Skip to content
VidiMaster it, module by module
Module 6/Probability Spaces & Rules

Probability Spaces: Sample Space, Events & Axioms

Every probability question begins with a probability space (Ω,F,P)(\Omega,\mathcal{F},\mathbb{P}) (Definition 10.1): a sample space Ω\Omega holding every possible outcome ω\omega, a collection of events F\mathcal{F} (the subsets of Ω\Omega we assign probabilities to), and a probability measure P\mathbb{P} obeying three axioms — every probability lies in [0,1][0,1], the certain event has P(Ω)=1\mathbb{P}(\Omega)=1 (and the impossible event P(∅)=0\mathbb{P}(\varnothing)=0), and the probabilities of disjoint events add. This lesson translates experiments into the language of sets, unpacks the axioms and their first consequences (Remark 10.1), and — when Ω\Omega is finite with equally likely outcomes — collapses probability into pure counting via P(A)=∣A∣∣Ω∣\mathbb{P}(A)=\tfrac{|A|}{|\Omega|} (Proposition 10.1). We put the machinery to work on a fair die (Example 10.1), a pair of distinguishable dice (Example 10.2), and an urn draw (Example 10.3).

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

An experiment has sample space Ω={ω1,ω2,ω3}\Omega=\{\omega_1,\omega_2,\omega_3\}. Someone proposes the assignment P({ω1})=0.5\mathbb{P}(\{\omega_1\})=0.5, P({ω2})=0.3\mathbb{P}(\{\omega_2\})=0.3, P({ω3})=0.3\mathbb{P}(\{\omega_3\})=0.3. True or false: this is a valid probability measure.

Which of the following is one of the axioms that every probability measure P\mathbb{P} must satisfy (Definition 10.1)?

What you’ll be able to do

  • State Definition 10.1: a probability space is a triple (Ω,F,P)(\Omega,\mathcal{F},\mathbb{P}) — a sample space Ω\Omega of outcomes ω\omega, a collection of events F\mathcal{F} (subsets of Ω\Omega), and a probability measure P:F→R\mathbb{P}:\mathcal{F}\to\mathbb{R} satisfying the three axioms.
  • List the three axioms and read off their meaning: 0≤P(A)≤10\le\mathbb{P}(A)\le 1, P(Ω)=1\mathbb{P}(\Omega)=1 with P(∅)=0\mathbb{P}(\varnothing)=0, and countable additivity P ⁣(⋃i=1∞Ai)=∑i=1∞P(Ai)\mathbb{P}\!\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty}\mathbb{P}(A_i) for pairwise disjoint events.
  • Translate between experiments and set theory — outcomes as points ω∈Ω\omega\in\Omega, events as subsets, and compound events A∪BA\cup B, A∩BA\cap B (both occur), Ac=Ω∖AA^c=\Omega\setminus A (does not occur) — and recognise mutually exclusive events as those with A∩B=∅A\cap B=\varnothing (Remark 10.1).
  • Apply finite additivity P(A1∪⋯∪An)=P(A1)+⋯+P(An)\mathbb{P}(A_1\cup\cdots\cup A_n)=\mathbb{P}(A_1)+\cdots+\mathbb{P}(A_n) for disjoint events to compute an event's probability by summing over its outcomes (Example 10.1).
  • Use Proposition 10.1: when Ω\Omega is finite with equally likely outcomes, P(A)=∣A∣/∣Ω∣\mathbb{P}(A)=|A|/|\Omega|, applying it to dice (Examples 10.1–10.2) and urns (Example 10.3) by counting favourable outcomes.

