Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms: Worked examples

Formula Two consequences we'll reuse (proved here, from A1–A3)

These are not extra axioms — they follow from A1–A3, and we prove them once so the page stands alone.

(T0) The impossible event has probability $0$ : $P (\emptyset) = 0$ . Take the disjoint sequence $A_{1} = Ω, A_{2} = A_{3} = \dots = \emptyset$ ; their union is $Ω$ . By A3, $P (Ω) = P (Ω) + \sum_{i \geq 2} P (\emptyset)$ , so $\sum_{i \geq 2} P (\emptyset) = 0$ . Each term is $\geq 0$ by A1, so each must be $0$ . Hence $P (\emptyset) = 0$ . $■$

(T1) Continuity from above: if $A_{1} \supseteq A_{2} \supseteq \dots$ shrink to $A = ⋂_{n} A_{n}$ , then $P (A_{n}) \to P (A)$ . (T2) Continuity from below: if $B_{1} \subseteq B_{2} \subseteq \dots$ grow to $B = ⋃_{n} B_{n}$ , then $P (B_{n}) \to P (B)$ . Why they hold: both are packaged forms of A3. For (T2), write the growing sequence as disjoint "rings" $B_{1}, (B_{2} ∖ B_{1}), (B_{3} ∖ B_{2}), \dots$ ; A3 turns $P (B)$ into the infinite sum of ring-probabilities, and the partial sum of the first $n$ rings is exactly $P (B_{n})$ — so the partial sums (which converge to the total by definition of an infinite sum) are $P (B_{n}) \to P (B)$ . (T1) follows from (T2) by taking complements. This is why "countable" in A3 matters — it's what lets probability pass through infinite limits.

The scenario matrix

Every probability-space question lives in one of these cells. Each worked example below is tagged with the cell it hits.

Cell	Scenario class	What's tricky about it	Example
C1	Finite space, disjoint events	Plain additivity (A3)	Ex 1
C2	Finite space, overlapping events	Must subtract the overlap (inclusion–exclusion)	Ex 2
C3	Complement shortcut	"At least one" is easier via $1 - P (none)$	Ex 3
C4	Degenerate events: $\emptyset$ and $Ω$	The certain and impossible edges	Ex 4
C4′	Finite space, non-uniform atoms	Add atomic weights $\sum_{ω \in A} P ({ω})$ , not counts	Ex 4b
C5	Continuous space, intervals	Length = probability, single points have length $0$	Ex 5
C6	Limiting sequence (shrinking and growing)	Continuity of $P$ from above and below	Ex 6
C7	Real-world word problem	Translate English → $Ω, F, P$	Ex 7
C8	Exam twist: three overlapping events	General inclusion–exclusion, sign bookkeeping	Ex 8

Before we start, one symbol we lean on constantly:

Ex 1 — Finite space, disjoint events [C1]

Worked example Two cards, no overlap

A card is drawn from a standard $52$ -card deck; every card equally likely, so $Ω$ has $52$ outcomes and $P (A) = ∣ A ∣/52$ . Let $A = {the card is a King}$ and $B = {the card is a Queen}$ . Find $P (A \cup B)$ = probability the card is a King or a Queen.

Forecast: Do you expect to subtract anything here? Guess a fraction.

Step 1 — Count each event. There are $4$ Kings and $4$ Queens, so $∣ A ∣ = 4$ , $∣ B ∣ = 4$ , giving $P (A) = \frac{4}{52}$ and $P (B) = \frac{4}{52}$ . Why this step? On an equally-likely finite space, probability is just counting.

Step 2 — Check disjointness. A single card cannot be both a King and a Queen, so $A \cap B = \emptyset$ . This is cell C1: disjoint. Why this step? Only when $A \cap B = \emptyset$ may we use plain additivity — otherwise we double-count.

Step 3 — Add (finite additivity, derived from A3). $P (A \cup B) = P (A) + P (B) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13} .$ Why this step? Disjoint pieces' probabilities add — this is the padded-with-empties version of A3.

Verify: $A \cup B = {8 face-cards that are K or Q}$ , so $∣ A \cup B ∣ = 8$ and $8/52 = 2/13 \approx 0.1538$ . ✓ Nothing was subtracted, matching our disjoint forecast.

