Probability & Counting Principles

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Probability basics:

• P(event) = (number of favourable outcomes) / (total number of equally likely outcomes)
• P(success) + P(failure) = 1
• 0 ≤ P(event) ≤ 1

Counting principles:

• Addition principle: If task A can be done in m ways and task B in n ways (mutually exclusive), then A or B can be done in m + n ways
• Multiplication principle: If task A has m ways and task B has n ways, doing both A and B has m × n ways

Permutations (order matters): nPk = n! / (n − k)! Combinations (order doesn’t matter): nCk = n! / (k!(n − k)!)

⚡ Exam tip: Use combinations when selecting team members (Anna and Bob together = Bob and Anna), permutations when arranging in positions (first, second, third place). If unsure, ask: does swapping two selected items change the outcome? If no → combinations; if yes → permutations.

⚡ Exam tip: “At least one” probability problems: P(at least one) = 1 − P(none). Much easier than calculating all “at least one” cases separately.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Working with Probability

Example: A bag contains 4 red balls and 6 blue balls. Two balls are drawn without replacement.

• P(first red) = 4/10 = 2/5
• P(second red given first red) = 3/9 = 1/3
• P(both red) = (4/10) × (3/9) = 12/90 = 2/15

With replacement vs without replacement: With replacement means each draw is independent; without replacement means they’re dependent (conditional probability).

Combined Events

For independent events A and B: P(A and B) = P(A) × P(B)

For mutually exclusive events (cannot happen together): P(A or B) = P(A) + P(B)

For non-mutually exclusive events: P(A or B) = P(A) + P(B) − P(A and B)

Example: Drawing a king OR a heart from a 52-card deck.

• P(king) = 4/52 = 1/13
• P(heart) = 13/52 = 1/4
• P(king of hearts) = 1/52 (this was subtracted twice)
• P(king or heart) = 1/13 + 1/4 − 1/52 = (4 + 13 − 1)/52 = 16/52 = 4/13

Permutations — Arrangements

How many ways can 5 books be arranged on a shelf? 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.

How many 3-digit numbers can be formed from digits {1, 2, 3, 4} if digits cannot repeat?

• Hundreds digit: 4 choices
• Tens digit: 3 remaining choices
• Units digit: 2 remaining choices
• Total: 4 × 3 × 2 = 24 (which is 4P3 = 24)

Combinations — Selections

How many ways to choose 3 students from a class of 12 for a committee?

• 12C3 = 12! / (3! × 9!) = (12 × 11 × 10) / (3 × 2 × 1) = 220 ways.

Note: 5C2 = 5C3 = 10. In general, nCk = nC(n−k).

Common Mistakes to Avoid:

Mistake	Correct approach
Treating dependent events as independent	Without replacement → dependent; calculate conditional probability
Confusing permutations with combinations	Ask: does order matter? Is this a ranking/arrangement or just a selection?
Forgetting to divide by factorials	Permutations: divide by (n−k)!; Combinations: divide by k!(n−k)!
Not simplifying probability answers	Reduce fractions: 12/48 = 1/4
Double-counting in “or” problems	Subtract the intersection: P(A∪B) = P(A) + P(B) − P(A∩B)

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Conditional Probability and Bayes’ Theorem

P(A|B) means “probability of A given that B has occurred.” Formula: P(A|B) = P(A∩B) / P(B).

Example: In a class, 60% of students passed Mathematics, 55% passed Physics, and 40% passed both. Given that a student passed Physics, what’s the probability they also passed Mathematics?

• P(M|P) = P(M∩P) / P(P) = 0.40 / 0.55 = 8/11 ≈ 0.727

Bayes’ Theorem: P(A|B) = P(B|A) × P(A) / P(B)

For a disease affecting 1% of the population: P(disease) = 0.01. A test has 95% sensitivity (correct positive rate) and 90% specificity. P(positive|disease) = 0.95, P(negative|no disease) = 0.90, so P(positive|no disease) = 0.10. P(disease|positive) = (0.95 × 0.01) / (0.95 × 0.01 + 0.10 × 0.99) = 0.0095 / 0.1085 ≈ 0.0876 (8.76%). Despite the test being reasonably accurate, the probability is low because the disease is rare.

The Birthday Problem

How many people needed in a room for a >50% chance that at least two share a birthday? Counter-intuitively, only 23. With 23 people, P(no match) = 365/365 × 364/365 × … × 343/365 ≈ 0.493, so P(match) ≈ 0.507 > 0.5.

This is a classic “at least one” problem solved by computing 1 − P(none).

Permutations with Repeated Elements

How many distinct arrangements of the letters in “STATISTICS”? Total letters = 10, with repeats: S appears 3 times, T appears 3 times, I appears 3 times, A appears 1 time.

• Number of arrangements = 10! / (3! × 3! × 3! × 1!) = 168,168.

Circular Arrangements

Number of ways to arrange n distinct objects in a circle = (n − 1)! (fix one object as reference and arrange the rest relative to it).

At a round table with 6 people, how many seatings? (6 − 1)! = 120. However, arrangements that are rotations of each other are identical; reflections are different unless the table is unlabelled and symmetric.

The Binomial Distribution

For n independent trials, each with probability p of success: P(X = k) = nCk × p^k × (1 − p)^(n − k).

A biased coin has P(heads) = 0.6. Tossed 5 times, P(exactly 3 heads) = 5C3 × (0.6)³ × (0.4)² = 10 × 0.216 × 0.16 = 0.3456.

Expected value E(X) = np = 5 × 0.6 = 3 heads.

Counting Chords and Intersections

How many chords can be drawn connecting n points on a circle if no three chords intersect at a single interior point? Each chord is determined by choosing 2 points: nC2 chords. In a convex polygon with n vertices, the number of diagonals = nC2 − n = n(n − 3)/2.

Historical Context

Probability originated in games of chance. Girolamo Cardano (1501-1576) wrote Liber de Ludo Aleae on gambling probability. Fermat and Pascal corresponded in 1654 on the problem of points (how to divide stakes when a game is interrupted), considered the founding of probability theory. Jacob Bernoulli’s Ars Conjectandi (1713) established the law of large numbers. Thomas Bayes developed Bayes’ theorem (published 1763). Pascal’s triangle, known to Chinese mathematicians centuries earlier, encodes binomial coefficients.

Exam Pattern Analysis

UI probability questions frequently:

1. Ask for probability of compound events (with or without replacement)
2. Use counting principles in “how many ways” questions
3. Test conditional probability with two-way tables
4. Apply binomial probability for repeated independent trials
5. Use “at least one” complement approach

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