Elementary probability, and conditional-probability notation

An experiment is any process with an uncertain result — rolling a die, drawing a card, surveying a customer. Every possible result is an outcome, and the full set of outcomes is the sample space, written $\Omega$. An event is simply a subset of $\Omega$: 'the die shows an even number' picks out $\{2, 4, 6\}$ from $\Omega = \{1,2,3,4,5,6\}$. This is genuinely all spec 3.1 asks for as a starting definition — the specification's own unit description names 'the basic ideas of mathematical modelling as applied in probability and statistics' as the whole of this spec point, and the modelling idea itself is usually just as basic: calling a die 'fair,' or a coin 'unbiased,' is an ASSUMPTION about the real object being modelled, not a fact that could be proven from the die itself — a die could always, in principle, be very slightly weighted. The research bank behind this lesson finds this assumption showing up as ordinary phrasing embedded inside real questions ('assume the die is fair') rather than as a topic tested on its own — see this lesson's own header for the honest scope of that.

When every outcome in $\Omega$ is equally likely — a fair die, a well-shuffled deck, a random draw — probability becomes pure counting: $P(A) = \dfrac{n(A)}{n(\Omega)}$, the number of outcomes in the event divided by the total number of outcomes. This is often called classical probability, and almost every skill in this lesson, right through to conditional notation, is a variation on getting this one count right: how many outcomes are in $\Omega$, and how many of those are in the event actually being asked about.

Getting $n(\Omega)$ right depends on listing SYSTEMATICALLY, not by eye. Two genuinely different traps live here. First: when an experiment combines several independent choices — two dice, a coin and a die, three coin tosses — the sizes MULTIPLY, not add: two 4-sided dice give $4 \times 4 = 16$ outcomes, not $4 + 4 = 8$, because every one of the 4 first-roll results can be paired with every one of the 4 second-roll results. Second: outcomes built from more than one stage are usually ORDERED — rolling a 2 then a 6 is a different outcome from rolling a 6 then a 2, even though they'd be 'the same pair of numbers' if you stopped caring about order. Treating $(2,6)$ and $(6,2)$ as one outcome instead of two is exactly the kind of undercounting a systematic list — a table with one row per first result and one column per second — is built to prevent, because a table forces every combination to get its own cell.

Spec 3.2 gives events a small, precise vocabulary, verified directly against the real WST01 formula booklet for which parts of it are printed there and which aren't (WST01-verified-facts.md §2a). The complement $A'$ is 'everything in $\Omega$ that isn't in $A$' — $A$ and $A'$ between them always account for the whole sample space, so $P(A') = 1 - P(A)$. This rule is NOT printed on the formula sheet — the spec's own notation box states plainly that formulae 'expected to know' are left out of the booklet deliberately, and this is one of them; it costs nothing to memorise, but nothing will remind you of it in the exam either. Two events are mutually exclusive when they share no outcomes at all — $A \cap B = \emptyset$, so $P(A \cap B) = 0$ — a statement about whether the two SETS overlap, decided by what the events mean, before any probability is attached.

The general addition law, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, IS printed on the formula sheet, and it's worth reading as a translation rule as much as a formula: 'or' (in the everyday, inclusive sense — 'A or B or both') means $\cup$, and simply adding $P(A)$ and $P(B)$ double-counts whatever's shared, which the $-P(A \cap B)$ term exists purely to correct. For mutually exclusive events specifically, that correction term is already zero, so the law simplifies to $P(A \cup B) = P(A) + P(B)$ — a genuine shortcut, but only in that one special case; using the short version when events actually overlap is exactly the double-counting trap above.

Reading a word problem into the right symbol is the whole skill spec 3.2 is testing at this level, and it comes down to a short, fixed dictionary: 'or' (inclusive) $\to \cup$; 'and'/'both' $\to \cap$; 'not' $\to {}'$; and — the one that trips up notation more than any other — 'given that' $\to$ the conditioning bar, $\mid$. The event named IMMEDIATELY after 'given that' always goes on the RIGHT of the bar; the event actually being asked about goes on the LEFT. 'The probability a customer buys a magazine, given they've bought a newspaper' names the newspaper LAST in the sentence but writes it SECOND in the notation: $P(\text{magazine} \mid \text{newspaper})$ — the grammar of the sentence and the order of the symbols don't have to match, and assuming they do is exactly how the two events end up swapped.

Conditional probability itself, $P(A \mid B)$, is defined by restricting the sample space: once you know $B$ has happened, every outcome outside $B$ is no longer possible, so the relevant 'whole' shrinks from $\Omega$ down to $B$ itself. Read as counting — spec 3.1's classical-probability idea, carried straight into spec 3.2's notation — this gives the cleanest possible route to the formula: $P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$, the outcomes that are in BOTH $A$ and the shrunk space $B$, divided by the size of $B$ itself. Dividing top and bottom by $n(\Omega)$ turns that straight into the version on the formula sheet, $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ (requiring $P(B) > 0$) — the same statement, in probabilities instead of raw counts. Whenever a question hands you equally likely outcomes you can list, the counting version is usually the safer route: there's no formula to misremember, only a smaller list to count correctly.