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Kolmogorov axioms

There are a few ways to axiomatize the subject of probability. This is one way of doing it:

$$P(E) \in \mathbb, P(E) \ge 0$$ $$P(\Omega) = 1$$ $$P\left(\bigcup_{i=1}^nE_i\right) = \sum_{i=1}^nE_n \text{ , where all sets E_i are mutually exclusive}$$

Consequences

MathJax TeX Test Page By definition, mutually exclusive means the intersection is the empty set. $$\Omega \cup \emptyset = \Omega$$ $$\Omega \cap \emptyset = \emptyset$$ $$P(\Omega \cup \emptyset) = P(\Omega) + P(\emptyset) \text{ (by property 3})$$ $$P(\Omega \cup \emptyset) = P(\Omega) = 1 \text{ By property 1}$$ $$1 = 1 + P(\emptyset) \to \boxed{P(\emptyset) = 0}$$

For this second problem, let's start with a few facts: $$A \cap (B - A) = A \cap (B \cap A^c) = (A \cap A^c) \cap B = \emptyset$$ $$A \cup (B - A) = A \cup (B \cap A^c) = A \cup B$$ Now, let's make our assumptions. Let's say $A$ is completely contained in $B$. So, $A \cap B = A$ and $A \cup B = B$. $$p(A \cup (B - A)) = p(A) + p(B-A) \text{ (prop 3)}$$ $$p(B) = P(A) + p(B-A)$$ $$\text A \in B \to p(B) \geq P(A) \text{ (property 1. All probabilities are \geq 0)}$$ From this, we can prove something else. All sets are $\in \Omega$. Therefore, $p(S) \leq P(\Omega) = 1$. $$\forall S \in \Omega, \boxed{0 \leq p(S) \leq 1} \text{ (prop. 1)}$$