total favorable ways:

$6 \times 10 = 60$

Step 3: Probability

$P = \frac{\text{favorable ways}}{\text{total ways}} = \frac{60}{126} = \frac{10}{21}$