Probability Axioms & the Rules They Imply
Kolmogorov's three axioms pin down what a probability can be — and once you know them, the complement rule and inclusion–exclusion fall out for free.
By the end you'll be able to state the three axioms, derive the complement and inclusion–exclusion rules from them, and spot a probability assignment that breaks the rules.
Predict: if you drag P(A∩B) up, what happens to P(A∪B)? Then drag the slider below to check.
Drag the sliders to set P(A), P(B), and their overlap P(A∩B). The rectangle is the sample space S (total probability 1); the circles are events A and B. Hover or focus a region — A-only, B-only, the overlap, or neither — for its exact probability. The overlap slider is bounded by the axioms themselves: it can't exceed min(P(A), P(B)), and once P(A)+P(B) > 1 it can't fall below P(A)+P(B)−1 — otherwise P(A∪B) would exceed 1 or the "neither" region would go negative.
A probability is a set function \(P\) that assigns a number to every event in the sample space, obeying three rules: \(0 \le P(A) \le 1\), \(P(S) = 1\), and countable additivity for disjoint events. You don't have to separately memorize the complement rule \(P(A') = 1 - P(A)\) or the inclusion–exclusion rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) — both fall straight out of those three rules.
Picture \(P(A)\) as the slice of a pie chart that event A takes up. The whole pie — the sample space — is 1. If A and B overlap and you just add their slices, you double-count the overlapping wedge, so you subtract it back out once — that's inclusion–exclusion. And "not A" is whatever's left of the pie once you take A's slice out.
Here are Kolmogorov's three axioms. (1) \(0 \le P(A) \le 1\) for every event A. (2) \(P(S) = 1\) — the whole sample space has probability 1. (3) For pairwise disjoint events \(A_1, A_2, \ldots\), you get countable additivity: \(P\left(\bigcup_i A_i\right) = \sum_i P(A_i)\).
You can derive both headline rules from these three. Complement rule: A and its complement A′ are disjoint and together make up S, so axiom 3 gives \(P(A) + P(A') = 1\), hence \(P(A') = 1 - P(A)\). Inclusion–exclusion: split \(A \cup B\) into three disjoint pieces — A-only, B-only, and the overlap — and axiom 3 gives \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
Say event A is "a policyholder files an auto claim this year" and B is "the same policyholder files a home claim." These can overlap — a bad year can hit both lines. If you're pricing "any claim at all," you need \(P(A \cup B)\), and you have to subtract \(P(A \cap B)\) or you'll double-count customers who file both types and overprice the bundle.
Say \(P(A) = 0.6\), \(P(B) = 0.5\), \(P(A \cap B) = 0.2\) — the same numbers the interactive starts with. Plug into inclusion–exclusion: \(P(A \cup B) = 0.6 + 0.5 - 0.2 = \) 0.9.
Still with \(P(A) = 0.6\): find \(P(A')\) using the complement rule. You know \(P(A) + P(A') = 1\), so:
\(P(A') = 1 - \) ____
Reveal the answer
\(P(A') = 1 - 0.6 = \) 0.4 — check it against the P(A′) readout above the diagram when the sliders sit at their starting values.
More info — why the overlap slider has limits
Look back at the interactive: the P(A∩B) slider refuses to leave a certain range. That's not a UI quirk — it's the axioms enforcing themselves. If \(P(A \cap B)\) exceeded \(\min(P(A), P(B))\), the overlap would be "bigger" than one of the sets it sits inside, which is impossible. If it dropped below \(\max(0, P(A) + P(B) - 1)\), then \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) would exceed 1, breaking axiom 1. Try dragging both P(A) and P(B) above 0.5 and watch the overlap slider's minimum jump above zero — that's this exact bound in action. The Khan Academy links in Dive deeper below work through more examples of this.
Check your understanding
If P(A) = 0.6, P(B) = 0.5, and P(A ∩ B) = 0.2, what is P(A ∪ B)?
Events A and B are mutually exclusive (disjoint). Which pair of probabilities is IMPOSSIBLE under Kolmogorov's axioms?
P(A) = 0.7, P(B) = 0.4, P(A ∩ B) = 0.3. What is P((A ∪ B)′) — the probability neither A nor B occurs?
Recap
- Kolmogorov's three axioms: every probability is between 0 and 1; \(P(S) = 1\); and probabilities of disjoint events add (countable additivity).
- Complement rule: \(P(A') = 1 - P(A)\).
- Inclusion–exclusion: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
- Both rules are consequences of the axioms, not separate facts — if you forget one, rederive it from the three axioms above.
Dive deeper
- Khan Academy — Axiomatic approach to probability See Kolmogorov's three axioms stated and applied to simple examples.
- Khan Academy — Probability of the event 'not A' Drill the complement rule P(A′) = 1 − P(A) on worked problems.
- Khan Academy — Addition rule for probability Learn inclusion–exclusion and when to subtract the overlap.
Sources
- Kolmogorov's Axioms of Probability