Sample-Space Enumeration & Event Listing
How to spot it — Asks you to list outcomes, never to compute a number.
Definition
Problems that ask you to build a sample space explicitly and then write down which outcomes belong to named events. No probability is attached to anything — the deliverable is a set, not a number.
The underlying idea
An experiment's sample space 𝒮 is the set of all outcomes; an event is a subset of 𝒮. Before any probability can be assigned, you must be able to say what the outcomes are and how finely they are distinguished. The modelling decision — ordered vs unordered, fixed-length vs variable-length, distinguishable vs identical objects — is the entire content of these problems.
Recognition
Two independent paths. Read the left column when you can quote the page; read the right when the wording is unfamiliar but the situation is not.
Traps — where surface and structure disagree
Q6 looks like a product-rule count but is not. "A student examines these books in random order, stopping only when a second printing has been selected" means outcomes have different lengths (1, 2 or 3 books). Anyone who computes 5·4·3 has ignored the stopping rule. Stratify by length: 3 + 6 + 6 = 15.
Q2(e) — D′ is not A. "Exactly two vehicles go in the same direction" has complement "all three the same OR all three different", i.e. A ∪ B, not just A. Complements of 'exactly k' events always sweep up two or more cases.
Q7 has repeated symbols. Four A-slips and three B-slips are identical within type, so the count is C(7,3) = 35, not 7! = 5040. Treating identical objects as distinguishable is the single most common enumeration error.
Q5 is ordered, Q23 (Section 2.2) is unordered. Both involve 'assigning' or 'selecting', but persons are distinguishable (ordered triples, 3³ = 27) while a committee is not (unordered pairs, C(6,2) = 15). Read whether the identity of the position matters.
Solution recipe
- Decide what a single outcome records, and pick a compact notation (a string, a tuple, an ordered list).
- Decide whether order matters and whether objects are distinguishable. This fixes the counting rule:
• fixed-length strings over an alphabet of size k, length n → kⁿ outcomes
• distinct objects arranged → n! outcomes
• repeated symbols (a multiset) → C(n, k) style counts
• a stopping rule → outcomes of unequal length; stratify by length and list each stratum - Write out 𝒮 systematically (by length, then lexicographically) so nothing is missed.
- For each named event, filter 𝒮 by the stated condition.
- Apply ∪, ∩, ′ directly to the listed sets; look for containments (A ⊆ B) that collapse the work.
How the book escalates
Q1–Q5 are straight product-rule enumerations of increasing size (16, 27, 8, 16, 27). Q6 breaks the fixed-length assumption with a stopping rule. Q7 breaks the distinguishability assumption with a multiset, and its part (b) adds a path condition (the ballot problem) that no formula in Chapter 2 covers — it must be checked outcome by outcome.