1
The axioms are few
Non-negative, total 1, additive over disjoint events. The equally-likely model satisfies them, and so do many others.
Lesson · Discrete Probability · The model
The Probability Axioms
Non-negativity, total mass 1, and additivity over disjoint events. Everything else — the complement rule, monotonicity, the union bound — is derived from these three.
A1 P(E) ≥ 0
A2 P(S) = 1
A3 P(A ∪ B) = P(A) + P(B) when A ∩ B = ∅
everything else is a THEOREM
Three axioms
Core ideas
What this lesson establishes
The axioms are what make probability a mathematical subject rather than a collection of intuitions.
2
The familiar rules are consequences
P(Ē) = 1 − P(E) and P(A ∪ B) = P(A) + P(B) − P(A ∩ B) are proved, not assumed.
3
Additivity needs disjointness
For overlapping events the general rule is inclusion and exclusion, which is “The Inclusion-Exclusion Principle” in Combinatorics and Counting with probabilities in place of counts.
The material
The axioms and what follows
The three rules, then six theorems derived from them.
Definition · Probability space
A finite probability space is a sample space S with a function P on its subsets satisfying: (A1) P(E) ≥ 0 for every event E; (A2) P(S) = 1; (A3) P(A ∪ B) = P(A) + P(B) whenever A ∩ B = ∅.
Nothing here says what probability means. The axioms constrain how probabilities combine, and the equally-likely model of “Computing Probabilities” is one assignment satisfying them — not the only one.
Theorem · Consequences of the axioms
P(∅) = 0; P(Ē) = 1 − P(E); P(E) ≤ 1; if A ⊆ B then P(A) ≤ P(B); P(A ∪ B) = P(A) + P(B) − P(A ∩ B); and P(A ∪ B) ≤ P(A) + P(B).
P(∅) = 0: S and ∅ are disjoint with union S, so P(S) = P(S) + P(∅) by A3, giving P(∅) = 0.
P(Ē) = 1 − P(E): E and Ē are disjoint with union S, so A3 and A2 give P(E) + P(Ē) = 1.
P(E) ≤ 1: from the previous line and A1 applied to Ē.
Monotonicity. If A ⊆ B then B = A ∪ (B \ A) with the two disjoint, so P(B) = P(A) + P(B \ A) ≥ P(A) by A1.
Inclusion–exclusion. A ∪ B splits into the three disjoint pieces A \ B, A ∩ B, B \ A. Adding P(A) and P(B) counts A ∩ B twice, so subtracting it once gives the union.
Union bound. Immediate from the previous line and A1.
□
The union bound P(A ∪ B) ≤ P(A) + P(B) looks weak and is used constantly: it bounds the probability that anything goes wrong by the sum of the individual failure probabilities, with no independence assumption at all.
Assigning probabilities directly
For a finite S, any assignment of non-negative numbers to the outcomes summing to 1 defines a probability space, with P(E) = Σ_{s ∈ E} P(s). The equally-likely case gives each outcome 1/|S|; a biased coin gives P(H) = p and P(T) = 1 − p.
Example · A loaded die
Suppose P(6) = 1/3 and the other five faces share the rest equally, so each has 2/15. Check: 1/3 + 5 · 2/15 = 1/3 + 2/3 = 1. Then P(even) = 2/15 + 2/15 + 1/3 = 3/5, which the equally-likely model would have put at 1/2.
Notice that the axioms did all the work and the equally-likely assumption did none. Everything from “Conditional Probability” on is proved from the axioms and therefore holds for loaded dice too.
Interactive
Sample space, event, condition
Exact fractions, every outcome listed
Method
Working from the axioms
Use disjointness where you have it, inclusion and exclusion where you do not.
1
Check the events are disjoint before adding
A3 requires it. Adding overlapping probabilities over-counts exactly the intersection.
2
Use the complement when the event is large
P(E) = 1 − P(Ē) is often the shorter calculation, and it is a theorem rather than a trick.
3
Decompose into disjoint pieces
Any union splits into disjoint parts, and A3 then applies cleanly to the parts.
4
Bound when you cannot compute
The union bound needs no independence and is often enough — "the chance anything fails is at most the sum of the individual chances".
Axioms
Score 0 / 0Choose the answer you can justify.
Deriving inclusion and exclusion
From the axioms, with no counting.
Split A ∪ B into three DISJOINT pieces:
A ∪ B = (A \ B) ∪ (A ∩ B) ∪ (B \ A)
By A3 applied twice:
P(A ∪ B) = P(A\B) + P(A∩B) + P(B\A)
Also A = (A\B) ∪ (A∩B), disjoint, so P(A\B) = P(A) − P(A∩B),
and likewise P(B\A) = P(B) − P(A∩B).
Substituting:
P(A ∪ B) = [P(A) − P(A∩B)] + P(A∩B) + [P(B) − P(A∩B)]
= P(A) + P(B) − P(A∩B) ∎
No counting appeared. The rule holds for any probability assignment satisfying the axioms, including loaded dice and continuous distributions, which is exactly why the axiomatic route is worth taking.
Common mistakes
1
Adding probabilities of overlapping eventsA3 needs disjointness. Without it the sum exceeds the true probability by P(A ∩ B), and may exceed 1.
2
Confusing disjoint with independentDisjoint events with positive probability are maximally dependent: one occurring rules the other out.
3
Assuming the equally-likely model when computingThe axioms are general. A loaded die satisfies them and |E|/|S| is simply the wrong formula for it.
Check your understanding
✓
Finish when you can derive a rule rather than recall it.
Derive P(A ∪ B ∪ C) from the axioms by splitting into disjoint pieces. It is the three-set inclusion–exclusion formula, and deriving it is more useful than memorising the signs.
Note: Kolmogorov's axioms add countable additivity, which matters for infinite sample spaces and is what makes the geometric distribution of “The Geometric Distribution and Waiting Times” legitimate. For finite spaces the three rules here are the whole story.