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

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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?

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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 $P(A \cap B) = P(A) = P(\text{both girls}) = \frac{1}{4}$.

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  • $\begingroup$ Thank you for your clear explanation! I though something was wrong with my multiplication and I knew they were not independent but just couldn't figure anything else out. Never thought of this in terms of subsets to be honest. I really appreciate your help! $\endgroup$ – slazien Aug 14 '16 at 19:34
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    $\begingroup$ 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! $\endgroup$ – Matt Brems Aug 14 '16 at 19:36
  • $\begingroup$ Yes, I've seen that formula ahead of the page I'm currently at and Bayes' Theorem directly follows from it. Now I'll remember to think in terms of subsets, thanks! $\endgroup$ – slazien Aug 14 '16 at 19:39

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