Sample Space, Events & Axioms

Roll a die. The outcome can be 1, 2, 3, 4, 5, or 6. That whole list is the sample space S; one number is an outcome; “even” means the subset {2,4,6} — that’s an event. Probability is just a function that puts a number on each event.

Here’s the surprising part. Three short rules — the axioms — already pin down every probability identity you’ll ever use. Get them right and the rest of the chapter is mostly bookkeeping. (They’re also the sanity checks every model’s output must pass: a classifier that reports class probabilities summing to 1.3, or a negative probability, has violated an axiom — and is simply broken.)

The three axioms

1. Non-negativity: P(E) ≥ 0 for every event E.
2. Normalisation: P(S) = 1 — the total probability of the whole sample space is 1.
3. Additivity: if A and B are disjoint (no shared outcomes, A ∩ B = ∅), then P(A ∪ B) = P(A) + P(B).

What the axioms force to be true

These small rules already settle the everyday facts you reach for constantly:

• Empty event: P(∅) = 0. Since S and ∅ are disjoint and S ∪ ∅ = S, Axiom 3 gives P(S) = P(S) + P(∅), so P(∅) = 0.
• Complement rule: P(Aᶜ) = 1 − P(A). An event and its complement are disjoint and together fill S, so P(A) + P(Aᶜ) = P(S) = 1. Rearrange.
• Monotonicity: if A ⊆ B then P(A) ≤ P(B) — a bigger event can’t have smaller probability. (Split B into A and the disjoint leftover B ∩ Aᶜ, both ≥ 0.)
• Bounded: combining the above, 0 ≤ P(A) ≤ 1 for every event.

Inclusion-exclusion — when events overlap

Axiom 3 only adds probabilities for disjoint events. When A and B overlap, adding P(A) + P(B) counts the overlap A ∩ B twice. Subtract it once:

This is inclusion-exclusion: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). When the events are disjoint, P(A ∩ B) = 0 and it collapses straight back to Axiom 3 — so additivity is just the no-overlap special case.

Drag the two circles around. Watch P(A), P(B), and the overlap P(A ∩ B) update as you move them. Slide the circles apart and the overlap collapses to zero — that’s when additivity (Axiom 3) applies cleanly without subtracting anything.

How GATE asks this

Usually an MCQ asking which statements are valid consequences of the axioms (the complement rule is true; “always add probabilities” is a trap), or a short NAT that hands you P(A), P(B), and P(A ∩ B) and asks for P(A ∪ B) — a direct inclusion-exclusion plug-in. The numbers are easy; the marks are lost by forgetting to subtract the overlap.

Worked example

Given P(A) = 0.5, P(B) = 0.4, and P(A ∩ B) = 0.2, find P(A ∪ B).

Apply inclusion-exclusion directly:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
         = 0.5  + 0.4  − 0.2
         = 0.7

So P(A ∪ B) = 0.7. Contrast with the disjoint case: had the events been mutually exclusive, P(A ∩ B) would be 0 and P(A ∪ B) = 0.5 + 0.4 = 0.9. But here they overlap by 0.2 — ignoring that gives the wrong answer 0.9 instead of 0.7.

Quick check