Multiplication theorem on probability

Draw a card from a shuffled deck, set it aside, then draw a second card. Does the second draw "remember" the first? It does — once a king is removed, both the number of kings and the total cards left have changed, so the odds for the second draw are no longer what they were for the first. The multiplication theorem on probability is the tool that turns this everyday intuition into an exact formula: it rearranges the definition of conditional probability into a rule for the probability that two (or more) events happen together. It is the workhorse for every "without replacement" problem — balls from a bag, cards from a deck, defective items from a batch — and it also quietly explains why independent events multiply so simply. This article derives the theorem, extends it to three events, and works two classic without-replacement problems end to end.

The core insight

The probability that two events both occur equals the probability of the first times the conditional probability of the second given the first has already happened: $P(A\cap B) = P(A)\,P(B\mid A)$ .

The core idea

From conditional probability to a product rule. The starting point is the definition of conditional probability itself, applied in both directions. For events $A, B$ of a sample space $S$ with $P(A) \neq 0$ and $P(B) \neq 0$ :

P(A\mid B) = \frac{P(A\cap B)}{P(B)}, \qquad P(B\mid A) = \frac{P(A\cap B)}{P(A)}.

where: $P(A\mid B)$ is the probability of $A$ given $B$ has occurred, $P(A\cap B)$ is the probability both $A$ and $B$ occur, and $P(A), P(B)$ are the individual (unconditional) probabilities. Cross-multiplying each equation isolates $P(A\cap B)$ two different ways, and both must agree — that agreement is the multiplication theorem.

The theorem itself.

P(A\cap B) = P(A)\,P(B\mid A) = P(B)\,P(A\mid B).

In words: the probability that both events happen equals the probability of the first, times the probability of the second given the first has already occurred. Either event can play the role of "first" — the formula is symmetric, and both routes must give the same number.

Extension to three (or more) events. For events $A, B, C$ with $P(A \cap B) \neq 0$ :

P(A\cap B\cap C) = P(A)\,P(B\mid A)\,P(C\mid A\cap B).

where: $P(C\mid A\cap B)$ is the probability of $C$ given that both $A$ and $B$ have already occurred. The pattern chains outward — multiply the probability of the first event by the conditional probability of each later event given everything before it. This is exactly how "three cards drawn in a row" or "three items picked one after another" problems are solved.

Why it matters for without-replacement draws. The theorem earns its keep whenever removing one item changes the composition of what remains — a ball not put back, a card not reshuffled in, a defective item not returned to the batch. The conditional probabilities $P(B\mid A)$ , $P(C\mid A\cap B)$ , and so on are exactly the "recount the shrinking pool" probabilities a student computes at each step.

Connection to independence. If $A$ and $B$ are independent, then by definition $P(B\mid A) = P(B)$ — knowing $A$ happened tells you nothing new about $B$ . Substituting this into the multiplication theorem collapses it to the familiar product rule:

P(A\cap B) = P(A)\,P(B).

So the multiplication theorem is the more general result — it holds whether or not the events are independent — and the simple product rule for independent events is just the special case where every conditional probability equals its unconditional counterpart. "With replacement" experiments are typically independent (each draw resets the pool); "without replacement" experiments typically are not, which is exactly why conditioning is unavoidable there.

Figure

Two draws from a bag showing how removing one red ball changes the count for the second draw

Derivation ADeriving the multiplication theorem from conditional probability

The goal is to show, starting only from the standard definition of conditional probability $P(A\mid B) = P(A\cap B)/P(B)$ for $P(B) \neq 0$ , that the probability of two events occurring together can always be written as a product of an unconditional probability and a conditional one, in either order, and that this single relation extends by the same reasoning to any number of events chained together through successive conditioning.

Step 1. Write down the definition of conditional probability for $B$ given $A$ : $P(B\mid A) = \dfrac{P(A\cap B)}{P(A)}$ , valid whenever $P(A) \neq 0$ . This is not an assumption — it is simply how conditional probability is defined, restricting attention to the outcomes inside $A$ and asking what fraction of those also lie in $B$ .

Step 2. Multiply both sides by $P(A)$ to clear the fraction: $P(A)\,P(B\mid A) = P(A\cap B)$ . Rearranging gives $P(A\cap B) = P(A)\,P(B\mid A)$ , the multiplication theorem in its usual form.

Step 3. Repeat the same algebra starting from the mirror definition $P(A\mid B) = P(A\cap B)/P(B)$ , valid when $P(B) \neq 0$ . Multiplying through by $P(B)$ gives $P(A\cap B) = P(B)\,P(A\mid B)$ . Both expressions equal the same quantity $P(A\cap B)$ , so $P(A)\,P(B\mid A) = P(B)\,P(A\mid B)$ — the theorem is symmetric in $A$ and $B$ , and either can be treated as "first."