Ex 2 — Finite space, overlapping events [C2]

Worked example Even OR greater-than-3 on a die (the classic overlap)

Roll a fair die: $Ω = {1, 2, 3, 4, 5, 6}$ , $P (A) = ∣ A ∣/6$ . Let $A = {even} = {2, 4, 6}$ and $B = {> 3} = {4, 5, 6}$ . Find $P (A \cup B)$ .

Forecast: If you naively did $\frac{3}{6} + \frac{3}{6} = 1$ , you'd claim the event is certain. Is it? (Outcome $1$ is neither even nor $> 3$ ...) So the naive answer must be wrong. Guess the true value.

Step 1 — Find the overlap. $A \cap B = {4, 6}$ , so $P (A \cap B) = \frac{2}{6}$ . Why this step? $4$ and $6$ are counted once in $A$ and again in $B$ — inclusion–exclusion exists to remove exactly this double count.

Step 2 — Apply inclusion–exclusion. $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} .$ Why this step? Subtracting $P (A \cap B)$ corrects the overcount.

Verify: $A \cup B = {2, 4, 5, 6}$ directly, $∣ A \cup B ∣ = 4$ , so $4/6 = 2/3 \approx 0.667$ . ✓ Not $1$ — outcome $1$ is correctly excluded.

See Inclusion–exclusion principle for the general pattern extended in Ex 8.

Ex 3 — The complement shortcut [C3]

Worked example "At least one six" in four rolls

Roll a fair die four times independently. Let $E = {at least one roll shows a 6}$ . Find $P (E)$ .

Forecast: Directly, "at least one" splits into exactly-one, exactly-two, exactly-three, exactly-four — messy. Guess whether the answer is above or below $\frac{1}{2}$ .

Step 1 — Complement the event. "At least one six" is the opposite of "no six in any roll". So $E^{c} = {no six on all four rolls}$ . Why this step? The complement rule $P (E) = 1 - P (E^{c})$ (from A2 + additivity) turns a 4-way sum into a single easy product.

Step 2 — Probability of no six on one roll. Five of six faces are "not six", so $P (no six, one roll) = \frac{5}{6}$ .

Step 3 — All four rolls avoid six. The rolls are independent (defined above: independent events' joint probability multiplies), so $P (E^{c}) = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = (\frac{5}{6})^{4} = \frac{625}{1296} .$ Why this step? Because $P (no six on all four) = P (roll 1 no six) \dots P (roll 4 no six)$ by mutual independence — that lets us avoid enumerating outcomes.

Step 4 — Complement back. $P (E) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518.$

Verify: $625 + 671 = 1296$ ✓ (probabilities of complementary events sum to $1$ ). Value $\approx 0.518 > \frac{1}{2}$ , matching intuition that four chances make a six more-likely-than-not.

Ex 4 — Degenerate events: $\emptyset$ and $Ω$ [C4]

Worked example The impossible and the certain

On the die space $Ω = {1, \dots, 6}$ , evaluate: (a) $P (roll a 7)$ , (b) $P (roll \leq 6)$ , (c) $P (roll is 2 and 5)$ .

Forecast: Which of these is $0$ , which is $1$ ? Guess all three before reading.

Step 1 — (a) Roll a 7. No face equals $7$ , so this event is $\emptyset$ . By theorem (T0) proved above (from A1 + A3), $P (\emptyset) = 0$ . Why this step? $\emptyset$ contains no outcomes; (T0) forces its probability to be $0$ .

Step 2 — (b) Roll $\leq 6$ . Every face satisfies $\leq 6$ , so this event is the whole space $Ω$ . By axiom A2, $P (Ω) = 1$ . Why this step? A2 defines the total mass as $1$ : something must happen.

Step 3 — (c) Roll is $2$ AND $5$ . A single roll can't be two numbers at once, so ${2} \cap {5} = \emptyset$ , hence probability $0$ by (T0). Why this step? "AND" is intersection; disjoint singletons intersect to the impossible event.

Verify: $P (\emptyset) = 0/6 = 0$ ✓, $P (Ω) = 6/6 = 1$ ✓, and ${2} \cap {5} = \emptyset \Rightarrow 0$ ✓. These are the two edges of the range $[0, 1]$ .

