How can I generalize the multiplication rule of probability for more than two events?

Question

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1. What the question is asking

We know the multiplication rule for two events

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

The student wants to know how this rule extends when we have many events
(A_{1},A_{2},\dots ,A_{n}).
In other words, we must show (or derive) that

[ \boxed{\; P\Bigl(\bigcap_{i=1}^{n} A_{i}\Bigr)= P(A_{1})\; P(A_{2}\mid A_{1})\; P(A_{3}\mid A_{1}\cap A_{2})\; \cdots\; P\bigl(A_{n}\mid A_{1}\cap\cdots\cap A_{n-1}\bigr) \;} ]

and we want an intuitive, step‑by‑step explanation, not just a terse “by induction” proof.

2. Derivation from the definition of conditional probability

2.1 The definition we start from

For any two events (X) and (Y) with (P(Y)>0),

[ P(X\mid Y)=\frac{P(X\cap Y)}{P(Y)}\qquad\Longleftrightarrow\qquad P(X\cap Y)=P(Y)\,P(X\mid Y). \tag{1} ]

Equation (1) is exactly the two‑event multiplication rule.

2.2 Three events

Take three events (A_{1},A_{2},A_{3}) (assume each intersection we use has positive probability).
Apply (1) to the pair

[ X = A_{3},\qquad Y = A_{1}\cap A_{2}. ]

Then

[ P(A_{3}\cap A_{1}\cap A_{2}) = P(A_{1}\cap A_{2})\;P\bigl(A_{3}\mid A_{1}\cap A_{2}\bigr). \tag{2} ]

But we still have the factor (P(A_{1}\cap A_{2})).
Apply (1) again, now with

[ X = A_{2},\qquad Y = A_{1}. ]

[ P(A_{1}\cap A_{2}) = P(A_{1})\;P(A_{2}\mid A_{1}). \tag{3} ]

Insert (3) into (2):

[ \begin{aligned} P(A_{1}\cap A_{2}\cap A_{3}) &= \bigl[ P(A_{1})\,P(A_{2}\mid A_{1}) \bigr]\; P\bigl(A_{3}\mid A_{1}\cap A_{2}\bigr)\[2mm] &= P(A_{1})\;P(A_{2}\mid A_{1})\;P(A_{3}\mid A_{1}\cap A_{2}). \end{aligned} ]

Thus the rule holds for three events.

2.3 The pattern becomes clear

When we added the third event we:

Isolated the last event using (1) with the whole previous intersection as the conditioning set.
Repeated the same step for the remaining intersection.

If we keep doing this, each new event is “peeled off” from the right‑hand side and appears as a conditional probability given everything that has already been peeled off.

2.4 Formal induction (optional but short)

Define

[ Q_n = P\Bigl(\bigcap_{i=1}^{n} A_i\Bigr). ]

Base case (n=2): by (1) we have
(Q_2 = P(A_1)P(A_2\mid A_1)).

Inductive step: Assume the formula holds for (n-1) events:

[ Q_{\,n-1}=P(A_1)\,P(A_2\mid A_1)\cdots P\bigl(A_{n-1}\mid A_1\cap\cdots\cap A_{n-2}\bigr). ]

Now apply (1) to the pair

[ X = A_n,\qquad Y = \bigcap_{i=1}^{n-1}A_i . ]

[ Q_n = P\Bigl(\bigcap_{i=1}^{n-1}A_i\Bigr)\; P\bigl(A_n \mid \bigcap_{i=1}^{n-1}A_i\bigr) = Q_{\,n-1}\; P\bigl(A_n \mid A_1\cap\cdots\cap A_{n-1}\bigr). ]

Replace (Q_{\,n-1}) with the induction hypothesis and we obtain exactly the product with (n) factors. Hence the formula is true for all (n).

3. Intuitive interpretation

Imagine performing a sequence of experiments:

First we check whether (A_{1}) occurs. Its probability is (P(A_{1})).
Given that (A_{1}) has happened, we now ask whether (A_{2}) happens. The chance of this is (P(A_{2}\mid A_{1})).
Given that both (A_{1}) and (A_{2}) have happened, we ask about (A_{3}); its chance is (P(A_{3}\mid A_{1}\cap A_{2})).
Continue until the (n)‑th step.

The overall probability that all steps succeed (i.e., that the intersection of all events occurs) is the product of the step‑by‑step conditional probabilities. This is exactly the chain rule we derived.

4. Final answer

[ \boxed{ P\Bigl(\bigcap_{i=1}^{n} A_{i}\Bigr)= P(A_{1})\; P(A_{2}\mid A_{1})\; P(A_{3}\mid A_{1}\cap A_{2})\; \cdots\; P\bigl(A_{n}\mid A_{1}\cap\cdots\cap A_{n-1}\bigr) } ]

provided each conditioning event has positive probability (otherwise the corresponding conditional probability is undefined).

5. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Treating the order as irrelevant	The product depends on the order of the events; (P(A_2	A_1)) is generally not equal to (P(A_1	A_2)).	Write the product with a clear order and keep the conditioning set as the previous events.
Multiplying unconditional probabilities	Assuming independence when it isn’t given gives (P(A_1\cap\cdots\cap A_n)=\prod_i P(A_i)).	Only replace a conditional term by an unconditional one if you have proved the events are independent.
Omitting a conditioning intersection	Writing (P(A_3	A_2)) instead of (P(A_3	A_1\cap A_2)) forgets the information that (A_1) already occurred.	Remember the conditioning set always contains all earlier events.
Dividing by zero	Using the definition (P(B	A)=P(A\cap B)/P(A)) when (P(A)=0) is illegal.	The rule is valid only when each conditioning event (the intersection of earlier events) has positive probability.
Assuming the formula “proves” independence	The chain rule holds for any events, independent or not.	Independence is an extra property: if the events are independent, every conditional factor collapses to an unconditional probability.

Keeping these points in mind will help you apply the multiplication (chain) rule correctly in any probability problem.

Original question: How can I generalize the multiplication rule of probability for more than two events? on Mathematics Stack Exchange, licensed CC BY-SA.