In your course

· MATH2015 · Linear Algebra & Probability
§10.1 Probability Space
  • Definition 10.1Probability space (Ω,F,P)(\Omega,\mathcal{F},\mathbb{P}) and the three axioms
    A sample space Ω\Omega, a collection of events F\mathcal{F}, and a measure P\mathbb{P} with 0≤P(A)≤10\le\mathbb{P}(A)\le 1, P(Ω)=1\mathbb{P}(\Omega)=1, P(∅)=0\mathbb{P}(\varnothing)=0, and countable additivity P ⁣(⋃iAi)=∑iP(Ai)\mathbb{P}\!\left(\bigcup_i A_i\right)=\sum_i\mathbb{P}(A_i) for pairwise disjoint AiA_i.
  • Remark 10.1Empty event, mutually exclusive events, finite additivity
    P(∅)=0\mathbb{P}(\varnothing)=0; A,BA,B are disjoint when A∩B=∅A\cap B=\varnothing; and P(A1∪⋯∪An)=P(A1)+⋯+P(An)\mathbb{P}(A_1\cup\cdots\cup A_n)=\mathbb{P}(A_1)+\cdots+\mathbb{P}(A_n) for disjoint events.
  • Example 10.1Rolling a fair die: Ω={1,…,6}\Omega=\{1,\dots,6\}, P(even)=12\mathbb{P}(\text{even})=\tfrac12
  • Remark 10.2A loaded die — probabilities need not be uniform (P({6})=27\mathbb{P}(\{6\})=\tfrac27)
  • Example 10.2A pair of fair dice: ∣Ω∣=36|\Omega|=36, P(sum=8)=536\mathbb{P}(\text{sum}=8)=\tfrac{5}{36}
  • Proposition 10.1Probability by counting
    If Ω\Omega is finite with equally likely outcomes, then P(A)=∣A∣/∣Ω∣\mathbb{P}(A)=|A|/|\Omega|.
  • Example 10.3Drawing from an urn: counting equally likely subsets with (nk)\binom{n}{k}
Grounded in §10.1. The OCR glyphs are restored to Ω\Omega (sample space), ω\omega (outcome), ∅\varnothing (empty event), ∈\in, ⊆\subseteq, ∪\cup, ∩\cap, AcA^c, P\mathbb{P} and ≤\le. Proposition 10.1 (probability by counting) and Example 10.3 (urn) are the standard equally-likely results of this section; the urn here uses 66 red and 44 white balls as a representative instance.
1

The probability space $(\Omega,\mathcal{F},\mathbb{P})$

A probability model (Definition 10.1) formalises a random experiment with three ingredients. The sample space Ω\Omega is the set of all possible outcomes; its elements are called sample points (or outcomes) and are usually written ω\omega. For a single roll of a die, Ω={1,2,3,4,5,6}\Omega=\{1,2,3,4,5,6\} and a typical sample point is ω=4\omega=4. The events form a collection F\mathcal{F} of subsets of Ω\Omega: an event A∈FA\in\mathcal{F} is a set of outcomes, and we say AA occurs when the result of the experiment lies in AA. 'The roll is even' is the event A={2,4,6}⊆ΩA=\{2,4,6\}\subseteq\Omega. Finally, the probability measure P:F→R\mathbb{P}:\mathcal{F}\to\mathbb{R} assigns each event a number measuring how likely it is. The triple (Ω,F,P)(\Omega,\mathcal{F},\mathbb{P}) is the probability space — the complete description of the experiment, and the object over which every later theorem is stated.

2

The three axioms of $\mathbb{P}$

A function P\mathbb{P} earns the name probability measure only if it obeys three axioms (Definition 10.1). (1) Range: 0≤P(A)≤10\le\mathbb{P}(A)\le 1 for every event AA — probabilities are never negative and never exceed 11. (2) Normalisation: P(Ω)=1\mathbb{P}(\Omega)=1 (something in Ω\Omega is certain to happen) and P(∅)=0\mathbb{P}(\varnothing)=0 (the empty event, 'nothing happens', is impossible). (3) Countable additivity: if A1,A2,A3,…A_1,A_2,A_3,\dots are pairwise disjoint (meaning Ai∩Aj=∅A_i\cap A_j=\varnothing whenever i≠ji\neq j), then P ⁣(⋃i=1∞Ai)=∑i=1∞P(Ai).\mathbb{P}\!\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty}\mathbb{P}(A_i). Additivity is the workhorse: the probability of a union of non-overlapping events is the sum of their probabilities. Taking all but finitely many AiA_i empty gives the finite version (Remark 10.1), P(A1∪⋯∪An)=P(A1)+⋯+P(An)\mathbb{P}(A_1\cup\cdots\cup A_n)=\mathbb{P}(A_1)+\cdots+\mathbb{P}(A_n) — the rule that lets us add the probabilities of individual outcomes to get the probability of an event.

