Prove $P(A|B)=P(A\cap B)/P(B)$?

I've been doing some learning for Native Bayes classification. I came across this formula, but I'm having trouble remembering it because I don't know how to get this formula. Can anyone explain how to get $P(A|B)={\frac {P(A\cap B)}{P(B)}}$?

• $$P(A\mid B) = \frac{P(A\cap B)}{P(B)}$$ is the universally accepted definition of what $P(A\mid B)$ means, and there is no way to "get this formula" unless you have some alternative definition of the meaning of $P(A\mid B)$ from which we could arrive at $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$. Nov 10 '16 at 16:57
• My answer here has some pictures that might help. Nov 10 '16 at 18:49

As already pointed out in the comment, the statement $$P(A|B):={\frac {P(A\cap B)}{P(B)}}}$$ is the very definition of the conditional probability. Check for example here on wikipedia for an overview.
• how would you write such an expression, if you would condition on the whole event space $\Omega$ - what would you expect as an outcome
• and for the peace of mind, show that $P(\cdot|B):={\frac {P(\cdot\cap B)}{P(B)}}$ is indeed a probability measure itself