TITLE: Help with conditional probability?
QUESTION [5 upvotes]: I've got to show that:
$$\mathbb{P} (A | A \cap B) = \frac{ \mathbb{P}(A)}{ \mathbb{P} (A \cap B)}$$
I'm not sure how to get to this.
Surely the probability of A occurring given A and B occurs is 1?
Or, by the equation...
$$\mathbb{P} (A | A \cap B) = \frac{ \mathbb{P} (A \cap A \cap B)}{\mathbb{P} (A \cap B)}$$
Thanks for your help

REPLY [2 votes]: Yet another explanation is that the typo(s) in question are that $\cup$ was replaced
by $\cap$ (or an u in \cup got changed to an a to give \cap for it is certainly
true that
$$P(A \mid A\cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)}.$$