All regular [[Probability Axioms]] apply to [[Conditional Probability]] as well. - *Non-negativity:* All conditional probabilities are non-negative. For that it just needs to be ensured that the condition has a positive probability $\mathbf P(B) >0$. $ \mathbf P(A \vert B) \ge 0 $ - *Normalization:* Within the conditioned [[Sample Space]], all probabilities sum to $1$. $ \mathbf P(\Omega \vert B) = \frac{\mathbf P(\Omega \cap B)}{\mathbf P(B)} = \frac{\mathbf P(B)}{\mathbf P(B)}=1 $ - *Additivity:* When two [[Events]] are disjoint $A \cap C= \emptyset$, they are also disjoint in any possible conditional sample space. $ \mathbf P(A \cup C \vert B)=\mathbf P(A \vert B) + \mathbf P(C\vert B) $ Let us prove that additivity also works in the conditional case: $ \begin{align} \mathbf P(A \cup C \vert B) &= \frac{\mathbf P\big((A \cup C) \cap B\big)}{\mathbf P(B)} \tag{1}\\[6pt] &= \frac{\mathbf P(A \cap B) \cup \mathbf P (B \cap C)}{\mathbf P(B)} \tag{2}\\[6pt] &= \frac{\mathbf P(A \cap B)}{\mathbf P(B)} \cup \frac{\mathbf P(B \cap C)}{\mathbf P(B)} \tag{3}\\[6pt] &=\mathbf P(A \vert B) + \mathbf P(C \vert B) \tag{4} \end{align} $ where: - (1) We express the conditional probability in the form of Bayes theorem. - (3) The numerator can be split into two terms.