# Conditional probability - union of events

I'm going through Blitzstein's book Introduction to Probability and on p. 45 there is this exercise:

Now, NOT using Bayes' Rule, $P(A|B) = \frac{P(A \cap B)}{P(B)}$. For the exercise, $P(A \cap B) = P(both \hphantom\ girls, at \hphantom\ least \hphantom\ one \hphantom\ girl)$, since the comma means AND. This is then: $P(both \hphantom\ girls)*P(at \hphantom\ least \hphantom\ one \hphantom\ girl) = \frac{1}{4} * \frac{3}{4} = \frac{3}{16}$. This is different from the $\frac{1}{4}$ in the exercise. Why is that?

You are incorrect in saying that $P(A \cap B) = P(A)P(B)$ here. This only holds if $A$ and $B$ are independent. In this case, since $A$ is both girls and $B$ is at least one girl, it is clear that $A$ is a subset of $B$ and thus $A \cap B$ is simply $A$. It follows that $P(A \cap B) = P(A) = P(\text{both girls}) = \frac{1}{4}$.
• Not a problem. The general rule for events $A$ and $B$ is that $P(A\cap B)=P(A|B)*P(B)$. (If it's easier, you can swap the $A$ and $B$.) However, in this case recognizing that one event is contained within the other significantly simplifies the intersection. I actually did a similar problem a few weeks ago, which gave me the idea. :) Glad to help! – Matt Brems Aug 14 '16 at 19:36