# $\frac{1}{4}$ numerator of the boy or girl paradox problem

I have a question regarding the $$\frac{1}{4}$$ numerator derived from the boy or girl paradox problem. I will be using the "Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?" here.

I've read through Wiki and the Q&A posted here and understand the general concept of the boy or girl paradox. However, the equation is a little fuzzy to me using conditional probability.
Let;
$$S = \{(B_1,B_2),(B_1,G_2),(G_1,B_2),(G_1,G_2)\}$$

$$A = \{(G_1,G_2)\}$$ and $$B = \{(G_1,G_2),(G_1,B_2)\}$$

$$P(A) = \frac{1}{4}$$ and $$P(B) = \frac{2}{4}$$

Using the conditional probability approach, shouldn't the equation be:

$$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{\frac{1}{4} \cdot \frac{2}{4}}{\frac{2}{4}}$$

I understand the intuition of using $$\frac{1}{4}$$ as the numerator and it makes sense, however, how is the multiplication rule applied here, what am I missing?

• You've used $P(A \cap B) = P(A)P(B)$ in the numerator, but this is only true if the events are independent. (in this case, they are certainly not) The numerator should just be $P(A)$ since $A$ is the event of "the older child is a daughter AND the younger child is a daughter." Apr 19, 2022 at 13:28
• Note that $P(A\cap B)=P(A)$ since $A\subset B$.
• $P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \dfrac{\frac{1}{4}}{\frac{2}{4}} =\frac12$ Apr 19, 2022 at 13:54
$$P(A \cap B) = P(\{(G_1,G_2)\} \cap \{(G_1,G_2), (G_1,B_2)\})=P(\{(G_1,G_2)\})=\frac{1}{4} \neq \frac{1}{4} \cdot\frac{1}{2}=P(A) \cdot P(B).$$