In a family with 3 children, what is the probability that they have 2 boys and 1 girl?

Solution 1:

No, the possible outcomes are $$\mathsf{BBB, \boxed{\mathsf{BBG}}, \boxed{\mathsf{BGB}}, \boxed{\mathsf{GBB}}, BGG,\mathsf{GBG}, GGB, GGG}$$ wherein $3$ meet the requirement. There are $8$ possible outcomes, all equally likely (if we assume each gender is equally likely). Hence the choice is $3/8$.

We can also think about it in at least one more way: You identified all the possible ways to get 2 boys and one girl. Since the events are disjoint, we can add up the probabilities \begin{align*} P(\text{2 boys, 1 girl}) &= P(\mathsf{BBG})+P(\mathsf{BGB})+P(\mathsf{GBB}) \\ &= \frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2} \\ &= \frac{3}{8}. \end{align*}

Solution 2:

Since there are two genders and three children, that's $2^3=8$ permutations. You had three permutations. They would be BBB, BBG, BGB, BGG, GBB, GBG, GGB, and GGG. Out of those, the permutations with two boys and one girl are BBG, BGB, and GBB. The answer is the ratio of how many choices fit your conditions to the total number of possibilities, $\frac{3}{8}$.