The Three Axioms of Probability

May 2026 · 14 min read

We have a sample space $S$ and a collection of events (subsets of $S$). The next step is to assign each event a number that captures “how likely” it is. Kolmogorov's 1933 framework says: pick any function $P$ from events to real numbers that satisfies just three rules, and you've got a legitimate probability. Every theorem we prove afterwards uses only those three rules.

Motivating the Axioms: Relative Frequency

Imagine repeating an experiment $n$ times under identical conditions. For each event $E$, let $n(E)$ be the number of trials in which $E$ occurred. The relative frequency of $E$ is

$\displaystyle f_n(E) = \frac{n(E)}{n}.$

Three things are obvious about $f_n$ for any finite $n$:

1. $f_n(E) \ge 0$, because counts are non-negative.
2. $f_n(S) = n/n = 1$, because some outcome happens every trial.
3. If $E$ and $F$ are mutually exclusive then $n(E \cup F) = n(E) + n(F)$, so $f_n(E \cup F) = f_n(E) + f_n(F)$.

Empirically, as $n \to \infty$, $f_n(E)$ seems to settle to a stable limit which we want to call $P(E)$. Whatever value it settles on, it should inherit those three properties. Kolmogorov's axioms are exactly that wishlist, written in mathematical form (and strengthened from finite to countable additivity to handle infinite sample spaces).

The Three Axioms

A probability is a function $P$ that assigns to each event $E \subseteq S$ a real number $P(E)$ satisfying:

That is the entire foundation. Every formula in the rest of probability is downstream of these three statements.

Finite Additivity Is Free

If you only have finitely many disjoint events $E_1, \ldots, E_n$, you can pad the sequence with $E_{n+1} = E_{n+2} = \cdots = \emptyset$. Since the empty event has probability $0$ (proved next post), the infinite sum collapses to the finite one and we get

So countable additivity is genuinely stronger than finite additivity — it says the rule extends to infinite sequences as well.

Why Countable Additivity?

Why not stop at finite additivity? Two reasons.

1. Infinite sample spaces force it. Toss a fair coin until you see the first head. The sample space is $\{H, TH, TTH, TTTH, \ldots\}$ — countably infinite. Each elementary outcome “first head on toss $k$” has probability $1/2^k$. To recover $P(S) = 1$ we need

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{2^k} = 1.$

Without countable additivity, that infinite sum would not be guaranteed to equal $P(S)$, and the model would be inconsistent.

2. Continuity, which we prove in Part 6. Countable additivity is exactly equivalent to: $\lim_n P(E_n) = P(\lim_n E_n)$ for monotone sequences of events. Without that, we couldn't even say that the probability of “a coin eventually shows heads” equals $\lim_n P(\text{heads in first } n \text{ tosses})$.

Why $P(E) \le 1$

We marked Axiom 1 as $0 \le P(E) \le 1$. The lower bound is part of the axiom. For the upper bound: $S = E \cup E^c$ with $E$ and $E^c$ disjoint, so by finite additivity $1 = P(S) = P(E) + P(E^c)$. Both summands are non-negative, so each is at most $1$. Done.

Two Interpretations of $P$

The axioms tell us what $P$ is mathematically (a measure of total mass $1$) but not what $P(E) = 0.3$ means. There are two main interpretations, both consistent with the axioms:

• Frequentist. $P(E)$ is the long-run fraction of trials in which $E$ would occur if we repeated the experiment indefinitely. This is what the relative-frequency motivation captures.
• Subjective / Bayesian. $P(E)$ is a rational agent's degree of belief in $E$. We will see in Part 7 (the Dutch-book argument) that any coherent system of betting odds must satisfy the same three axioms.

Different schools of thought, identical mathematics.

