Probability

Two different single-outcome events share no outcome, so \(\{x_i\} \cap \{x_j\} = \emptyset\) whenever \(i \neq j\): they do not overlap. Every outcome of \(U\) is one of the \(x_i\), so \(\{x_1\} \cup \{x_2\} \cup \dots \cup \{x_n\} = U\): together they cover the sample space. They therefore form a partition. It is the finest one, because any event is a union of single-outcome events — no partition cuts \(U\) into smaller pieces.

Everything rests on the equivalent formulation of the definition: every outcome of \(U\) belongs to exactly one of \(E_1, E_2, \dots, E_n\).

• The pieces do not overlap. Take \(i \neq j\). An outcome of \((B \cap E_i) \cap (B \cap E_j)\) would belong to \(E_i\) and to \(E_j\), whereas it belongs to at most one of the two. There is no such outcome: $$(B \cap E_i) \cap (B \cap E_j) = \emptyset.$$
• The pieces give back \(B\). An outcome of any \(B \cap E_i\) is an outcome of \(B\). Conversely, let \(x\) be an outcome of \(B\): it is an outcome of \(U\), so it belongs to exactly one \(E_i\); being in \(B\) as well, it is in \(B \cap E_i\). Every outcome of \(B\) therefore lies in one of the pieces: $$(B \cap E_1) \cup (B \cap E_2) \cup \dots \cup (B \cap E_n) = B.$$

The pieces are pairwise mutually exclusive and cover \(B\): they form a partition of \(B\).

The events of a partition are pairwise mutually exclusive and together cover the whole sample space. Adding their probabilities therefore gives: $$\begin{aligned}[t] P(E_1) + \dots + P(E_n) &= P(E_1 \cup \dots \cup E_n) && \text{(Axiom 3 --- pairwise mutually exclusive)}\\ &= P(U) && \text{(the events cover \(U\))}\\ &= 1 && \text{(Axiom 2).} \end{aligned}$$ The statement about the outcomes is this same result applied to the finest partition \(\{x_1\}, \{x_2\}, \dots, \{x_n\}\).

The impossible event first. The events \(U\) and \(\emptyset\) have no outcome in common, and \(U \cup \emptyset = U\), so Axiom 3 applies: $$\begin{aligned}[t] P(U) &= P(U \cup \emptyset) && \text{(\(U = U \cup \emptyset\))}\\ &= P(U) + P(\emptyset) && \text{(Axiom 3, since \(U \cap \emptyset = \emptyset\))}, \end{aligned}$$ hence \(P(\emptyset) = 0\). Each outcome. Let \(p\) be the common probability of the outcomes. The singletons \(\{x_1\}, \{x_2\}, \dots, \{x_n\}\) form the finest partition of \(U\), so by the Proposition “Sum over a Partition” their probabilities add up to \(1\): $$\begin{aligned}[t] 1 &= P(\{x_1\}) + P(\{x_2\}) + \dots + P(\{x_n\}) && \text{(Sum over a Partition)}\\ &= n \times p && \text{(\(n\) equal terms)}, \end{aligned}$$ hence \(p = \dfrac{1}{n}\). Each event. If \(E = \emptyset\), then \(P(E) = 0 = \dfrac{0}{n} = \dfrac{\Card{E}}{\Card{U}}\). Otherwise, \(E\) is the union of the \(\Card{E}\) singletons of its outcomes, which are pairwise mutually exclusive, and Axiom 3, applied step by step, adds their probabilities: $$\begin{aligned}[t] P(E) &= \underbrace{\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}}_{\Card{E} \text{ terms}} && \text{(Axiom 3, step by step)}\\ &= \Card{E} \times \frac{1}{n}\\ &= \frac{\Card{E}}{\Card{U}} && \text{(\(n = \Card{U}\))}. \end{aligned}$$ So the formula of the Definition “Equally Likely” is not an extra rule: when all the outcomes have the same probability, it follows from the axioms.

