Axiomatic approach to probability

The three axioms. Let $S$ be the sample space of a random experiment with a finite number of outcomes. A rule assigning a real number $P(A)$ to every event (subset) $A\subseteq S$ is a probability function if:

and, for pairwise mutually exclusive events $A_1, A_2, \dots, A_n$ (so $A_i \cap A_j = \varnothing$ for $i \ne j$),

where: $P(A)$ is the probability assigned to event $A$, $S$ is the whole sample space (the "sure event"), and mutually exclusive means the events share no outcome.

Two immediate consequences. Since $S \cup \varnothing = S$ with $S \cap \varnothing = \varnothing$, additivity gives $P(S) = P(S) + P(\varnothing)$, so $P(\varnothing) = 0$ — the impossible event always has probability zero. Since $A \cup A' = S$ with $A \cap A' = \varnothing$, the same reasoning gives the complement rule, $P(A') = 1 - P(A)$.

Equally likely outcomes recovered. If $S = \{\omega_1, \dots, \omega_n\}$ and every simple event $\{\omega_i\}$ is assigned the same probability $p$, axioms two and three force $np = 1$, so $p = 1/n$. For any event $A$ built from $m$ of these outcomes, additivity gives the classical formula students already know, now derived rather than assumed:

Monotonicity and bounds. If $A \subseteq B$, writing $B = A \cup (B \setminus A)$ as a disjoint union and applying additivity shows $P(A) \le P(B)$; combined with axiom one, every event satisfies $0 \le P(A) \le 1$. The figure below shows how two overlapping events split the sample space into three disjoint pieces — the geometric picture behind the addition rule derived next.

A card is drawn at random from a well-shuffled deck of 52 playing cards. Let $A$ = "the card is a king" and $B$ = "the card is a heart". Find $P(A \cup B)$, and verify the complement rule using $A$.

Step 1. Total outcomes: $S$ has $n = 52$ equally likely outcomes, so by the equally-likely formula each simple event has probability $1/52$.

Step 2. Individual probabilities: 4 kings give $P(A) = 4/52 = 1/13$; 13 hearts give $P(B) = 13/52 = 1/4$; the single "king of hearts" gives $P(A \cap B) = 1/52$.

Step 3. Apply the derived addition rule (the events overlap, since one card is both a king and a heart):

Step 4. Complement check on $A$: $A'$ = "not a king" has $52 - 4 = 48$ outcomes, so $P(A') = 48/52 = 12/13$, and indeed $P(A) + P(A') = 1/13 + 12/13 = 1$, confirming the complement rule.

The probability of drawing a king or a heart is $\mathbf{4/13}$, and $P(A) + P(A') = 1$ checks out exactly.

Competitive stretch

Three coins are tossed together; sample space $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$, $n = 8$, each outcome equally likely with probability $1/8$. For event $C$ = "at least two heads" = $\{HHH, HHT, HTH, THH\}$, $P(C) = 4/8 = 1/2$; its complement $C'$ = "at most one head" also has 4 outcomes, so $P(C') = 1/2$, again giving $P(C) + P(C') = 1$ — the same axiom-derived rule holding regardless of how the sample space is built.

• Applying $P(A \cup B) = P(A) + P(B)$ (axiom three) even when $A$ and $B$ are not mutually exclusive — the $-P(A \cap B)$ correction is mandatory whenever the events can occur together.
• Using $P(A) = m/n$ when outcomes in $S$ are not equally likely; the shortcut needs that assumption, though the axioms themselves hold regardless.
• Treating $P(\varnothing) = 0$ as a separate assumption rather than a derived consequence of the three axioms.
• Confusing $P(A')$ with $P(A \cap B')$ or similar — the complement rule $P(A') = 1 - P(A)$ only ever concerns the complement of one event with respect to $S$.

Ask the tutor: if you want to see the addition rule derivation redone for three overlapping events instead of two, or want a fresh worked example with dice instead of cards, ask the AI tutor to walk through it step by step.