Sample spaces, events, and axioms of probability: Worked examples

Intuition What this page is

The parent note gave you the rules. This page stress-tests them against every kind of situation you can meet: overlapping events, disjoint events, complements, the impossible event, the certain event, non-uniform (data-driven) spaces, and exam traps. If you can work all cells below, no probability question at this level can surprise you.

A quick reminder of the notation, so no symbol is used before it is named:

$Ω$ (capital omega) = the sample space, the set of every possible outcome. Picture a big box; each outcome is one dot inside it.
An event $E$ = any collection of dots, i.e. a subset written $E \subseteq Ω$ .
$∣ E ∣$ = the count of dots in $E$ (its size).
$\cup$ ("union") = OR — dots in $A$ or $B$ or both.
$\cap$ ("intersection") = AND — dots in both $A$ and $B$ .
$A^{c}$ ("complement of $A$ ") = NOT — every dot outside $A$ .
$\emptyset$ = the empty set, a box with no dots (the impossible event).
$A ∖ B$ ("A minus B") = dots in $A$ that are not in $B$ .
$P (E)$ = a number in $[0, 1]$ telling us how much of the total certainty sits on $E$ .

The scenario matrix

Every problem this topic can throw falls into one of these case classes. The examples below are labelled with the cell they cover.

Cell	Case class	What makes it tricky	Covered by
C1	Disjoint events (Axiom 3 directly)	events cannot overlap → just add	Ex 1
C2	Overlapping events (inclusion–exclusion)	must subtract the overlap once	Ex 2
C3	Complement / "at least one"	count the opposite, subtract from 1	Ex 3
C4	Degenerate: $\emptyset$ and $Ω$	probability exactly 0 and exactly 1	Ex 4
C5	Non-uniform space (data / ML)	outcomes NOT equally likely	Ex 5
C6	Monotonicity / subset bound	$A \subseteq B \Rightarrow P (A) \leq P (B)$	Ex 6
C7	Real-world word problem	translate English → sets first	Ex 7
C8	Exam twist: three events / limiting behaviour	inclusion–exclusion for 3 sets, and a limit	Ex 8

Figure 1 is our master picture of these ideas: a sample space $Ω$ (the outer box) holding two events $A$ and $B$ that overlap. The red lens is exactly the overlap $A \cap B$ — the part that inclusion–exclusion (Example 2) must subtract so it is counted only once. Keep this picture in mind: every worked example below is a special case of "which dots are shared, which are separate?"

Figure — Sample spaces, events, and axioms of probability

Example 1 — Disjoint events (cell C1)

Forecast: guess the answer before reading on. Add up the pieces?

Check overlap. $A \cap B = {1, 2} \cap {5} = \emptyset$ . In the Figure 1 picture, these are two circles with no red lens between them. Why this step? Axiom 3 (add probabilities) only applies when events cannot both happen. We must confirm no shared dot before adding.
Single-outcome probabilities. Fair die ⇒ each face has $P ({ω}) = \frac{1}{6}$ . So $P (A) = \frac{2}{6}$ , $P (B) = \frac{1}{6}$ . Why this step? Classical probability: $P (E) = ∣ E ∣/∣Ω∣$ holds because every outcome is equally likely.
Add. $P (A \cup B) = P (A) + P (B) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ . Why this step? Disjoint ⇒ Axiom 3 gives a clean sum, no correction.

Example 2 — Overlapping events (cell C2)

Forecast: if you just add $P (A) + P (B)$ you will be wrong — why?

Figure 2 shows the full $6 \times 6$ grid of dice outcomes: the black strip is event $A$ (first die $= 6$ ), the red cells are event $B$ (sum $= 10$ ), and the one shaded red cell where they meet is the overlap $A \cap B$ — the concrete version of the red lens in Figure 1.

