Probability — classical, empirical, axiomatic (Kolmogorov axioms): Worked examples

Everything below uses only ideas already built in the parent:

Outcome, sample space $S$ , event $A$ — a possible result, the set of all results, and a subset of that set.
$P (A) = \frac{m}{n}$ (classical): favourable count over equally-likely total.
$P (A) \approx \frac{f}{N}$ (empirical): how often it happened in $N$ tries.
The three axioms (No–Sure–Split) and the rules squeezed out of them: complement $P (A^{c}) = 1 - P (A)$ , addition $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .

If a symbol is new, I define it the moment it appears.

The scenario matrix

Think of this table as a checklist. Every probability question you meet falls into one of these cells. The goal of the page is to hit all of them.

Cell	What makes it distinctive	Which tool answers it	Example
C1 — disjoint	events cannot happen together	axiom 3: just add	Ex 1
C2 — overlapping	events can co-occur	addition rule (subtract overlap)	Ex 2
C3 — complement / "at least one"	many favourable cases, one easy opposite	$P (A^{c}) = 1 - P (A)$	Ex 3
C4 — no symmetry	outcomes not equally likely	empirical $f / N$	Ex 4
C5 — degenerate: certain / impossible	$P = 1$ or $P = 0$ boundary	axioms 1–2 + derivations	Ex 5
C6 — counting-heavy	"favourable" needs combinations	Permutations & Combinations + classical	Ex 6
C7 — real-world word problem	translate English → sets first	set algebra + addition rule	Ex 7
C8 — limiting behaviour / continuous	$P = 0$ but still possible; $N \to \infty$	axioms in infinite spaces + Law of Large Numbers	Ex 8
C9 — exam twist	looks like one cell, is another	pick the right cell deliberately	Ex 9

We build the events in Set Theory — Union, Intersection, Complement language throughout, so "and/or/not" become $\cap / \cup /^{c}$ .

Example 1 — Cell C1 (disjoint events)

Worked example One die: "roll a 2

or roll a 5" A fair six-sided die is rolled once. Find $P (2 or 5)$ .

Forecast: two outcomes out of six — guess a number before reading on. Is it more or less than a half?

Write the sample space. $S = {1, 2, 3, 4, 5, 6}$ , all equally likely, so $n = 6$ . Why this step? Classical probability is only legal when the $n$ outcomes are equally likely — a fair die guarantees that (symmetry of a cube).
Name the events. $A = {2}$ , $B = {5}$ . Check overlap: $A \cap B = \emptyset$ (a single roll can't be both 2 and 5). Why this step? Because they are disjoint, axiom 3 lets us just add — no subtraction needed. This is Cell C1.
Add. $P (A \cup B) = P (A) + P (B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ . Why this step? Axiom 3 (Split): disjoint events add exactly.

Verify: favourable outcomes are ${2, 5}$ , so directly $m / n = 2/6 = 1/3$ . ✓ Matches. And $1/3 < 1/2$ , sensible.

Example 2 — Cell C2 (overlapping events)

Worked example One card: "a face card

or a spade" Draw one card from a standard 52-card deck. A face card is a Jack, Queen, or King. Find $P (face \cup spade)$ .

Forecast: there are 12 face cards and 13 spades — is the answer $\frac{12 + 13}{52} = \frac{25}{52}$ ? Guess, then watch the trap.

Individual counts. $P (F) = \frac{12}{52}$ (4 suits × 3 faces), $P (S_{♠}) = \frac{13}{52}$ . Why this step? 52 equally likely cards → classical counting is valid.
Find the overlap. Cards that are both face and spade: J♠, Q♠, K♠ → $P (F \cap S_{♠}) = \frac{3}{52}$ . Why this step? This is Cell C2 — the events share outcomes, so naive addition double-counts them.
Apply the addition rule. $P (F \cup S_{♠}) = \frac{12}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26} .$ Why subtract? The 3 spade-faces were counted once inside $P (F)$ and again inside $P (S)$ ; remove one copy so each outcome is counted exactly once.

Verify: list the union directly: 12 faces + (13 spades − 3 already counted) = 22 distinct cards. $22/52 = 11/26$ . ✓ The naive $25/52$ was wrong by exactly the $3/52$ overlap.

