Probability Foundations
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Probability Axioms

Every event has a probability. For example, P(A) means “the probability of event A happening.” A rule that assigns such probabilities to all events is called a probability law. All probability laws must satisfy three fundamental axioms, formalized by Andrey Kolmogorov in 1933.


The Three Axioms

Axiom 1: Non-negativity

For any event A:

P(A)≥0P(A) \geq 0

Probability is always a non-negative number. You cannot have a “negative chance” of something happening.


Axiom 2: Normalization

The probability of the entire sample space Ω:

P(Ω)=1P(\Omega) = 1

Some outcome must occur — the total probability across all possible outcomes equals 1 (or 100%).

Normalization means scaling all weights so the total equals 1. Imagine each outcome has a “weight” — some measure of how likely it is. These weights could sum to any number: 16, 100, even ∞. Dividing each weight by the total turns raw weights into proper probabilities.

Example: A biased dice has outcome weights 2, 3, 1, 4, 2, 4 (total = 16). Normalizing by dividing each weight by 16 gives probabilities: 2/16, 3/16, 1/16, 4/16, 2/16, 4/16. Now P(Ω) = 1.


Axiom 3: Additivity

If events A and B are mutually exclusive (disjoint, meaning A ∩ B = ∅):

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

In words: if two events cannot happen at the same time, the probability that either one happens equals the sum of their individual probabilities.

Example: A blood donor has blood type A with probability 0.36, and type B with probability 0.11. Since a person can only have one blood type, these are disjoint:

P(A or B)=0.36+0.11=0.47P(\text{A or B}) = 0.36 + 0.11 = 0.47

This extends beyond two events. For any countable sequence of pairwise disjoint events A₁, A₂, A₃, …:

P(⋃i=1∞Ai)=∑i=1∞P(Ai)P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)

Here each Aᵢ is an event, and the subscript i just tells you which one — A₁ is the first, A₂ is the second, and so on. The big union ∪ means “any of these events occurs,” and the sum Σ adds up all their probabilities. Pairwise disjoint means any two of them share no outcomes (A₁ ∩ A₂ = ∅, A₁ ∩ A₃ = ∅, A₂ ∩ A₃ = ∅, etc.).

Example: A student’s test score falls into exactly one category: below 60 (probability 0.15), 60–79 (0.45), 80–89 (0.25), or 90–100 (0.15). These are pairwise disjoint — a score can’t be in two categories at once. By additivity:

P(score≥80)=P(80–89)+P(90–100)=0.25+0.15=0.40P(\text{score} \geq 80) = P(80\text{–}89) + P(90\text{–}100) = 0.25 + 0.15 = 0.40


Derived Properties

From these three simple axioms, we can derive many important properties:

Complement Rule

P(Ac)=1−P(A)P(A^c) = 1 - P(A)

The probability that A does not happen equals 1 minus the probability that A does happen.

Proof: A and Aᶜ are disjoint, and A ∪ Aᶜ = Ω. By Axioms 2 and 3:

P(A)+P(Ac)=P(Ω)=1P(A) + P(A^c) = P(\Omega) = 1


Probability Bounds

0≤P(A)≤10 \leq P(A) \leq 1

Every probability is between 0 and 1.

Proof:

  • Lower bound P(A) ≥ 0: this is Axiom 1 directly — all probabilities are non-negative.
  • Upper bound P(A) ≤ 1: by the complement rule, P(A) = 1 − P(Aᶜ). Since Aᶜ is also an event, Axiom 1 gives P(Aᶜ) ≥ 0. Subtracting a non-negative number from 1 can only make it smaller or equal, so P(A) ≤ 1.

Inclusion-Exclusion Principle

For any two events A and B (not necessarily disjoint):

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

We subtract P(A ∩ B) because it was counted twice—once in P(A) and once in P(B).


Monotonicity

If A ⊆ B (every outcome in A is also in B):

P(A)≤P(B)P(A) \leq P(B)

In words: a larger event is at least as likely as a smaller one.

Proof: We can split B into two disjoint pieces — the outcomes in A, and the outcomes in B that are not in A (written B \ A). Since these are disjoint:

P(B)=P(A)+P(B∖A)P(B) = P(A) + P(B \setminus A)

By Axiom 1, P(B \ A) ≥ 0, so P(B) ≥ P(A).


Empty Set

P(∅)=0P(\emptyset) = 0

The impossible event has probability zero.

Proof: The empty set ∅ and the sample space Ω are disjoint (they share no outcomes), and their union is Ω itself (adding nothing changes nothing). By Axiom 3:

P(Ω∪∅)=P(Ω)+P(∅)P(\Omega \cup \emptyset) = P(\Omega) + P(\emptyset)

But Ω ∪ ∅ = Ω, so the left side is just P(Ω). This gives:

P(Ω)=P(Ω)+P(∅)P(\Omega) = P(\Omega) + P(\emptyset)

Subtracting P(Ω) from both sides: P(∅) = 0.


Example: Fair Dice

Consider rolling a fair six-sided dice. The sample space is:

Ω={1,2,3,4,5,6}\Omega = \{1, 2, 3, 4, 5, 6\}

Each outcome has probability 1/6. Let A = “roll an even number” = {2, 4, 6}.

Using the axioms:

P(A)=P({2})+P({4})+P({6})=16+16+16=12P(A) = P(\{2\}) + P(\{4\}) + P(\{6\}) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}

This uses Axiom 3 (additivity) since {2}, {4}, {6} are disjoint.


Key Terms

Probability LawA function P assigning numbers to events satisfying the axioms
EventA subset of the sample space
AxiomA fundamental rule accepted without proof
Kolmogorov AxiomsThe three axioms defining probability theory
NormalizationScaling weights so the total equals 1
Countable AdditivityProbabilities of disjoint events add up, even for infinitely many