Building a Probability Model

To specify a probability you only need to assign weights $p_s$ to each elementary outcome $s \in S$ (when $S$ is countable) such that

• $p_s \ge 0$ for every $s$ (Axiom 1),
• $\sum_{s \in S} p_s = 1$ (Axiom 2),
• and define $P(E) = \sum_{s \in E} p_s$ (then Axiom 3 falls out automatically).

For example, a fair die has $p_s = 1/6$ for each face; a biased coin has $p_H = 0.7$, $p_T = 0.3$; a Poisson model on $\{0, 1, 2, \ldots\}$ has $p_k = e^{-\lambda} \lambda^k / k!$. All three respect the axioms.

Visual: Building Total Probability to 1

Step through the elementary outcomes of an experiment, watching how the probability mass accumulates. The total reaches exactly $1$ — that is Axiom 2 in action.

Non-examples (what the axioms rule out)

Watching what the axioms forbid sharpens the intuition.

• Assign $P(\{1\}) = 0.6$ and $P(\{2\}) = 0.6$ on a coin with $S = \{1, 2\}$. Then $P(S) = 1.2 \ne 1$ — violates Axiom 2.
• Try to put a uniform probability on the integers $\mathbb{Z}$ with $P(\{k\}) = c$ for every $k$. If $c = 0$, then $P(\mathbb{Z}) = 0$ by countable additivity, not $1$. If $c > 0$, the sum is $\infty$, not $1$. Either way it violates Axiom 2 + 3 jointly — uniform distributions on countably infinite sets are impossible.
• $P(E) = -0.1$ for some event — violates Axiom 1.
• $P(E \cup F) = P(E) - P(F)$ for disjoint $E, F$ — violates Axiom 3.

Worked Example

An urn contains 3 red, 2 white, and 5 black balls. One is drawn uniformly at random.

1. $S = \{R_1, R_2, R_3, W_1, W_2, B_1, \ldots, B_5\}$, $|S| = 10$.
2. Each elementary outcome has probability $1/10$ (Axiom 1 satisfied; Axiom 2 satisfied because $10 \cdot 1/10 = 1$).
3. For event $E$ = "red", $E = \{R_1, R_2, R_3\}$. By Axiom 3 applied to the three disjoint singletons, $P(E) = 3/10$.
4. For event $F$ = "white", $P(F) = 2/10$.
5. $E$ and $F$ are disjoint, so $P(E \cup F) = P(E) + P(F) = 5/10$.

Every step cited an axiom. That is the discipline this chapter is teaching.

Common Mistakes

• Using additivity on overlapping events. $P(A \cup B) = P(A) + P(B)$ requires $A \cap B = \emptyset$. We fix this in Part 4 with inclusion–exclusion.
• Assuming all outcomes are equally likely. Axiom 1 only says $P \ge 0$; it does not say all elementary events are equal. Equal probabilities are a modelling choice (Part 5), not an axiom.
• Using countable additivity on uncountably many events. The axiom needs a sequence $E_1, E_2, \ldots$ — not an arbitrary indexed family.
• Forgetting that the third axiom requires pairwise disjointness, not just adjacent disjointness.

Practice

1. Show that any function $P$ defined on a finite sample space by $P(E) = \sum_{s \in E} p_s$ for non-negative weights $p_s$ summing to $1$ satisfies all three axioms.
2. An experiment has $S = \{a, b, c, d\}$ with $P(\{a\}) = 0.2$, $P(\{b\}) = 0.3$, $P(\{c\}) = x$, $P(\{d\}) = 0.1$. Find $x$.
3. Toss a fair coin until the first head appears. Verify Axiom 2 by summing the geometric series $\sum_{k \ge 1} 1/2^k$.
4. Disprove that there exists a probability $P$ on $\mathbb{N}$ with $P(\{k\}) = c$ constant in $k$.
5. Suppose $P(A) = 0.5$, $P(B) = 0.4$, $A \cap B = \emptyset$. Use the axioms to find $P(A \cup B)$, $P((A \cup B)^c)$, and bound $P(A \cap C)$ for an arbitrary event $C$.