Ex 4b — Finite space with non-uniform atoms [C4′]

Worked example A loaded (biased) die

A die is weighted so its six faces are not equally likely. The atomic probabilities — the probability of each single outcome $ω$ , written $P ({ω})$ — are: $P ({1}) = 0.1, P ({2}) = 0.1, P ({3}) = 0.1, P ({4}) = 0.2, P ({5}) = 0.2, P ({6}) = 0.3.$ Find (a) $P (Ω)$ as a sanity check, (b) $P (A)$ for $A = {even} = {2, 4, 6}$ , (c) $P (B)$ for $B = {> 3} = {4, 5, 6}$ , (d) $P (A \cup B)$ .

Forecast: The counting formula $∣ A ∣/6$ is now wrong — why? Guess whether $P (even)$ is bigger or smaller than $\frac{1}{2}$ given face $6$ is heavy.

Step 1 — Check the atoms sum to 1. $0.1 + 0.1 + 0.1 + 0.2 + 0.2 + 0.3 = 1.0 = P (Ω)$ . Why this step? A valid measure needs A2: the total mass over $Ω$ must be exactly $1$ . If atoms don't sum to $1$ , it isn't a probability measure.

Step 2 — Probability of an event = sum of its atoms. For any $A$ , since the singletons ${ω}$ are disjoint and their union is $A$ , additivity (A3) gives $P (A) = \sum_{ω \in A} P ({ω}) .$ Why this step? This is the general discrete rule; the uniform formula $∣ A ∣/∣Ω∣$ is just its special case when every atom equals $1/∣Ω∣$ .

Step 3 — (b) Even. $P (A) = P ({2}) + P ({4}) + P ({6}) = 0.1 + 0.2 + 0.3 = 0.6.$

Step 4 — (c) Greater than 3. $P (B) = P ({4}) + P ({5}) + P ({6}) = 0.2 + 0.2 + 0.3 = 0.7.$

Step 5 — (d) Union via inclusion–exclusion. Overlap $A \cap B = {4, 6}$ , so $P (A \cap B) = 0.2 + 0.3 = 0.5$ . Then $P (A \cup B) = 0.6 + 0.7 - 0.5 = 0.8.$ Why this step? Inclusion–exclusion is a theorem about any measure, uniform or not — it never assumed equal weights.

Verify: Directly, $A \cup B = {2, 4, 5, 6}$ , so $P = 0.1 + 0.2 + 0.2 + 0.3 = 0.8$ . ✓ And $P (even) = 0.6 > \frac{1}{2}$ because the heavy faces $4, 6$ are even — matching the forecast.

Ex 5 — Continuous space, intervals [C5]

Figure — Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms

Figure s01 — what you're looking at: the grey bar is the whole stick $Ω = [0, 1]$ . The two green blocks are the intervals $[0.2, 0.5]$ (width $0.3$ ) and $[0.7, 0.9]$ (width $0.2$ ); their widths are their probabilities. The single red dot at $0.4$ marks the point event ${0.4}$ — it has zero width, so probability $0$ . The tick labels along the bottom read off the endpoints $0, 0.2, 0.4, 0.5, 0.7, 0.9, 1$ .

Step 1 — (a) Length of one interval. $P ([0.2, 0.5]) = 0.5 - 0.2 = 0.3$ . Why this step? In the uniform model, probability is length — that's the definition of this $P$ .

Step 2 — (b) A single point. $P ({0.4}) = 0$ : a point is an interval of length $0.4 - 0.4 = 0$ . Why this step? Write ${0.4} = ⋂_{n} [0.4, 0.4 + \frac{1}{n}]$ . These intervals shrink, so by continuity from above (T1) proved at the top, $P ({0.4}) = lim_{n} P ([0.4, 0.4 + \frac{1}{n}]) = lim_{n} \frac{1}{n} = 0$ . Yet $0.4$ can occur — probability $0 \neq =$ impossible.

Step 3 — (c) Two disjoint intervals. They don't overlap ( $0.5 < 0.7$ ), so add their lengths: $P = 0.3 (0.5 - 0.2) + 0.2 (0.9 - 0.7) = 0.5.$ Why this step? Disjoint events add (finite additivity), exactly as in the finite case — the machinery is identical.