• Algebraic Proof
  By definition, an event \(E\) and its complement \(E'\) are mutually exclusive (\(E \cap E' = \emptyset\)) and their union is the entire sample space (\(E \cup E' = U\)).
  Using Axiom 3: \(P(E \cup E') = P(E) + P(E')\).
  Using Axiom 2: \(P(U) = 1\).
  Since \(E \cup E' = U\), we can equate their probabilities: $$ P(E) + P(E') = P(U) = 1. $$ Rearranging the formula gives the complement rule: $$P(E') = 1 - P(E).$$ In particular, the complement of the sample space \(U\) is the impossible event, \(U' = \emptyset\), so $$\begin{aligned}[t] P(\emptyset) &= P(U') && \text{(\(\emptyset\) is the complement of \(U\))}\\ &= 1 - P(U) && \text{(complement rule)}\\ &= 1 - 1 && \text{(Axiom 2)}\\ &= 0. \end{aligned}$$ The impossible event has probability \(0\).
• Geometric Illustration
  The total area of the sample space, \(\textcolor{olive}{P(U)}\), , is the sum of the area of the event, \(P(E)\), and the area of its complement, \(P(E')\).
  So, $$\textcolor{colordef}{P(E)} + \textcolor{colorprop}{P(E')} = \textcolor{olive}{P(U)}.$$ Since \(\textcolor{olive}{P(U) = 1}\) (from Axiom 2), we have: $$\textcolor{colordef}{P(E)} + \textcolor{colorprop}{P(E')} = 1.$$

The event \(A\) and its complement \(A'\) form the two-piece partition of \(U\). The Proposition “Partition Induced on an Event” therefore cuts \(B\) into two mutually exclusive pieces, \(B \cap A\) and \(B \cap A'\): $$\begin{aligned}[t] B &= (B \cap A) \cup (B \cap A') && \text{(the partition \(\{A, A'\}\) cuts \(B\))}\\ &= A \cup (B \cap A') && \text{(\(B \cap A = A\), since \(A \subseteq B\))} \end{aligned}$$ The two pieces are mutually exclusive, so Axiom 3 applies, and Axiom 1 closes the argument: $$\begin{aligned}[t] P(B) &= P(A) + P(B \cap A') && \text{(Axiom 3)}\\ &\geq P(A) && \text{(Axiom 1: \(P(B \cap A') \geq 0\)).} \end{aligned}$$ So \(A \subseteq B\) implies \(P(A) \leq P(B)\). The hypothesis \(A \subseteq B\) was used exactly once: to replace \(B \cap A\) by \(A\).

The lower bound is Axiom 1. For the upper bound, every event is a part of the sample space, so \(E \subseteq U\), and monotonicity applies: $$\begin{aligned}[t] P(E) &\leq P(U) && \text{(monotonicity, since \(E \subseteq U\))}\\ &= 1 && \text{(Axiom 2).} \end{aligned}$$

The Venn diagram below shows the sample space \(U\) with two intersecting events, \(E\) and \(F\). The region represents the event \(E \cap F\); its probability \(P(E \cap F)\) is counted twice in \(P(E) + P(F)\). To correct for this, we must subtract \(P(E \cap F)\) once.
Therefore, the total probability of the union is $$P(E \cup F) = P(E) + P(F) - P(E \cap F).$$ The same count, written with events. The union \(E \cup F\) is made of the outcomes of \(E\) together with the outcomes of \(F\) that are not in \(E\): \(E \cup F = E \cup (F \cap E')\), two events with no outcome in common. In the same way, \(F = (F \cap E) \cup (F \cap E')\), two events with no outcome in common, so \(P(F) = P(F \cap E) + P(F \cap E')\). Hence $$\begin{aligned}[t] P(E \cup F) &= P(E) + P(F \cap E') && \text{(\(E\) and \(F \cap E'\) have no outcome in common)}\\ &= P(E) + P(F) - P(F \cap E) && \text{(\(P(F \cap E') = P(F) - P(F \cap E)\))}, \end{aligned}$$ and since \(F \cap E = E \cap F\), this is the addition law.