Count $A$ . First die $= 6$ with any second die: $(6, 1), \dots, (6, 6)$ ⇒ $∣ A ∣ = 6$ , so $P (A) = \frac{6}{36}$ . Why this step? Enumerate favourable dots so the classical formula applies.
Count $B$ . Sum $10$ : $(4, 6), (5, 5), (6, 4)$ ⇒ $∣ B ∣ = 3$ , $P (B) = \frac{3}{36}$ . Why this step? Same enumeration; these are the red cells lying along the sum-10 diagonal in Figure 2.
Count the overlap $A \cap B$ . "first die 6 AND sum 10" needs second die $= 4$ : only $(6, 4)$ . So $P (A \cap B) = \frac{1}{36}$ . Why this step? $(6, 4)$ is the single cell that Figure 2 shows in both the black strip and the red diagonal — it would be counted twice if we naively added.
Inclusion–exclusion. $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{6}{36} + \frac{3}{36} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9} .$ Why this step? Subtracting the overlap once removes the double-count. This is derived from the axioms, not a new rule.

Example 3 — Complement / "at least one" (cell C3)

Forecast: the direct count is fiddly. Is there a lazier route?

Name the complement. Let $E =$ "at least one 6". Then $E^{c} =$ "no die shows a 6". Why this step? "At least one" is awkward to count directly; its opposite ("none") is a clean product.
Count $E^{c}$ . Each die avoids 6 in $5$ ways, independently laid out: $5 \times 5 = 25$ dots ⇒ $P (E^{c}) = \frac{25}{36}$ . Why this step? The "no 6" region is a clean $5 \times 5$ block in the grid — easy to size.
Complement rule. $P (E) = 1 - P (E^{c}) = 1 - \frac{25}{36} = \frac{11}{36}$ . Why this step? $E$ and $E^{c}$ are disjoint and fill $Ω$ , so $P (E) + P (E^{c}) = 1$ (Axioms 2 + 3).

Example 4 — Degenerate: impossible and certain (cell C4)

Forecast: two of these are "obvious" — but can you prove the 0?

(a) Impossible event. $7 \in / Ω$ , so the event is $\emptyset$ . $∣\emptyset∣ = 0$ ⇒ $P = \frac{0}{6} = 0$ . Why this step? An outcome outside the box contributes no dots.
(b) Certain event. "between 1 and 6" is all of $Ω$ . By Axiom 2, $P (Ω) = 1$ . Why this step? SOMETHING must happen — total certainty is 1 by definition.
(c) Prove $P (\emptyset) = 0$ . Note $Ω = Ω \cup \emptyset$ and $Ω \cap \emptyset = \emptyset$ (disjoint). Axiom 3: $P (Ω) = P (Ω) + P (\emptyset)$ . Subtract $P (Ω)$ from both sides ⇒ $P (\emptyset) = 0$ . Why this step? We never assume the impossible event has probability 0 — it follows from Axioms 2 and 3.

Example 5 — Non-uniform space, ML flavour (cell C5)

Forecast: the outcomes are not equally likely — does the classical formula still work?

Frequency = probability estimate. With data-driven (non-symmetric) spaces we use relative frequency: $P (spam) \approx \frac{300}{1000} = 0.3$ . Why this step? No symmetry justifies equal weights; instead each estimate is (count in event)/(total observations). This is maximum likelihood estimation (MLE) — the "most-supported-by-data" guess — for a Bernoulli space; see Maximum Likelihood Estimation.
Joint event count. "spam AND missed" is the intersection of two conditions: $30$ emails satisfy both, so $P (spam \cap missed) = \frac{30}{1000} = 0.03$ . Why this step? We count only the dots satisfying both conditions, over the same total 1000. Note this is a joint probability $P (\cap)$ , not a conditional $P (\cdot ∣ \cdot)$ — conditional probability rescales by a smaller total and is treated in Conditional Probability.
Consistency with axioms. $P (spam) + P (not-spam) = 0.3 + 0.7 = 1$ (Axiom 2 holds); every estimate $\geq 0$ (Axiom 1). Why this step? Even data-driven probabilities must obey Kolmogorov's axioms, or the model is incoherent.

Example 6 — Monotonicity / subset bound (cell C6)

Forecast: which is bigger, and why must it be bigger — not just "it looks bigger"?

