When can you multiply probabilities?

When you can multiply: the decision

You can multiply $P(A)$ by $P(B)$ only when the two events are independent, meaning one happening does not change the chance of the other. If you have not checked independence, that formula is not yours to use yet.

The rule that is always safe is the general multiplication rule:

Read it out loud: the chance both happen equals the chance of the first, times the chance of the second given that the first already happened. Write this down first, every time. If the events turn out to be independent, then $P(B \mid A) = P(B)$ and the rule collapses on its own to $P(A) \cdot P(B)$.

So the question is not "which of two formulas fits this problem". There is one formula. $P(A) \cdot P(B)$ is what it becomes once you have earned the right to simplify.

Why the general rule never fails

The Fall 2026 formula sheet gives conditional probability as

Multiply both sides by $P(B)$ and you get $P(A \cap B) = P(B) \cdot P(A \mid B)$. Relabel the two events and that is the general multiplication rule. The AP course does not print a separate multiplication formula, because it is this one rearranged.

Notice what the derivation assumed about the events: nothing. It is algebra applied to a definition, so it holds for any two events with $P(B) > 0$. (A conditional probability is undefined when you condition on an event that cannot happen.)

$P(A \cap B) = P(A) \cdot P(B)$ is a different kind of statement. It is not a definition, it is a claim about how the two events relate, and it is true only when they are independent. That asymmetry is the whole answer to the title question.

What counts as verified independence

Four things license the shortcut. Anything else does not.

• Physically separate random actions. Two dice, two spins, two coin flips, two machines filling bottles on separate lines. A die has no memory of the last roll.
• Sampling with replacement. You put the item back and re-randomize, so the second draw faces exactly the population the first one did. See with replacement vs without replacement.
• The problem tells you. Phrases like "assume the trials are independent" or "each patient responds independently of the others" are the exam handing you permission. Quote it in your justification.
• The numbers check out. Compare $P(A \cap B)$ with the product $P(A) \cdot P(B)$. If they match, the events are independent. Equivalently, check whether $P(B \mid A) = P(B)$. This is the test to run when a two-way table gives you the counts.

Two things that look like evidence and are not. First, the events feeling unrelated: "height" and "reading score" feel unrelated in a room of adults, but in a room of children both rise with age, so they are dependent. Second, the events being mutually exclusive. Disjoint events with nonzero probability are strongly dependent, never independent, which is worked through in disjoint vs independent events.

The same problem worked both ways

Draw two cards from a standard 52-card deck and ask for the probability that both are hearts. The answer depends entirely on whether the first card goes back.

The first card changes the deck. After a heart comes out, 12 hearts remain among 51 cards, so $P(B \mid A) = \frac{12}{51}$, not $\frac{13}{52}$. Using $P(A) \cdot P(B)$ here gives an answer 6.25 percent too high.

That error grows as the pool shrinks. In a class of 10 students, 4 of them seniors, the chance that 2 students chosen at random (without replacement) are both seniors is $\frac{4}{10} \times \frac{3}{9} = \frac{2}{15} \approx 0.1333$. The shortcut gives $0.4^2 = 0.16$, which is 20 percent too high. Picking two students from a class is the setting where this mistake is made most often, because nothing in the wording announces that the draws are linked.

Sampling without replacement, and the 10 percent condition

Almost every real sample is drawn without replacement. You do not survey the same person twice. So strictly speaking, the draws are never independent, and the honest formula is always $P(A) \cdot P(B \mid A)$.

The AP course gives you one licensed approximation. When you sample without replacement, if the sample size $n$ is at most 10 percent of the population size $N$, the probabilities barely shift from draw to draw and you may treat the draws as approximately independent. That is the 10 percent condition, and it is why the binomial model, which assumes independent trials, can be applied to survey data at all.

Read that condition for what it is: a tolerance for inference, not blanket permission to multiply. Drawing 2 cards from 52 satisfies it on paper, yet the second probability still slides from $0.25$ to about $0.2353$, enough to move the answer by 6.25 percent. Drawing 2 people from a town of 40,000 shifts the second probability by less than $0.0001$, which no answer you write would ever show. When a question asks for an exact probability from a small pool, condition. Save the approximation for the sampling distributions and inference conditions later in the course.

Mistakes to avoid

• Multiplying $P(A) \cdot P(B)$ for two draws from the same finite pool. Cards, students, marbles, and defective parts pulled from a crate are all dependent unless replaced.
• Multiplying when the question asks for "at least one" rather than "both". Multiplication answers "both happen"; addition, with the overlap subtracted, answers "at least one happens". If you are unsure which is being asked, read union vs intersection.
• Assuming independence because the events sound unconnected. Independence is a numerical property, so test it or cite the problem statement.
• Treating mutually exclusive as independent. If $A$ and $B$ cannot both occur, then $P(A \cap B) = 0$, which cannot equal $P(A) \cdot P(B)$ when both probabilities are above zero.
• Conditioning on the wrong event. $P(B \mid A)$ is not $P(A \mid B)$. In the sport and job survey worked below, $P(\text{job} \mid \text{sport}) = \frac{24}{120} = 0.2$ while $P(\text{sport} \mid \text{job}) = \frac{24}{50} = 0.48$, so reversing the two events more than doubles the answer.
• Forgetting to say why in free response. Writing $\frac{13}{52} \times \frac{12}{51}$ with no words leaves the reader to guess why the second fraction changed. One sentence naming that the first card was not replaced is what earns the justification point.

AP note: where this shows up on the exam

The multiplication rule lives in Unit 2. Conditional probability is topic 2.6, which is where $P(A \mid B)$ and the general multiplication rule come from, and independent events are topic 2.7, where $P(A \cap B) = P(A) \cdot P(B)$ appears. Unit 2 is 15 to 25 percent of the multiple-choice section.

The classic trap gives you a two-way table and an "and" question. The table already contains the joint count, so the correct move is often to read $P(A \cap B)$ straight off it rather than multiply anything. When a free-response question does want independence, it will usually ask you to justify it, and the expected justification is the comparison of $P(A \cap B)$ with $P(A) \cdot P(B)$, or of $P(A \mid B)$ with $P(A)$. Run a few reps on the conditional probability practice set.