Four persons are choosen at random from a group of 3 men, 2 women and 4 children. The probability of having 2 children out of 4 is :

$\frac{10}{21}$

Since we want exactly 2 of them to be children, we choose the 2 children first in ⁴C₂=6 ways

Choose the other 2 people from the men and women in ⁵C₂=10 ways

The universal set had ⁹C₄=126 number of ways

Hence, probability is 10×6/126 = 10/21