Multiplication Rule vs General Multiplication Rule

For independent events the rule is $P(A \cap B) = P(A)P(B)$, where $\cap$ is the intersection symbol, read A and B. Independence means learning that one event happened leaves the probability of the other unchanged, so the second factor never has to be adjusted. Flipping a fair coin and rolling a fair die qualify: listing all 12 equally likely pairs shows exactly one is heads with a 6, and $0.5 \times 1/6 = 1/12 \approx 0.0833$ agrees.

The product form is a shortcut, not the rule. The general multiplication rule $P(A \cap B) = P(A) \cdot P(B \mid A)$ holds for any two events, and the shortcut is what it collapses to when $P(B \mid A) = P(B)$. Write the general form first and let it simplify rather than choosing between two formulas.

The error worth naming: a shelf holds 8 batteries of which 2 are dead, so the chance that two batteries taken at random are both dead is $0.25 \times 0.25 = 0.0625$. Taken without replacement, the first battery removes itself from the shelf, so the second pick faces 1 dead among 7 and the answer is $\frac{2}{8} \cdot \frac{1}{7} = \frac{2}{56} \approx 0.0357$. The shortcut overstates it by 75 percent. Enumerating all 56 ordered pairs of distinct batteries confirms that 2 of them are dead pairs.

Mutually exclusive is not independent, and the product rule is at its worst there. If $P(A) = 0.3$, $P(B) = 0.5$, and the two events cannot both occur, then $P(A \cap B) = 0$ while the product returns 0.15.

The gap closes as the population grows. Take 2 batteries from 1,000 of which 250 are dead and the exact value is $\frac{250}{1000} \cdot \frac{249}{999} \approx 0.06231$ against 0.0625 from the shortcut, a difference of 0.3 percent. That is what the 10 percent condition formalizes: sampling without replacement may be treated as approximately independent when the sample is at most 10 percent of the population.

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.