General multiplication rule

The rule is $P(A \cap B) = P(A) \cdot P(B \mid A)$, where $\cap$ is the intersection sign, read "and", and $P(B \mid A)$ is read "the probability of B given A". It holds for any two events with $P(A) > 0$, independent or not, because the conditional factor already absorbs whatever the first event does to the second. The order is yours to choose: $P(B) \cdot P(A \mid B)$ returns the same product.

Draw two cards from a shuffled 52-card deck without replacing the first. The first is an ace with probability $4/52$. Given that it was, 3 aces are left among 51 cards, so the second is an ace with probability $3/51$. The product is $\frac{4}{52} \cdot \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.004525$. Enumerating all 2,652 ordered pairs of distinct cards finds exactly 12 that run ace then ace, which is the same answer arrived at by counting.

"There are 4 aces in 52 cards, so both cards are aces with probability $(4/52)^2 \approx 0.005917$." Squaring the unconditional probability answers a different question, the one where the first card goes back and the deck is reshuffled. Drawn without replacement the second card faces a smaller deck holding one fewer ace, so the true value is 0.004525 and the shortcut runs 31 percent high. The conditional factor is the entire content of the rule.

$P(B \mid A)$ is undefined when $P(A) = 0$, which is the one place the rule has nothing to say. Rearranged, it is the definition of conditional probability, $P(B \mid A) = P(A \cap B) / P(A)$, so the two statements are the same fact written twice. When the events happen to be independent, $P(B \mid A) = P(B)$ and the general rule collapses into the plain multiplication rule for independent events.

Conditional probability, and this rule with it, is topic 2.6.