Conditional probability

If we know that \(B\) occurred, the only possible outcomes are those of \(B\), and the probability of \(A\) must be measured only inside them. This explains the formula: out of the total \(P(B)\) we keep the part where \(A\) also occurs, namely \(P(A\cap B)\). That is why \(P(B)>0\) is required; conditioning on an impossible event makes no sense.

With a die, let \(B=\) «even» and \(A=\) «a 6». Here \(P(A\cap B)=1/6\) and \(P(B)=1/2\), so \(P(A\mid B)=1/3\): after discarding the odd faces, the 6 goes from one in six to one in three. Information may raise, lower or leave a probability unchanged.

Solving the definition gives the product rule \(P(A\cap B)=P(B)\,P(A\mid B)\), which computes probabilities of sequences: two kings in a row without replacement have probability \(\tfrac{4}{52}\cdot\tfrac{3}{51}\). Moreover, for fixed \(B\), \(P(\cdot\mid B)\) is a probability measure: it satisfies the same axioms, for example \(P(A\mid B)+P(A^c\mid B)=1\).

The most common mistake is to confuse \(P(A\mid B)\) with \(P(B\mid A)\); in general they differ, since their denominators differ. This concept prepares independence (when conditioning changes nothing) and total probability, which combines conditionals over distinct cases.