Where or adds, and multiplies. The question is only ever what the second factor should be. If the events are independent — the first leaves the second exactly as it found it — then $P(A \cap B) = P(A) \, P(B)$, read P of A-intersect-B equals P of A times P of B. Independence is a claim about the world, not a convenience; a scenario must earn it, usually with the words independent or with replacement.

When the first event changes the second, the honest form is $P(A \cap B) = P(A) \, P(B \mid A)$. The vertical bar is read given: $P(B \mid A)$ is the probability of B given that A has already happened. Order the letters carefully — the one AFTER the bar is the one you are told to assume, and the one BEFORE it is the one you are still chasing. Draw two red marbles from a bag of five without replacement and the second draw meets a smaller bag: one fewer red, one fewer marble. Both counts drop, which is why the answer is always below the with-replacement version.

Read that same relation solved for the bar and you get the definition of a conditional probability: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, P of A given B equals P of A-and-B over P of B. Conditioning does not change the overlap; it changes what the overlap is measured AGAINST. The world shrinks to B, and the same slice is now a bigger share of a smaller whole — which is why a conditional probability is never smaller than the joint one it came from.