Bayes' theorem is a powerful tool for problems like this. Let's use it to solve the problem.

Let ( B_1 ) be the event that the first child is a boy and ( B_2 ) be the event that the second child is a boy.

We want to find the probability that both children are boys given that at least one of them is a boy. In terms of events, we want to find:

[ P(B_1 \cap B_2 | B_1 \cup B_2) ]

Using Bayes' theorem:

[ P(B_1 \cap B_2 | B_1 \cup B_2) = \frac{P(B_1 \cup B_2 | B_1 \cap B_2) \times P(B_1 \cap B_2)}{P(B_1 \cup B_2)} ]

Now, let's break down each of these probabilities:

1. ( P(B_1 \cup B_2 | B_1 \cap B_2) ): If both children are boys, then the probability that at least one of them is a boy is 1. So, ( P(B_1 \cup B_2 | B_1 \cap B_2) = 1 ).

2. ( P(B_1 \cap B_2) ): This is the probability that both children are boys. Since the children's genders are independent and each has a ( \frac{1}{2} ) chance of being a boy, ( P(B_1 \cap B_2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} ).

3. ( P(B_1 \cup B_2) ): This is the probability that at least one child is a boy. From our initial list of combinations (BB, BG, GB, GG), three of them have at least one boy. So, ( P(B_1 \cup B_2) = \frac{3}{4} ).

Plugging these values into Bayes' theorem:

[ P(B_1 \cap B_2 | B_1 \cup B_2) = \frac{1 \times \frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} ]

So, using Bayes' theorem, we again find that the probability that both children are boys, given that at least one of them is a boy, is ( \frac{1}{3} ).