$P(A) \geq 0 \quad \text{for every event } A$ $P(S) = 1$ $ P\!\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i) \quad \text{where } A_1, A_2, \ldots \text{are mutually exclusive}$ The Axioms of Probability (Kolmogorov's axioms) are three foundational rules that any valid [[Probability]] $P$ must satisfy: 1. **Non-negativity**: Probability is > 0 2. **Normalization**: The probability of the sample space is 1. 3. **Countable Additivity**: The probability of the union of mutually exclusive events is the sum of the probability of each event. --- > [!example]- Axioms satisfied? {💡 Example} > A probability model assigns $P(A) = 0.3$, $P(B) = 0.5$, and $P(A \cup B) = 0.9$ where $A$ and $B$ are mutually exclusive. Does this violate the axioms of probability? > > > [!answer]- Answer > > If $A$ and $B$ are mutually exclusive, countable additivity requires $P(A \cup B) = P(A) + P(B) = 0.3 + 0.5 = 0.8$. Since the model states $P(A \cup B) = 0.9 \neq 0.8$, **yes, it violates the third axiom**.