> [!NOTE] Theorem (Law of Total Probability) > Let $(\Omega,\mathcal{F},\mathbb{P})$ be a [[Probability Space|probability space]]. Let $\mathbb{B}$ be a [[Countable Set|countable]] [[Partition of a Set|partition]] of the [[Sample Space|sample space]] $\Omega$ such that for all $B\in \mathbb{B},$ $\mathbb{P}(B)>0.$ Then for all $A\in \mathcal{F},$ $\mathbb{P}(A)= \sum_{B\in \mathbb{B}} \mathbb{P}(A \mid B) \mathbb{P}(B) $or equivalently, $\mathbb{P}(A)= \sum_{B\in \mathbb{B}}\mathbb{P}(A \cap B)$where $\mathbb{P}(A\mid B)$ denotes the [[Conditional Probability|conditional probability]] of $A$ given $B.$ ###### Proof Let $A\in \mathcal{F}.$ Then since $A \subset \Omega$ and $\mathbb{B}$ forms a partition of $\Omega,$ we have $A=A \cap \Omega = A \cap \bigcup_{B\in \mathbb{B}} B = \bigcup_{B\in \mathbb{B}} A \cap B$Since the elements of $\mathbb{B}$ are pairwise disjoint so are the $A\cap B$ for $B\in \mathbb{B}$. Thus by countable additivity of $\mathbb{P},$ indeed $\mathbb{P}(A)= \mathbb{P}\left( \bigcup_{B\in \mathbb{B}} A \cap B \right) = \sum_{B\in \mathbb{B}}\mathbb{P}(A \cap B) =\sum_{B\in \mathbb{B}} \mathbb{P}(A \mid B) \mathbb{P}(B). $ # Applications **Consequences**: [[Bayes' Theorem]] asserts that for all $A\in \mathcal{F}$ such that $\mathbb{P}(A)>0,$ $\mathbb{P}(B\mid A)= \frac{\mathbb{P}(A \mid B) \mathbb{P}(B)}{\sum_{D\in \mathbb{B}}\mathbb{P}(A\mid D)\mathbb{P}(D)}.$