Step 4. Extend to three events by conditioning twice. Treat $A \cap B$ as a single compound event and apply Step 2's reasoning to it and $C$ : $P((A\cap B)\cap C) = P(A\cap B)\,P(C\mid A\cap B)$ . Now expand $P(A\cap B)$ itself using Step 2: $P(A\cap B) = P(A)\,P(B\mid A)$ . Substituting gives $P(A\cap B\cap C) = P(A)\,P(B\mid A)\,P(C\mid A\cap B)$ , confirming that chaining conditional probabilities correctly reproduces the multi-event theorem, and the same substitution can be repeated for any further events.

Worked example

A bag contains 5 red and 3 black balls. Two balls are drawn one after another without replacement. Find the probability that both balls drawn are red.

Step 1. Define the events. Let $A$ = "first ball is red" and $B$ = "second ball is red." The quantity wanted is $P(A\cap B)$ .

Step 2. Find $P(A)$ . There are 8 balls in total, 5 of them red, so $P(A) = \dfrac{5}{8}$ .

Step 3. Find $P(B\mid A)$ . Given the first ball drawn was red and was not replaced, 7 balls remain, of which 4 are red: $P(B\mid A) = \dfrac{4}{7}$ .

Step 4. Apply the multiplication theorem.

P(A\cap B) = P(A)\,P(B\mid A) = \frac{5}{8}\times\frac{4}{7} = \frac{20}{56} = \frac{5}{14}.

The probability both balls drawn are red is $\dfrac{5}{14}$ .

Competitive stretch

Three cards are drawn successively, without replacement, from a well-shuffled standard deck of 52 cards. Let $A, B, C$ be "first card is a king," "second card is a king," "third card is a king." Then $P(A) = \dfrac{4}{52} = \dfrac{1}{13}$ , $P(B\mid A) = \dfrac{3}{51} = \dfrac{1}{17}$ (3 kings left in 51 cards), and $P(C\mid A\cap B) = \dfrac{2}{50} = \dfrac{1}{25}$ (2 kings left in 50 cards). By the three-event multiplication theorem:

P(A\cap B\cap C) = \frac{1}{13}\times\frac{1}{17}\times\frac{1}{25} = \frac{1}{5525}.

The probability all three cards drawn are kings is $\dfrac{1}{5525}$ .

Common traps

Reusing the first draw's probability for later draws even though sampling is without replacement — each conditional probability must reflect the pool as it stands after the earlier draws are removed.
Mixing up "with replacement" and "without replacement." With replacement, draws are independent and every draw uses the original probabilities; without replacement, later draws must be conditioned on earlier ones.
Updating only the numerator (or only the denominator) when computing $P(B\mid A)$ — after removing one red ball, both the red count and the total count decrease together.
Using the two-event formula for three or more events, instead of chaining conditional probabilities as $P(A)P(B\mid A)P(C\mid A\cap B)$ .

Ask the tutor: Stuck on how the probabilities should update at each draw in a bag-of-balls or deck-of-cards problem — ask the AI tutor to walk through the conditional probabilities step by step.

Why does the multiplication theorem need P(A) or P(B) to be nonzero?

Conditional probability $P(B\mid A) = P(A\cap B)/P(A)$ is a fraction with $P(A)$ in the denominator, so it is undefined if $P(A) = 0$ — you cannot meaningfully ask "given that an impossible event happened, what is the chance of B." The multiplication theorem is derived directly from this definition, so it inherits the same requirement. In practice, every event you actually condition on in a probability problem has a positive probability, so this is rarely an issue — it is mainly a technical condition to state when writing the theorem formally.

How is this different from just multiplying P(A) and P(B) directly?

Multiplying $P(A)$ and $P(B)$ directly only gives $P(A\cap B)$ when $A$ and $B$ are independent, meaning $A$ occurring does not change the chance of $B$ . The multiplication theorem is more general: it uses $P(B\mid A)$ , the probability of $B$ recalculated using the information that $A$ already happened. For without-replacement problems this recalculated probability is different from $P(B)$ because the pool of outcomes has changed, so plain multiplication $P(A)P(B)$ would give a wrong answer.

For the three-card king problem, could I condition on B and C instead of A and B for the last term?

Yes, as long as the conditioning event is everything that happened before the event you are computing. The standard chain conditions each new event on all previous ones: $P(C\mid A\cap B)$ for the third card uses both earlier results. You could reorder which card you call "first," "second," "third," but you cannot skip conditioning — the third draw's probability must always account for both cards already removed, not just one of them.

Multiplication theorem on probability

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads