> [!NOTE] Definition (Conditional Probability) > Let $(\Omega, \mathcal{F},\mathbb{P})$ be a [[Probability Space|probability space]]. Let $B\in F$ such that $\mathbb{P}(B)>0.$ > >Define $P_{B}:F\to \mathbb{R}$ by $P_{B}(A)= \mathbb{P}(A\mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$where $\mathbb{P}(A\mid B)$ is the [[Conditional Probability|conditional probability]] of $A$ given $B.$ > >Then $(\Omega, \mathcal{F},P_{B})$ is a probability space: that is, $P_{B}$ is well-defined ($P_{B}(A)$ exists for all $A\in \mathcal{F}$) and is a [[Probability Measure|probability measure]] on $(\Omega,\mathcal{F}).$ **Proof**: Let $A\in \mathcal{F}.$ Since $B\in \mathcal{F},$ $A \cap B \in \mathcal{F}$ as [[Event Spaces are Closed Under Finite Intersections]]. So $\mathbb{P}(A \cap B)$ is exists. Since $\mathbb{P}(B)>0,$ the ratio $P_{B}(A)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$ is well defined. Now we check that $P_{B}$ satisfies Kolmogorov axioms: (1) We have $A\cap B \subset B$ thus by [[Probability of Subset of Event is Less Than or Equal to Probability of Event|inequality for probability of subset]], $\mathbb{P}(A\cap B) \leq \mathbb{P} (B).$ It follows that $\mathbb{P}_{B}(A)\leq 1.$ On the other hand, since $\mathbb{P}(A \cap B)>0$ and $\mathbb{P}(B)>0,$ it follows that $P_{B}(A)>0.$ Therefore $P_{B}(A)\in[0,1].$ (2) $P_{B}(\Omega)= \frac{\mathbb{P}(\Omega \cap B)}{\mathbb{P}(B)}= \frac{\mathbb{P}(B)}{\mathbb{P}(B)} =1,$as required. (3) Let $(A_{n})_{n\geq 1}$ be a sequences of pairwise disjoint events. Then $ \begin{aligned} P_{B}\left(\bigcup_{n=1}^{\infty} A_n\right) & =\frac{1}{\mathbb{P}(B)} \mathbb{P}\left(\left(\bigcup_{n=1}^{\infty} A_n\right) \cap B\right) \\ & =\frac{1}{\mathbb{P}(B)} \mathbb{P}\left(\bigcup_{n=1}^{\infty}\left(A_n \cap B\right)\right) \end{aligned} $ Now, since $A_n \in \mathcal{F}$ and $B \in \mathcal{F}$ for all $n \geqslant 1$, it follows that $A_n \cap B \in \mathcal{F}$ for all $n \geqslant 1$. Moreover, the events $A_n \cap B$ are disjoint. Indeed, for $n \neq m$ $ \left(A_n \cap B\right) \cap\left(A_m \cap B\right) \subseteq A_n \cap A_m=\emptyset . $ By definition, the probability measure $\mathbb{P}$ is countably additive which gives $ P_{B}\left(\bigcup_{n=1}^{\infty} A_n\right)=\frac{1}{\mathbb{P}(B)} \sum_{n=1}^{\infty} \mathbb{P}\left(A_n \cap B\right)=\sum_{n=1}^{\infty} \frac{\mathbb{P}\left(A_n \cap B\right)}{\mathbb{P}(B)}=\sum_{n=1}^{\infty} \mathbb{P}\left(A_n \mid B\right)=\sum_{n=1}^{\infty} P_B\left(A_n\right) $