Family with three children. Chances of at least one boy and girl? Here's my exercise:
A family has three children. What's the probability of event $A \cup C$ where:
$A$- Family has children of both sexes.
$C$- Family has at most one girl.
Well I see two ways to look at this conundrum:
The first one would be to differentiate possibilities by triples, $B$ meaning boy, $D$ meaning girl. So we have eight possibilities: $BBB,BBD,BDB,DBB,BDD,DDB,DBD,DDD$. So the answer would be $P(A \cup C)=\frac{7}{8}$.
And this is correct according to solutions in my textbook. Yet my first solution was this, since they ask about if the family has two boys, and not about whether the first and third child are boys, then $BDB$ and $BBD$ are indistinguishable, then we have only four possibilities: one girl and two boys, three boys, three girls, one boy and two girls. Then $P(A \cup C)=\frac{3}{4}$. 
Why was my first intuition wrong?

You distinguish between four possibilities: $BBB, BBD, BDD, DDD$. In three out of four cases the event $A \cup C$ occurs. However this only implies $P(A \cup C) = \frac{3}{4}$ when each of the four cases are equally likely, which is not the case here. 