3

Building new events from old

Because events are sets, we combine them with set operations, and each has a plain-language meaning (Remark 10.1). For events A,BA,B: the union A∪BA\cup B is the event that AA occurs, or BB occurs, or both; the intersection A∩BA\cap B (also written ABAB) is the event that both occur; the complement Ac=Ω∖AA^c=\Omega\setminus A is the event that AA does not occur, collecting every outcome of Ω\Omega outside AA; and A∩BcA\cap B^c is the event that AA occurs but BB does not. Two events are mutually exclusive (disjoint) when A∩B=∅A\cap B=\varnothing — they cannot occur together — which is exactly the hypothesis the additivity axiom needs. Every event and its complement partition the sample space: A∪Ac=ΩA\cup A^c=\Omega and A∩Ac=∅A\cap A^c=\varnothing, so exactly one of AA, AcA^c occurs. For three or more events the distributive laws A∩(B∪C)=(A∩B)∪(A∩C)A\cap(B\cup C)=(A\cap B)\cup(A\cap C) and De Morgan's laws (A∪B)c=Ac∩Bc(A\cup B)^c=A^c\cap B^c, (A∩B)c=Ac∪Bc(A\cap B)^c=A^c\cup B^c rewrite compound events — the algebra behind nearly every probability rule to come.

4

Equally likely outcomes: probability by counting

When a sample space is finite and all its outcomes are equally likely, probability reduces to counting. If ∣Ω∣=n|\Omega|=n, the nn single-outcome events are disjoint and their union is Ω\Omega, so additivity together with P(Ω)=1\mathbb{P}(\Omega)=1 forces each one to have probability 1n\tfrac1n. Summing over the outcomes in an event AA then gives Proposition 10.1: P(A)=∣A∣∣Ω∣=number of outcomes favourable to Atotal number of outcomes.\mathbb{P}(A)=\frac{|A|}{|\Omega|}=\frac{\text{number of outcomes favourable to }A}{\text{total number of outcomes}}. This is the classical 'favourable over total' rule, and it turns probability questions into combinatorics — count ∣A∣|A| and ∣Ω∣|\Omega|. The crucial caveat is the hypothesis: the outcomes must be equally likely. A loaded die (Remark 10.2), where a six is twice as likely as any other face, is a perfectly valid probability measure, yet P({6})=27≠16\mathbb{P}(\{6\})=\tfrac27\neq\tfrac16, so the counting formula does not apply. Always check the symmetry (fair coins, balanced dice, well-shuffled cards, randomly drawn balls) before counting.

Definition 10.1 — Probability space and its axioms

A probability space is a triple (Ω,F,P)(\Omega,\mathcal{F},\mathbb{P}): a sample space Ω\Omega of outcomes ω\omega, a collection F\mathcal{F} of events (subsets of Ω\Omega), and a probability measure P:F→R\mathbb{P}:\mathcal{F}\to\mathbb{R} such that (1) 0≤P(A)≤10\le\mathbb{P}(A)\le 1 for all A∈FA\in\mathcal{F}; (2) P(Ω)=1\mathbb{P}(\Omega)=1 and P(∅)=0\mathbb{P}(\varnothing)=0; and (3) for pairwise disjoint A1,A2,…A_1,A_2,\dots (i.e. Ai∩Aj=∅A_i\cap A_j=\varnothing for i≠ji\neq j), P ⁣(⋃i=1∞Ai)=∑i=1∞P(Ai)\mathbb{P}\!\left(\bigcup_{i=1}^{\infty}A_i\right)=\sum_{i=1}^{\infty}\mathbb{P}(A_i).

Intuition. The three axioms encode the bare minimum we demand of 'chance': probabilities are fractions of certainty (between impossible, 00, and certain, 11); the whole sample space is certain, so it scores 11; and probabilities of non-overlapping events pile up additively. Everything else in probability — the complement rule, inclusion–exclusion, conditional probability — is derived from just these three rules.
Remark 10.1 — Disjoint events and finite additivity

The empty set ∅\varnothing is the event that nothing happens and always has P(∅)=0\mathbb{P}(\varnothing)=0. Events AA and BB are mutually exclusive (disjoint) when A∩B=∅A\cap B=\varnothing, meaning they cannot occur together. For finitely many pairwise disjoint events, additivity reads P(A1∪⋯∪An)=P(A1)+⋯+P(An)\mathbb{P}(A_1\cup\cdots\cup A_n)=\mathbb{P}(A_1)+\cdots+\mathbb{P}(A_n).

