Sample Space, Events & Axioms

Roll an ordinary die. Even before it lands, you know the chances: each of the six faces, one in six. And without anyone ever teaching you, you already trust two things about those chances — that no face has a negative chance, and that the six of them together make one whole. Those feel too obvious to say out loud. Yet it is exactly the obvious things, once written down carefully, that the rest of probability is built on.

Naming the pieces

Let us name the parts on this familiar die. The full list of what can happen — 1, 2, 3, 4, 5, 6 — is the sample space, written S. A single result, say a 4, is an outcome. And a question like “was it even?” picks out a part of that list, {2, 4, 6} — that part is an event. So an event is just a subset of the sample space, and probability is simply a rule that hands each event a number.

Three obvious demands

What must that rule obey to be sensible? Just three short demands — the axioms — and each is only one of the die-facts you already trusted, made general.

1. Non-negativity. Every event’s chance is at least zero: P(E) ≥ 0. No event happens a negative amount.
2. Normalisation. The whole sample space has chance one: P(S) = 1. Something on the list is certain to happen.
3. Additivity. If two events A and B cannot happen together — they share no outcome, A ∩ B = ∅ — then their chances simply add: P(A ∪ B) = P(A) + P(B).

One warning while that third axiom is fresh, because two words get swapped constantly here: disjoint is not the same as independent. Disjoint (mutually exclusive) means the two events cannot both happen. Independent — which the next lesson defines properly — means one happening tells you nothing about the other. Far from being the same idea, they are close to opposites: if A and B are disjoint and each has a real chance of occurring, then learning that A happened tells you B certainly did not, which is about as dependent as two events get.

What the three demands give you for free

Here is the quiet power of writing the obvious down. From these three rules alone, several everyday facts follow without any new assumption.

The empty event has chance zero, P(∅) = 0. Since S and ∅ share nothing and together are just S, additivity gives P(S) = P(S) + P(∅) — so P(∅) must be zero.

The complement rule, P(Aᶜ) = 1 − P(A), follows the same way: an event and its opposite share nothing and together fill S, so their chances add to one.

Monotonicity says a bigger event cannot carry a smaller chance — if A ⊆ B then P(A) ≤ P(B). It is worth watching that one fall out of the axioms rather than out of intuition. Cut B into the part inside A and the part outside it: B = A ∪ (B ∖ A), two pieces that share no outcome. Additivity gives P(B) = P(A) + P(B ∖ A), and Axiom 1 says that second term is at least zero — so P(B) cannot fall below P(A).

Put the three together and every chance is boxed into 0 ≤ P(A) ≤ 1. None of this was assumed; all of it was forced.

When events overlap

Additivity had a catch — it only adds chances for events that cannot happen together. So what do we do when they can?

Draw a single card from a deck. Let A be “it is a King” and B be “it is a Heart”. These events overlap, because one card — the King of Hearts — is both. If you simply add P(A) + P(B), that one card gets counted twice, once as a King and once as a Heart. To repair it, subtract the overlap exactly once:

This repair is inclusion-exclusion: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). And notice it contains additivity as a special case — when the events do not overlap, P(A ∩ B) = 0 and the subtraction does nothing, leaving Axiom 3 exactly. For three events the same correction continues — add the singles, take out each pairwise overlap, add the triple back: P(A∪B∪C) = ΣP(single) − ΣP(pair) + P(A∩B∩C).

You can feel the overlap directly. Drag the two circles below: slide them apart and the shared region vanishes, so the chances simply add; push them together and the overlap grows, and that is precisely the amount inclusion-exclusion subtracts.

A worked example

In a class, P(Maths) = 0.7, P(Physics) = 0.6, and P(both) = 0.5. Find the probability a student passes at least one of the two.

“At least one” is the union, so reach for inclusion-exclusion — and the missing piece the prediction hinted at is exactly P(both), the overlap:

P(Maths ∪ Physics) = P(Maths) + P(Physics) − P(both)
                   = 0.7 + 0.6 − 0.5
                   = 0.8

So 0.8. Without subtracting the 0.5 overlap you would have got 1.3 — a chance above one, which Axiom 2 forbids, and a clear signal you double-counted the students who passed both.

A question to carry forward

Inclusion-exclusion fixed the double-counting between two events by subtracting their overlap. Here is the thread for the next lesson: that overlap P(A ∩ B) is itself a probability worth understanding on its own — what does it really mean to ask for the chance of A given that B has already happened?

In one breath

• Sample space S = every outcome; an event is a subset of S; probability scores each event.
• Three axioms: (1) P(E) ≥ 0, (2) P(S) = 1, (3) disjoint A,B → P(A∪B) = P(A)+P(B).
• For free: P(∅)=0, complement P(Aᶜ)=1−P(A), monotonicity A⊆B ⇒ P(A)≤P(B), and 0 ≤ P(A) ≤ 1.
• Inclusion-exclusion (overlap): P(A∪B) = P(A)+P(B)−P(A∩B); additivity is the no-overlap case. Three events: add singles − pairs + triple.
• The marks are lost by forgetting to subtract the overlap, or assuming equally-likely outcomes when explicit probabilities are given.

Practice