Verify: $0.3 \leq 1$ ✓, single point $0$ ✓, and $0.3 + 0.2 = 0.5 \leq 1$ ✓. The two green bars together span half the stick.

Ex 6 — Limiting sequences: shrinking and growing [C6]

Figure s02 — what you're looking at: the red nested boxes on top are the shrinking $A_{n} = [0, 1/ n]$ for $n = 1, 2, 4, 8$ ; each is half-transparent, and you can see them closing down onto the black dot at $0$ (that dot is $⋂_{n} A_{n} = {0}$ ). The blue nested boxes below are the growing $B_{n} = [0, 1 - 1/ n]$ ; each is longer than the last, creeping rightward toward the open end at $1$ (their union is $[0, 1)$ ). Each box is labelled with its probability $P (A_{n}) = 1/ n$ or $P (B_{n}) = 1 - 1/ n$ .

Step 1 — (a) shrinking, number limit. $P (A_{n}) = \frac{1}{n} \to 0$ . Why this step? Length of each box; the numbers march to $0$ .

Step 2 — (a) shrinking, set limit. Every $A_{n}$ contains $0$ ; any $x > 0$ leaves once $\frac{1}{n} < x$ . So $⋂_{n} A_{n} = {0}$ and $P ({0}) = 0$ . Why this step? We compare the set limit to the number limit. Continuity from above (T1): since $A_{n} \supseteq A_{n + 1}$ shrink to ${0}$ , $P (A_{n}) \to P ({0})$ . Both routes give $0$ . ✓

Step 3 — (b) growing, number limit. $P (B_{n}) = 1 - \frac{1}{n} \to 1$ . Why this step? Length of each growing box tends to the full stick's length as $\frac{1}{n} \to 0$ .

Step 4 — (b) growing, set limit. Every point $x < 1$ eventually lands inside once $1 - \frac{1}{n} > x$ ; only $x = 1$ is never included (no $n$ makes $1 - \frac{1}{n}$ reach $1$ ). So $⋃_{n} B_{n} = [0, 1)$ , and $P ([0, 1)) = 1 - 0 = 1$ (length of the half-open interval). Why this step? We again compare the set limit to the number limit.

Step 5 — (b) reconcile via continuity from below. Because $B_{n} \subseteq B_{n + 1}$ grow up to $B = [0, 1)$ , continuity from below (T2) says $lim_{n} P (B_{n}) = P (B)$ . The left side is $lim_{n} (1 - \frac{1}{n}) = 1$ ; the right side is $P ([0, 1)) = 1$ . They agree. ✓

Step 6 — why both directions work at all. (T1) and (T2) were both derived from countable additivity (A3) at the top of the page — the "countable" is exactly what lets probability pass through infinite limits. Finite additivity alone could do neither.

Verify: (a) $\frac{1}{n} \to 0$ and $P ({0}) = 0$ agree. (b) $1 - \frac{1}{n} \to 1$ and $P ([0, 1)) = 1$ agree. ✓

Ex 7 — Real-world word problem [C7]

Worked example Newspaper subscriptions (build

$Ω, F, P$ from English) In a town, $60%$ read paper $A$ , $40%$ read paper $B$ , and $25%$ read both. Pick a resident uniformly. Find the probability the resident reads (a) at least one paper, (b) neither paper, (c) exactly one paper.

Forecast: Guess (b) before computing — is "neither" larger or smaller than $\frac{1}{4}$ ?

Step 1 — Set up the space. $Ω =$ all residents, $P (X) =$ fraction of residents in $X$ . Translate: $P (A) = 0.60$ , $P (B) = 0.40$ , $P (A \cap B) = 0.25$ . Why this step? A word problem is a probability space in disguise; naming $Ω, F, P$ makes the axioms usable. ( $Ω$ is finite, so $F = 2^{Ω}$ — every group of residents is an event.)

Step 2 — (a) At least one = union (inclusion–exclusion). $P (A \cup B) = 0.60 + 0.40 - 0.25 = 0.75.$ Why this step? "Both" is the overlap we must subtract, cell C2 logic reused.

Step 3 — (b) Neither = complement of union. $P ((A \cup B)^{c}) = 1 - 0.75 = 0.25.$ Why this step? Complement rule; "neither" is the opposite of "at least one".

