Deriving Basic Probability Rules from the Axioms

Why Basic Probability Rules Need Proof

A probability model assigns numbers to events, but those numbers cannot be assigned arbitrarily: they must satisfy the probability axioms. Several rules that seem obvious, such as an event having a probability no greater than one, are consequences of those axioms rather than additional assumptions. Deriving the rules makes clear exactly which properties of the model justify them.

This tutorial works with finite probability models. The central arguments use only the total probability of the sample space, nonnegativity, and additivity for disjoint events. The focus is on three consequences: the empty event has probability zero, event probabilities are bounded by zero and one, and inclusion of events implies monotonicity of probability.

Finite Probability Axioms and Set Decomposition

Let \(\Omega\) be a finite sample space, and let \(\mathcal{F}\) be the collection of events allowed by the model. Assume that \(\Omega\), the empty event \(\varnothing\), and the complements of events belong to \(\mathcal{F}\), and that disjoint unions of events in \(\mathcal{F}\) are allowed. A probability assignment \(P\) on \(\mathcal{F}\) satisfies the following axioms.

For finite models, a probability assignment can also be specified by giving each outcome \(\omega\in\Omega\) a nonnegative weight \(p(\omega)\) such that the weights sum to one. The probability of an event is the sum of the weights of its outcomes. The axiomatic arguments below do not depend on a particular list of outcome weights, but this representation will make the examples concrete.

The proofs repeatedly use set decompositions. If \(A\) and \(B\) are disjoint, their union consists of two nonoverlapping parts. If \(A\subseteq B\), then \(B\) consists of \(A\) together with the part of \(B\) outside \(A\), namely \(B\setminus A\). In each case, the pieces are disjoint, so the additivity axiom applies.

Three Consequences of the Axioms

Proof that the empty event has probability zero. The sample space and the empty event are disjoint, and their union is the sample space: \(\Omega\cup\varnothing=\Omega\). Additivity and normalization therefore give

Since \(P(\Omega)=1\), this equality says \(1=1+P(\varnothing)\), which implies \(P(\varnothing)=0\). This argument uses the empty event as a set, not an outcome. It does not require assigning a separate probability to an impossible outcome.

Proof of the probability bounds. Nonnegativity is an axiom, so \(P(A)\geq 0\). To obtain the upper bound, use the event \(A\) and its complement \(A^c=\Omega\setminus A\). They are disjoint and their union is \(\Omega\). Thus

The axiom of nonnegativity gives \(P(A^c)\geq 0\). Rearranging the displayed equality yields \(P(A)=1-P(A^c)\leq 1\). Together, \(0\leq P(A)\leq 1\).

Proof of monotonicity. Suppose \(A\subseteq B\). Then \(B\) is the disjoint union of \(A\) and \(B\setminus A\):

By additivity, \(P(B)=P(A)+P(B\setminus A)\). Nonnegativity gives \(P(B\setminus A)\geq 0\), so \(P(B)\geq P(A)\). This proves all three claims. \(\square\)

Each proof has the same structure: express a set as a union of disjoint pieces, use the axiom on that union, and then use nonnegativity or normalization. The result is not an independent probability law; it follows from the axioms.

Worked Examples in Finite Models

Why the Rules Matter—and a Common Pitfall

These rules are useful whenever probabilities are compared or simplified. The bounds provide a basic check on a calculation: a value below zero or above one cannot be an event probability. Monotonicity gives a consistency check: an event that contains another event cannot have a smaller probability. In practical finite models, the rules can identify an arithmetic or event-definition error before more complicated calculations are attempted.

A common mistake is to write \(P(B)=P(A)+P(B\setminus A)\) whenever \(A\subseteq B\), without checking the sets involved. The equality is justified because \(B=A\cup(B\setminus A)\) and those two pieces are disjoint. By contrast, \(A\) and \(B\) themselves are not disjoint when \(A\subseteq B\), unless \(A=\varnothing\). Applying additivity directly to \(A\) and \(B\) would therefore be invalid.

A related mistake is to claim that probability is monotone simply because probability is nonnegative. Nonnegativity alone compares \(P(B\setminus A)\) with zero; the set decomposition and additivity are what connect that probability to \(P(B)-P(A)\). Each ingredient in the proof is needed.