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Lesson 39 — Classical Probability

Sample space, events, Kolmogorov axioms. Classical probability: favorable cases over possible cases. Complement, addition, conditional, and independence. Simple Bayes.

Used in: 1.º ano do EM (15–16 anos) · Equiv. Math B japonês · Equiv. Stochastik Klasse 11 alemã · Equiv. H2 Math Statistics (Singapura)

P(A) = \frac{|A|}{|\Omega|}

The classical probability of an event $A$ is the number of favorable outcomes divided by the total number of outcomes in the sample space $\Omega$ — valid when all outcomes are equally likely. The value always lies in $[0, 1]$ : zero means impossible, one means certain.

Choose your door

Rigorous notation, full derivation, hypotheses

Definitions and Axioms

Sample space and events

Kolmogorov's Axioms (1933)

Classical probability

Properties derived from the axioms

Conditional probability

Independence

Bayes' Theorem

"Bayes' theorem is a tool for updating beliefs in light of new evidence. The prior $P(A)$ is updated to the posterior $P(A \mid B)$ when we observe $B$ ." — Grinstead-Snell, Introduction to Probability, Chapter 4

Worked Examples

Example· Classical probability with two dice

Problem. Two fair six-sided dice are rolled simultaneously. Calculate the probability that their sum equals 8.

Strategy. Construct the sample space $\Omega$ of ordered pairs $(i, j)$ with $i, j \in \{1,2,3,4,5,6\}$ and identify the pairs summing to 8.

Solution.

Sample space: $|\Omega| = 36$ .
Event $A$ = sum equals 8: pairs $(2,6),(3,5),(4,4),(5,3),(6,2)$ , so $|A| = 5$ .
Probability:

P(A) = \frac{|A|}{|\Omega|} = \frac{5}{36}

Verification. Sum 8 is less likely than sum 7 (6 pairs) and more likely than sum 9 (4 pairs) — consistent with the triangular distribution of sums.

Source. OpenIntro Statistics, 4th ed., §3.1 — Diez, Çetinkaya-Rundel, Barr · CC-BY-SA.

Example· Complement and inclusion–exclusion in a deck

Problem. A card is drawn at random from a standard 52-card deck. Calculate: (a) $P(\text{king or hearts})$ ; (b) $P(\text{not a king})$ .

Strategy. For (a), apply inclusion–exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ , since the king of hearts belongs to both events. For (b), use complement.

Solution.

$A = \{\text{king}\}$ , $B = \{\text{hearts}\}$ . $|A| = 4$ , $|B| = 13$ , $|A \cap B| = 1$ .
(a) Inclusion–exclusion:

P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \approx 0.308

(b) Complement:

P(\text{not a king}) = 1 - \frac{4}{52} = \frac{48}{52} = \frac{12}{13} \approx 0.923

Verification. In (a): $4 + 13 - 1 = 16$ cards satisfy "king or hearts". Correct.

Source. OpenStax Statistics, §3.3 — Illowsky, Dean · CC-BY.

Example· Conditional probability in an urn without replacement

Problem. An urn contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. Calculate: (a) $P(\text{second is blue} \mid \text{first was red})$ ; (b) $P(\text{both blue})$ .

Strategy. For (a), after drawing one red ball, the urn has 7 balls remaining, 3 blue. For (b), apply the multiplication rule.

Solution.

(a) After drawing 1 red, 7 balls remain: 4 red and 3 blue.

P(\text{2nd blue} \mid \text{1st red}) = \frac{3}{7}

(b) Multiplication rule:

P(\text{both blue}) = \frac{3}{8} \cdot \frac{2}{7} = \frac{6}{56} = \frac{3}{28}

Verification via combinatorics: $\binom{3}{2}/\binom{8}{2} = 3/28$ . Matches.

Source. OpenIntro Statistics, 4th ed., §3.2, example 3.28 — CC-BY-SA.

Example· Birthday paradox — complement

Problem. In a room with 3 people, what is the probability that at least two share a birthday (365 equally likely days)?

Strategy. Calculate the complement: all 3 people have birthdays on different days.

Solution.

Complement (all days distinct):

P(\text{all distinct}) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} = \frac{365 \cdot 364 \cdot 363}{365^3}

Probability asked for:

P(\text{at least 2 equal}) = 1 - \frac{365 \cdot 364 \cdot 363}{365^3} \approx 1 - 0.9918 = 0.0082

Only 0.82% for 3 people. For 23 people, this value exceeds 50% — the famous birthday paradox.

Source. OpenIntro Statistics, 4th ed., §3.1, exercise 3.5 — CC-BY-SA.

Example· Bayes' theorem — medical diagnosis

Problem. A disease affects 1% of the population ( $P(D) = 0.01$ ). A test has sensitivity $P(+ \mid D) = 0.90$ and specificity $P(- \mid D^c) = 0.95$ . A person tests positive. What is the probability they have the disease?

Strategy. Apply Bayes with the law of total probability in the denominator.

Solution.

Total probability of positive test:

P(+) = 0.90 \times 0.01 + 0.05 \times 0.99 = 0.009 + 0.0495 = 0.0585

Bayes:

P(D \mid +) = \frac{0.90 \times 0.01}{0.0585} = \frac{0.009}{0.0585} \approx 0.154

Only 15.4% probability of disease with positive test — the effect of low prevalence: most positives are false positives ( $0.0495/0.0585 \approx 84.6\%$ ).

Verification: $P(D \mid +) + P(D^c \mid +) = 0.154 + 0.846 = 1$ . Consistent.

Source. OpenStax Statistics, §3.2, example 3.16 — CC-BY. Adapted from OpenIntro Statistics §3.2.

Exercise list

44 exercises · 11 with worked solution (25%)

Application 27Understanding 7Modeling 7Challenge 2Proof 1

Sources

OpenIntro Statistics, 4th ed. — Diez, Çetinkaya-Rundel, Barr · 2019 · EN · CC-BY-SA · Chapter 3: Probability (§3.1–§3.3). Primary source.
OpenStax Statistics — Illowsky, Dean · 2022 · EN · CC-BY · Chapter 3: Probability topics (§3.1–§3.5).
Introduction to Probability — Grinstead, Snell · Dartmouth · EN · GNU FDL · Chapters 1–4 (sample spaces, independence, conditional, Bayes).