Step 4 — (c) Exactly one — derive it set-theoretically first. "Reads exactly one paper" means "in $A$ but not $B$ , or in $B$ but not $A$ ": ${exactly one} = (A ∖ B) ⊔ (B ∖ A),$ a disjoint union (the two halves share no resident). By additivity, $P (exactly one) = P (A ∖ B) + P (B ∖ A) .$ Now use $P (A ∖ B) = P (A) - P (A \cap B)$ and $P (B ∖ A) = P (B) - P (A \cap B)$ (each set minus its shared overlap). Adding: $P (exactly one) = (P (A) - P (A \cap B)) + (P (B) - P (A \cap B)) = P (A) + P (B) - 2 P (A \cap B) .$ Since $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , this equals $P (A \cup B) - P (A \cap B)$ . Numerically: $P (exactly one) = 0.75 - 0.25 = 0.50.$ Why this step? Building it from the disjoint pieces $A ∖ B$ and $B ∖ A$ shows why the shortcut $P (A \cup B) - P (A \cap B)$ works — the overlap is removed once because "exactly one" excludes the "both" region entirely.

Verify: The four disjoint regions (both, only $A$ , only $B$ , neither) must sum to $1$ : $0.25 + (0.60 - 0.25) + (0.40 - 0.25) + 0.25 = 0.25 + 0.35 + 0.15 + 0.25 = 1.00. ✓$ "Neither" $= 0.25$ exactly. ✓

Ex 8 — Exam twist: three overlapping events [C8]

Worked example Three-set inclusion–exclusion (sign bookkeeping)

Among students: $P (Maths = M) = 0.5$ , $P (Physics = P h) = 0.4$ , $P (Chem = C) = 0.3$ ; pairwise $P (M \cap P h) = 0.2$ , $P (M \cap C) = 0.15$ , $P (P h \cap C) = 0.1$ ; all three $P (M \cap P h \cap C) = 0.05$ . Find $P (M \cup P h \cup C)$ = probability a student studies at least one of the three.

Forecast: Naively adding $0.5 + 0.4 + 0.3 = 1.2 > 1$ — impossible! So we must subtract. Guess the corrected value.

Step 1 — Write the three-set formula. $P (M \cup P h \cup C) = \sum P (singles) - \sum P (pairs) + P (triple) .$ Why this step? Pairs get double-counted by the singles, so subtract them; but the triple was then subtracted too many times, so add it back once. This alternating-sign pattern is the general Inclusion–exclusion principle.

Step 2 — Plug in with correct signs. $= (0.5 + 0.4 + 0.3) - (0.2 + 0.15 + 0.1) + 0.05.$ Why this step? $+$ singles, $-$ pairs, $+$ triple — get one sign wrong and the answer breaks the $[0, 1]$ range.

Step 3 — Arithmetic. $= 1.2 - 0.45 + 0.05 = 0.80.$

Verify: $0.80 \in [0, 1]$ ✓ (no longer the impossible $1.2$ ). Sanity: it must be $\geq$ the largest single event $P (M) = 0.5$ and $\leq$ their sum $1.2$ ; indeed $0.5 \leq 0.80 \leq 1.2$ . ✓

Recall

Recall Which cell, which tool?

Disjoint finite events — add directly? ::: Yes: $P (A) + P (B)$ (C1). Overlapping events — what to do? ::: Subtract $P (A \cap B)$ : inclusion–exclusion (C2, C8). Finite space but outcomes NOT equally likely? ::: $P (A) = \sum_{ω \in A} P ({ω})$ , add the atoms (C4′). "At least one of many" — fastest route? ::: Complement: $1 - P (none)$ (C3). $P (\emptyset)$ and $P (Ω)$ ? ::: $0$ and $1$ (C4). Probability of a single point on $[0, 1]$ ? ::: $0$ , but not impossible (C5). Shrinking $A_{n} ↓ A$ — which continuity? ::: From above (T1): $P (A_{n}) \to P (A)$ (C6). Growing $B_{n} ↑ B$ — which continuity? ::: From below (T2): $P (B_{n}) \to P (B)$ (C6). Independent events — how combine? ::: Multiply: $P (A \cap B) = P (A) P (B)$ (C3). Three-set formula sign pattern? ::: $+$ singles $-$ pairs $+$ triple (C8).