Intuition. Finite additivity is the countable axiom with all but finitely many events empty (each contributing 00). It is what lets you compute P(A)\mathbb{P}(A) for a finite event by adding the probabilities of its individual outcomes, since distinct outcomes are automatically disjoint. Disjointness is essential: for overlapping events this double-counts the shared outcomes.
Remark 10.2 — Probabilities need not be uniform

A probability measure is not required to give each outcome equal probability. For a die believed to be loaded so that a six is twice as likely as any other face, one uses P({1})=⋯=P({5})=17\mathbb{P}(\{1\})=\cdots=\mathbb{P}(\{5\})=\tfrac17 and P({6})=27\mathbb{P}(\{6\})=\tfrac27.

Intuition. All the axioms ask is that the outcome probabilities be nonnegative and sum to P(Ω)=1\mathbb{P}(\Omega)=1; here 5⋅17+27=15\cdot\tfrac17+\tfrac27=1, so this is a legitimate measure. Uniformity is an extra modelling assumption (justified by physical symmetry), and it is precisely that assumption which Proposition 10.1 needs — without it, the counting formula ∣A∣/∣Ω∣|A|/|\Omega| fails.
Proposition 10.1 — Probability by counting

If the sample space Ω\Omega is finite and all outcomes are equally likely, then for every event AA, P(A)=∣A∣∣Ω∣,\mathbb{P}(A)=\frac{|A|}{|\Omega|}, the number of outcomes in AA divided by the total number of outcomes.

Intuition. With ∣Ω∣=n|\Omega|=n equally likely outcomes, each singleton has probability 1n\tfrac1n (they are disjoint, union to Ω\Omega, and share the total probability 11 equally). By finite additivity P(A)\mathbb{P}(A) is the sum of 1n\tfrac1n over the ∣A∣|A| outcomes in AA, namely ∣A∣/n|A|/n. This is the classical 'favourable outcomes over total outcomes' recipe — valid only under the equal-likelihood hypothesis.

Worked examples

Example 1

Example 10.1 (a fair die). A standard six-sided die is rolled once. Write down the sample space, justify the probability of each outcome, and compute the probability of the event A={the outcome is even}A=\{\text{the outcome is even}\}.

  1. 1

    Sample space. The outcomes are the faces, so Ω={1,2,3,4,5,6}\Omega=\{1,2,3,4,5,6\} with ∣Ω∣=6|\Omega|=6; a sample point is an integer ω\omega between 11 and 66.

  2. 2

    Equally likely outcomes. The die is fair, so by symmetry every face is equally likely. The six singletons {1},…,{6}\{1\},\dots,\{6\} are disjoint and union to Ω\Omega, so by additivity 6⋅P({k})=P(Ω)=16\cdot\mathbb{P}(\{k\})=\mathbb{P}(\Omega)=1, giving P({1})=⋯=P({6})=16\mathbb{P}(\{1\})=\cdots=\mathbb{P}(\{6\})=\tfrac16.

  3. 3

    Identify the event. 'The outcome is even' collects the even faces: A={2,4,6}A=\{2,4,6\}, so ∣A∣=3|A|=3.

  4. 4

    Add the outcome probabilities (finite additivity). P(A)=P({2})+P({4})+P({6})=16+16+16=36=12.\mathbb{P}(A)=\mathbb{P}(\{2\})+\mathbb{P}(\{4\})+\mathbb{P}(\{6\})=\tfrac16+\tfrac16+\tfrac16=\tfrac36=\tfrac12. Equivalently, by Proposition 10.1, P(A)=∣A∣/∣Ω∣=3/6=12\mathbb{P}(A)=|A|/|\Omega|=3/6=\tfrac12.

Answer. Ω={1,2,3,4,5,6}\Omega=\{1,2,3,4,5,6\}, each outcome has probability 16\tfrac16, and P(A)=36=12=0.5\mathbb{P}(A)=\tfrac{3}{6}=\tfrac12=0.5.
Example 2

Example 10.2 (a pair of fair dice). A blue die and a red die are rolled. Treating the dice as distinguishable, describe the sample space and its size, then find the probability of the event D={the two dice sum to 8}D=\{\text{the two dice sum to }8\}.

Example 3

Example 10.3 (an urn). An urn contains 66 red and 44 white balls, identical apart from colour. Two balls are drawn at random without replacement. Find the probability that both drawn balls are red.