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Mathematics · Ch 14 — Probability

Axiomatic (Set-Theoretic) Probability

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Axiomatic (Set-Theoretic) Probability

Axiomatic (Set-Theoretic) Probability

Earlier classes may have introduced probability informally as number of favourable outcomestotal number of outcomes\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}} for experiments with equally likely outcomes. The axiomatic approach, due to the mathematician A. N. Kolmogorov, puts probability on a rigorous footing that works for any random experiment, not only ones with equally likely outcomes, and it is built entirely on the set-theoretic language of sample spaces and events developed above.

The Three Axioms

Let SS be the sample space of a random experiment. A probability function PP assigns to every event E⊆SE \subseteq S a real number P(E)P(E), subject to three axioms:

Axiom 1 (Non-negativity): P(E)≥0P(E) \ge 0 for every event EE.

Axiom 2 (Certainty): P(S)=1P(S) = 1.

Axiom 3 (Additivity): If EE and FF are mutually exclusive events (E∩F=ϕE \cap F = \phi), then P(E∪F)=P(E)+P(F)P(E \cup F) = P(E) + P(F). More generally, for any finite collection of pairwise mutually exclusive events E1,…,EnE_1, \dots, E_n, P(E1∪E2∪⋯∪En)=P(E1)+P(E2)+⋯+P(En)P(E_1 \cup E_2 \cup \cdots \cup E_n) = P(E_1) + P(E_2) + \cdots + P(E_n).

Any assignment of numbers to events that satisfies all three axioms is a valid probability function — the axioms constrain what is allowed, without forcing one particular formula.

Probability of Equally Likely Outcomes

When a sample space S={ω1,ω2,…,ωn}S = \{\omega_1, \omega_2, \dots, \omega_n\} has nn outcomes that are all equally likely, consistency with Axioms 2 and 3 forces each simple event to have probability 1n\dfrac{1}{n}: since the nn simple events {ω1},…,{ωn}\{\omega_1\}, \dots, \{\omega_n\} are pairwise mutually exclusive and their union is SS, Axiom 3 gives P(ω1)+⋯+P(ωn)=P(S)=1P(\omega_1) + \cdots + P(\omega_n) = P(S) = 1; equal likelihood then forces every term to equal 1n\frac{1}{n}. For any event EE built from mm of these outcomes, applying Axiom 3 again gives

P(E)=mn=n(E)n(S),P(E) = \frac{m}{n} = \frac{n(E)}{n(S)},

recovering the familiar classical formula as a consequence of the axioms, not as a separate definition. …