Figure — Probability — classical, empirical, axiomatic (Kolmogorov axioms)

Example 3 — Cell C3 (complement, "at least one")

Worked example Four coin tosses: "at least one head"

Toss a fair coin 4 times. Find $P (at least one head)$ .

Forecast: "at least one" out of four tries feels almost certain. Above or below $0.9$ ?

Spot the many-cases problem. "At least one head" means 1, 2, 3, or 4 heads — lots of cases. Why this step? Counting all of them is tedious; the complement is a single case. This is Cell C3.
Take the complement. The opposite of "at least one head" is "no heads" = all tails = $T T T T$ . Why this step? $P (A^{c}) = 1 - P (A)$ (proved from the axioms) lets us compute the hard event via the easy one.
Probability of the easy event. Each toss is independent with $P (T) = \frac{1}{2}$ , so $P (T T T T) = (\frac{1}{2})^{4} = \frac{1}{16}$ . Why this step? Four independent tails multiply (see Independence of Events).
Subtract. $P (at least one head) = 1 - \frac{1}{16} = \frac{15}{16} = 0.9375$ .

Verify: total outcomes $= 2^{4} = 16$ , all equally likely; only $T T T T$ has zero heads, so $15$ of $16$ have at least one head → $15/16$ . ✓ And $0.9375 > 0.9$ , matching the forecast.

Example 4 — Cell C4 (no symmetry, empirical)

Worked example Biased spinner

A spinner has no symmetry (uneven weighting). It is spun $N = 400$ times and lands on red $f = 148$ times. Estimate $P (red)$ , and say how far off you might be.

Forecast: roughly what fraction? And can you use $m / n$ here?

Reject classical. There is no symmetry argument, so we cannot claim outcomes are equally likely. Why this step? Classical $m / n$ requires equally-likely outcomes; without symmetry it is illegal. This is Cell C4.
Use relative frequency. $P (red) \approx \frac{f}{N} = \frac{148}{400} = 0.37.$ Why this step? Empirical probability: measure the fraction over many trials.
Why trust it. The Law of Large Numbers says $f / N$ settles toward the true value as $N$ grows; $400$ trials is fairly large, so $0.37$ is a reasonable estimate — not exact.

Verify: $148/400 = 0.37$ exactly, and $0 \leq 0.37 \leq 1$ so it is a valid probability. ✓

Example 5 — Cell C5 (degenerate: certain and impossible)

Example 6 — Cell C6 (counting-heavy, combinations)

Worked example Committee: 2 chosen from 5, both women

A group has 3 men and 2 women. Two people are chosen at random (order doesn't matter). Find $P (both women)$ .

Forecast: only 2 women exist and you pick 2 people — is it a small chance? Guess a fraction.

Total ways to choose 2 of 5. Order-free selection is a combination, written $(2 5)$ = "5 choose 2" = the number of 2-element subsets. $(2 5) = \frac{5 !}{2 ! 3 !} = 10.$ Why this step? Each 2-person committee is equally likely, so classical $m / n$ applies once we count committees, not ordered pairs. See Permutations & Combinations. This is Cell C6.
Favourable committees. Both women: choose 2 of the 2 women → $(2 2) = 1$ . Why this step? $m$ = number of committees satisfying the event.
Divide. $P (both women) = \frac{m}{n} = \frac{1}{10} = 0.1$ .

Verify: enumerate the 10 committees; exactly one is ${W_{1}, W_{2}}$ , so $1/10$ . ✓ Small chance, as forecast.

Example 7 — Cell C7 (real-world word problem)

Worked example Survey: newspaper and internet

In a town, $60%$ of people read a newspaper, $50%$ read news online, and $20%$ do both. Pick a person at random. Find the probability they read news in at least one of the two ways, and the probability they read news in neither way.

Forecast: if you just add $60% + 50% = 110%$ something is wrong — probabilities can't exceed $1$ . What fixes it?

