Probability and Permutations

Probability and Permutations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Probability quantifies how likely an event is, on a scale of 0 (impossible) to 1 (certain). For equally likely outcomes, P(E) = favorable outcomes ÷ total outcomes. A permutation is an ordered arrangement counted by nPr = n! / (n−r)!, while a combination is an unordered selection counted by nCr = n! / [r!·(n−r)!]. The complement rule P(Aᶜ) = 1 − P(A) is the fastest way to handle “at least one” problems. HAT-UG typically tests 1–3 MCQs: pick the permutation when a ranking or sequence is specified (1st, 2nd, 3rd), pick the combination when a committee or team is chosen and order is irrelevant.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Definitions and Foundations

The sample space is the set of every possible outcome of a random experiment; an event is any subset of that space. Classical probability applies when each outcome is equally likely. Mutually exclusive events cannot occur together (e.g., rolling a 2 and a 5 on one die), so P(A or B) = P(A) + P(B). When events can overlap, the addition rule becomes P(A ∪ B) = P(A) + P(B) − P(A ∩ B), subtracting the intersection to avoid double-counting.

Independent and Conditional Probability

Two events are independent when one does not influence the other, giving P(A ∩ B) = P(A)·P(B). Conditional probability P(A|B) = P(A ∩ B) / P(B) measures the chance of A given that B has occurred; it is the conceptual seed of Bayes’ theorem. The complement rule P(Aᶜ) = 1 − P(A) converts a messy “at least one” question into a single subtraction.

Counting: Permutations vs Combinations

The fundamental counting principle multiplies choices across stages. Permutations nPr count ordered arrangements of r items chosen from n distinct items. Combinations nCr count unordered selections. Note that nCr = nPr / r!. For arrangements around a circle where rotations are equivalent, use (n−1)!; for a necklace that can also be flipped, divide further by 2.

Scenario	Formula
Ordered line-up of r from n	n! / (n−r)!
Unordered committee of r from n	n! / [r!(n−r)!]
Circular seating of n	(n−1)!

Exam-Style Patterns

HAT-UG MCQs use dice, cards (52-card deck, 4 suits, 13 ranks), coins, and committee selection. Watch for trap wording: “arrange” and “rank” signal permutations; “select” and “choose” signal combinations. “At least one” almost always means use the complement — compute the probability of zero occurrences and subtract from 1.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Refinements

When items repeat, plain factorial overcounts. For arrangements of a word with repeated letters, divide n! by the factorial of each repetition count: arrangements of “MISSISSIPPI” = 11! / (1!·4!·4!·2!). Conditional probability requires the reduced sample space: P(king on 2nd draw | ace on 1st) is computed on a deck of 51, not 52. Bayes’ theorem extends this to invert conditions: P(A|B) = P(B|A)·P(A) / P(B).

Common Mistakes

1. Using nCr when a rank is given (1st prize, 2nd prize are distinguishable — use nPr).
2. Forgetting to subtract P(A ∩ B) in the addition rule for non-exclusive events.
3. Treating “draw two cards” as a single trial when computing probability; sample space is 52·51 = 2652 ordered pairs.
4. Dividing nCr numerator and denominator incorrectly — r and n−r are not interchangeable.
5. Using P(A∩B) = P(A)·P(B) for dependent events (e.g., cards without replacement).

Connection to Adjacent Topics

Probability and counting fuse naturally with set theory (Venn diagrams encode the addition rule) and data sufficiency questions in HAT-UG. A solid grip on nCr also speeds up binomial expansion coefficients and basic binomial probability P(X = k) = nCr·pᵏ·(1−p)ⁿ⁻ᵏ.

Worked Micro-Example

A bag holds 5 red and 3 blue balls. Two balls are drawn without replacement. Probability that the first is red and the second is blue: P = (5/8) · (3/7) = 15/56 ≈ 0.268.

Practice Prompts

1. How many 4-digit PINs can be formed from digits 0–9 if no digit may repeat? (Answer: 5040)
2. From 6 men and 4 women, a 3-person committee is chosen at random. What is the probability it contains exactly 2 women? (Answer: nCr(4,2)·nCr(6,1) / nCr(10,3) = 36/120 = 3/10)

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