 Why this definition? A helpful heuristic is to think of the probability of \(B\) given \(A\) as something like the portion of the \(A\)-possibilities that are also \(B\)-possibilities. For example, the probability of rolling a high number (4, 5, or 6) on a six-sided die given that the roll is even is 2/3. Why? There are 3 even possibilities (2, 4, 6), so \(p(A) = 3/6\). Of those 3 possibilities, 2 are also high numbers (4, 6), so \(p(B \wedge A) = 2/6\). Thus  \[p(B\mid A) = \frac{p(B \wedge A)}{p(A)} = \frac{2/6}{3/6} = 2/3.\]   Generalizing this idea, we start with the quantity of \(A\)-possibilities as a sort of baseline by putting \(p(A)\) in the denominator. Then we consider how many of those are also \(B\)-possibilities by putting \(p(B \wedge A)\) in the numerator.