Probability Axioms

Sample spaces and events

A sample space $\Omega$ is the set of all possible outcomes of a random experiment. A probability measure assigns to each subset $E \subseteq \Omega$ (called an event) a number $P(E)$ satisfying three axioms stated by Kolmogorov (1933):

1. Non-negativity: $P(E) \geq 0$ for every event $E$.
2. Normalization: $P(\Omega) = 1$.
3. Countable additivity: if $E_1, E_2, \ldots$ are mutually exclusive (disjoint) events, then $P(E_1 \cup E_2 \cup \cdots) = P(E_1) + P(E_2) + \cdots$

From these three axioms every other probability rule can be derived — nothing extra needs to be assumed.

Derived rules

Complement rule. Because $E$ and its complement $E^c$ partition $\Omega$, axioms 2 and 3 give

$P(E^c) = 1 - P(E).$

Addition rule. For any two events $A$ and $B$ (not necessarily disjoint),

$P(A \cup B) = P(A) + P(B) - P(A \cap B).$

The subtraction corrects for double-counting the overlap $A \cap B$. When $A$ and $B$ are mutually exclusive $P(A \cap B) = 0$ and the rule reduces to axiom 3.

Example. Roll a fair six-sided die, so $\Omega = \{1,2,3,4,5,6\}$ and each outcome has probability $\frac{1}{6}$ by symmetry (normalization and equal weights). Let $A = \{1,2,3\}$ and $B = \{3,4,5\}$. Then $A \cap B = \{3\}$, so

$P(A \cup B) = \frac{3}{6} + \frac{3}{6} - \frac{1}{6} = \frac{5}{6}.$

Why this matters for quantum mechanics

Quantum measurement outcomes are inherently probabilistic — the Born rule says the probability of measuring outcome $k$ is $p_k = |\langle k | \psi \rangle|^2$. For these probabilities to be consistent, the set $\{p_k\}$ must satisfy exactly Kolmogorov's axioms: each $p_k \geq 0$, and the sum over all possible outcomes equals 1 (normalization of $|\psi\rangle$ guarantees this). Understanding the axioms is therefore a prerequisite for reasoning about quantum measurement.

Try it

This is a numerical exercise — your code should return a number, not a circuit. A fair six-sided die is rolled once. Let $A = \{1,2,3\}$ and $B = \{3,4,5\}$. Compute $P(A \cup B)$ using the addition rule.

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