Show the subset. Every outcome in $A$ (sum 11) also has sum at least 11, so $A \subseteq B$ ; and $B$ has the extra dot $(6, 6)$ . Why this step? Monotonicity needs a genuine subset relation, established by inspecting the dots.
Split $B$ into disjoint pieces. Write $B = A \cup (B ∖ A)$ , where $B ∖ A = {(6, 6)}$ is "the extra stuff in $B$ not in $A$ ". These two pieces share no dot: $A \cap (B ∖ A) = \emptyset$ . Why this step? To use Axiom 3 (add probabilities) we must first break $B$ into non-overlapping parts.
Apply the axioms. By Axiom 3, $P (B) = P (A) + P (B ∖ A)$ . By Axiom 1, $P (B ∖ A) \geq 0$ . Adding a non-negative amount cannot decrease the value, so $P (B) \geq P (A)$ . Why this step? This is the whole proof of monotonicity — it is forced by additivity plus non-negativity, not assumed.
Numbers. $P (A) = \frac{2}{36} = \frac{1}{18}$ , $P (B) = \frac{3}{36} = \frac{1}{12}$ , and the extra piece $P (B ∖ A) = \frac{1}{36} \geq 0$ . Why this step? Concrete counts confirm the strict gap here: $\frac{1}{18} < \frac{1}{12}$ .

Example 7 — Real-world word problem (cell C7)

Forecast: English says "either… or…". Which set operation is that, and do the events overlap?

Translate to sets. "either sick or flagged" $= D \cup F$ . Given numbers: $P (D) = 0.08$ , $P (F) = 0.30$ , $P (D \cap F) = 0.06$ . Why this step? Word problems fail when solved in English; convert to $\cup, \cap$ first.
Overlap is non-zero. $P (D \cap F) = 0.06 \neq = 0$ , so the events overlap — inclusion–exclusion is required. Why this step? If we forgot this we'd double-count the flagged-and-sick patients.
Apply inclusion–exclusion. $P (D \cup F) = P (D) + P (F) - P (D \cap F) = 0.08 + 0.30 - 0.06 = 0.32.$ Why this step? Subtracting the overlap once gives the correct union.

Example 8 — Exam twist: three events + a limit (cell C8)

Forecast: part (a) has three overlaps and one triple-overlap — miss one and you're off.

(a) Sizes. $∣ A ∣ = 4, ∣ B ∣ = 3, ∣ C ∣ = 3$ , over $∣Ω∣ = 9$ . Why this step? List favourable outcomes so the classical formula applies uniformly.
(a) Pairwise overlaps. $A \cap B = {6}$ ( $∣ = ∣1$ ); $A \cap C = {8}$ ( $1$ ); $B \cap C = {9}$ ( $1$ ). Triple $A \cap B \cap C = \emptyset$ ( $0$ ). Why this step? The 3-set rule subtracts every pair once, then adds back the triple to fix over-subtraction.
(a) Three-set inclusion–exclusion. $P (A \cup B \cup C) = \frac{4 + 3 + 3 - 1 - 1 - 1 + 0}{9} = \frac{7}{9} .$ Why this step? $+$ singles $-$ pairs $+$ triple is the pattern that lands each dot with net weight exactly 1.
(b) Limit via complement. $P (no head in n) = (1/2)^{n}$ . So $P (\geq 1 head) = 1 - (1/2)^{n}$ . As $n \to \infty$ , $(1/2)^{n} \to 0$ , so the probability $\to 1$ . Why this step? "At least one" again begs the complement; the limiting behaviour shows the probability approaches but is defined by the axioms at every finite $n$ .

Recall Self-test

When may you use $P (A \cup B) = P (A) + P (B)$ with no correction? ::: Only when $A \cap B = \emptyset$ (disjoint), i.e. Axiom 3 applies directly. Why subtract $P (A \cap B)$ in inclusion–exclusion? ::: To remove the overlap that was counted in both $P (A)$ and $P (B)$ (avoid double-counting). How do you prove $P (\emptyset) = 0$ ? ::: $Ω = Ω \cup \emptyset$ disjoint ⇒ $P (Ω) = P (Ω) + P (\emptyset)$ ⇒ $P (\emptyset) = 0$ . Why does $A \subseteq B$ force $P (A) \leq P (B)$ ? ::: Write $B = A \cup (B ∖ A)$ disjoint; Axiom 3 gives $P (B) = P (A) + P (B ∖ A)$ , and $P (B ∖ A) \geq 0$ by Axiom 1. Why can't you assume equally likely outcomes for weather? ::: Equal weights need symmetry; asymmetric real-world spaces need data or a model instead.

See also: Random Variables, Probability Distributions, Classification Metrics, Entropy and Information.