Translate to sets. Let $A$ = "reads newspaper", $B$ = "reads online". Given $P (A) = 0.60$ , $P (B) = 0.50$ , $P (A \cap B) = 0.20$ . Why this step? English "both" = intersection; "at least one of the two" = union. Word → set algebra. This is Cell C7.
Addition rule. $P (A \cup B) = 0.60 + 0.50 - 0.20 = 0.90$ . Why subtract? The $20%$ who do both were counted inside both $60%$ and $50%$ ; remove one copy so the total stays $\leq 1$ . Now $0.90$ is legal.
"Neither" is the complement. $P ((A \cup B)^{c}) = 1 - P (A \cup B) = 1 - 0.90 = 0.10$ . Why this step? "Neither" is the opposite of "at least one"; use the complement rule.

Verify: the four disjoint regions must sum to 1: only-newspaper $= 0.60 - 0.20 = 0.40$ ; only-online $= 0.50 - 0.20 = 0.30$ ; both $= 0.20$ ; neither $= 0.10$ . Sum $= 0.40 + 0.30 + 0.20 + 0.10 = 1.00$ . ✓

Example 8 — Cell C8 (limiting / continuous: $P = 0$ yet possible)

Worked example Dartboard: hitting an exact point

A dart lands at a uniformly random point on a circular board of radius $10$ cm. Find (a) the probability it lands in the inner "bullseye" region of radius $1$ cm, and (b) the probability it hits one exact mathematical point.

Forecast: for (b), the point is "hittable" — the dart does land somewhere. So is its probability zero or positive?

Continuous means area, not counting. With infinitely many landing points, "equally likely" translates to "probability $\propto$ area". Why this step? We can't count $n$ outcomes; the axioms still hold, but favourable is measured by area. This is Cell C8.
Part (a): area ratio. $P (bullseye) = \frac{π ( 1 ) ^{2}}{π ( 10 ) ^{2}} = \frac{1}{100} = 0.01.$ Why this step? Uniform distribution → probability is the fraction of total area.
Part (b): a single point has zero area. Its area is $0$ , so $P (exact point) = \frac{0}{π ( 10 ) ^{2}} = 0$ . Why this step? Yet the dart does land at some point — so here $P = 0$ does not mean impossible. In infinite/continuous sample spaces the axioms permit this; only in finite spaces does $P = 0 \Leftrightarrow$ impossible.

Verify: $π (1)^{2} / π (10)^{2} = 1/100 = 0.01$ ✓, and $0/ (π \cdot 100) = 0$ ✓. The limiting lesson: as a region shrinks to a point, its probability $\to 0$ continuously.

Example 9 — Cell C9 (exam twist: looks disjoint, isn't)

Worked example Two dice: "sum is 7

or the two dice are equal" Roll two fair dice. Let $A$ = "sum $= 7$ ", $B$ = "doubles (both dice equal)". Find $P (A \cup B)$ .

Forecast: many students add $P (A) + P (B)$ instantly. Is that safe here? Look for a shared outcome first.

Sample space. Ordered pairs, $n = 36$ equally likely outcomes. Why this step? Ordering keeps outcomes equally likely (parent's two-dice logic).
Count each event. $A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$ → $P (A) = \frac{6}{36}$ . $B = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}$ → $P (B) = \frac{6}{36}$ .
Check the overlap — the twist. Can a pair be both a sum-of-7 and doubles? Doubles give sums $2, 4, 6, 8, 10, 12$ — never 7. So $A \cap B = \emptyset$ . Why this step? This is Cell C9: it looks like an overlap problem, but the events turn out disjoint. You must verify disjointness, not assume it either way.
Add (now justified). $P (A \cup B) = \frac{6}{36} + \frac{6}{36} = \frac{12}{36} = \frac{1}{3}$ . Why this step? Since $A \cap B = \emptyset$ , axiom 3 gives plain addition — no subtraction because the overlap is genuinely empty.

Verify: the union has $6 + 6 = 12$ distinct pairs (no repeats, since disjoint), so $12/36 = 1/3$ . ✓ The lesson: always check the intersection; here the subtraction term is legitimately $0$ .

Recall Which cell, which tool?

Disjoint → just add ::: axiom 3 Overlapping → subtract the intersection ::: addition rule "At least one" → complement ::: $1 - P (A^{c})$ No symmetry → measure it ::: empirical $f / N$ Order-free selection → combinations ::: $(r n)$ then $m / n$ Continuous space → area ratio, and $P = 0$ can still be possible ::: measure not count