What is the probability of having both girls given that at least one of them is girl? I'm reading the textbook entitled Harvard Statistics by Joseph K. Blitskein. In conditional probability context, I'm a little bit confused of calculating the intersection of event A and B $P(A\cap B)$. The problem is given below;

Example 2.2.5 (Elder is a girl vs. at least one girl). A family has two children, and it is known that at least one is a girl. What is the probability that both are girls, given this information? What if it is known that the elder child is a girl?

For the solution (based on textbook):
First, let the following events equal to A and B (just to keep it short): 
A = event of having both girls, B = event of at least one girl
$$P(A|B)=\frac{P(A∩B)}{P(B)}=\frac{1/4}{3/4}=1/3$$
Intuitively, the answer is correct but I was wondering on why $P(A∩B)=1/4$. Since $P(A∩B)=P(A)P(B)$, which in this case $P(A)=1/4$ and $P(B)=3/4$. Just to give you a reference, the possible combinations of having two children are ${\{GG, BG, GB, BB}\}$. 
Anyone can shed me a light regarding with my problem? 

$A$ and $B$ should be defined as events and not probabilities. $A$ is the event of having both girls and $B$ is the event of having at least one girl. By definition, you can see that they are not independent events. Hence $P(A,B) = P(A)P(B)$ does not hold. When we estimate $P(A,B)$, only one combination works $(GG)$ out of the four possible combinations. Hence the probability of both having both girls and having at least one girl is $P(A,B)=1/4$.
