There are 3 children in each of 3 different families. If two children are chosen at random from the children in these families, find the probability of each of the following events. (a) The two children come from the same family. (b) The two children come from different families.

Step 1 ：Let's find the total number of ways to choose 2 children from all the families. There are 9 children in total, so the total number of ways is \(\binom{9}{2} = 36\).

Step 2 ：For the two children to come from the same family, we can choose 2 children from any of the 3 families. So, the number of ways to choose 2 children from the same family is \(\binom{3}{2} + \binom{3}{2} + \binom{3}{2} = 3 + 3 + 3 = 9\).

Step 3 ：Thus, the probability of the two children coming from the same family is \(\frac{9}{36} = \frac{1}{4} = 0.25\). So, \(\boxed{0.25}\) is the probability of the two children coming from the same family.

Step 4 ：For the two children to come from different families, we can choose 1 child from each of the 3 families. So, the number of ways to choose 2 children from different families is \(3 \times 3 \times 3 - 9 = 27\).

Step 5 ：Thus, the probability of the two children coming from different families is \(\frac{27}{36} = \frac{3}{4} = 0.75\). So, \(\boxed{0.75}\) is the probability of the two children coming from different families.