> [!NOTE] Theorem (Chain Rule or Multiplication Rule) > Let $(\Omega,\mathcal{F},\mathbb{P})$ be a [[Probability Space|probability space]]. Let $A_{1},\dots, A_{n}\in \mathcal{F}$ such that $\mathbb{P}(\cap_{i=1}^{n-1} A_{i})>0.$ Then $\begin{align} \mathbb{P}\left(A_1 \cap A_2 \cap \ldots \cap A_n\right) & =\mathbb{P}\left(A_1\right) \mathbb{P}\left(A_2 \mid A_1\right) \mathbb{P}\left(A_3 \mid A_1 \cap A_2\right) \cdots \mathbb{P}\left(A_n \mid A_1 \cap \cdots \cap A_{n-1}\right) \\ & =\mathbb{P}\left(A_1\right) \prod_{j=1}^{n-1} \mathbb{P}\left(A_{j+1} \bigg | \bigcap_{i=1}^{j} A_{i} \right) \end{align}$where $\mathbb{P}(A\mid B)$ denotes the [[Conditional Probability|conditional probability]] of $A$ given $B.$ **Proof**: Note that for $k=1,\dots,n-1$ we have $A_{1} \cap A_{2} \cap \dots \cap A_{n} \subset A_{1} \cap A_{2} \cap \dots \cap A_{k}.$ Hence by [[Probability of Subset of Event]], $\mathbb{P}(A_{1} \cap A_{2} \cap \dots \cap A_{k}) \geq \mathbb{P}(A_{1} \cap A_{2} \cap \dots \cap A_{n} )>0 .$This ensures that all the conditional probabilities at the right hand side are well-defined. The result follows by the direct application of the definition of conditional probability on the RHS: $\begin{aligned}&\mathbb{P}(A_1)\mathbb{P}(A_2|A_1)\mathbb{P}(A_3|A_1\cap A_2)\ldots\mathbb{P}(A_n|A_1\cap\cdots\cap A_{n-1})\\&=\mathbb{P}(A_1)\frac{\mathbb{P}(A_1\cap A_2)}{\mathbb{P}(A_1)}\frac{\mathbb{P}(A_3\cap A_1\cap A_2)}{\mathbb{P}(A_1\cap A_2)}\ldots\frac{\mathbb{P}(A_1\cap\cdots\cap A_n)}{\mathbb{P}(A_1\cap\cdots\cap A_{n-1})}\\&=\mathbb{P}(A_1\cap\cdots\cap A_n).\end{aligned}$