Deep Dive: Sampling Tree Diagrams Venn Diagrams

#CODEX Deep-Dive — Sampling, Tree Diagrams, and Venn Diagrams (WST01, spec 3.4)

Provenance (2026-09-29): Read/checked for this document — research/veridian/WST01-verified-facts.md in full, grepped specifically for "sampl", "tree diagram", "venn", "replacement", "product law", "sum law", "addition law" and "formula booklet" (§1 spec transcription, §2a formula-booklet contents, §3 source archive, §4 mark-scheme conventions, §5 Section 3 trap catalogue, §7 topic-richness ranking, §8's dedicated sampling-tree-diagrams-venn-diagrams.ts audit, and the two elementary-probability-and-conditional-notation.ts / conditional-probability-independence-vs-mutually-exclusive.ts audits for their Venn-diagram anchors); the live lesson file content/veridian/curriculum/wst01/sampling-tree-diagrams-venn-diagrams.ts in full (header, all block types, all six MCQs, both worked-chains, the trap-taxonomy); content/veridian/technique-hub/wst01/codex-formula-and-method-bank.md, forge-exam-technique-guide.md and signal-frequency-and-boundaries.md (grepped for this topic); content/veridian/question-bank/wst01.ts (grepped — no item on this topic exists there yet). No new PDF fetches were made. Every arithmetic claim and micro-example below was independently checked with exact-fraction arithmetic (Python's fractions.Fraction) before being written in.

#Scope — what this adds, and what it deliberately does not repeat

The live lesson already drills the single richest confirmed trap on this topic in depth — the without-replacement "reused fraction" error, on both a one-container chain and a genuine two-container transfer chain — plus a Venn-diagram marked solution, an independence-vs-assumed-independence trap, and six MCQs. The formula bank already tabulates the product/sum law as "technique, not a formula." The frequency doc already gives this pairing's series count. None of that is reproduced here.

This document instead does four things the existing material doesn't: (1) derives, rather than just asserts, why a tree diagram and a Venn diagram are two pictures of the same two underlying laws, and why the two-event addition law follows directly from disjoint regions rather than being a formula to memorise; (2) states the general, provable reason the lesson's own two-container transfer example works — that a container's post-transfer composition depends only on which items were moved in, not the order — rather than leaving it as an empirical coincidence of one set of numbers; (3) treats three-event Venn diagrams as a genuine structural sub-case of this spec point, using two real anchors that are filed elsewhere in this repo under 3.2/3.3 but are, on the diagram-construction side, real 3.4 exam angles in their own right; (4) assembles every real exam angle this specific topic (sampling, tree diagrams, Venn diagrams — not the adjacent conditional-probability/independence pairing, which has its own deep-dive) has actually been tested from, with the honest limits of that evidence stated plainly.

#1. The formal model: one sample space, two ways to partition it

Sample space and events (spec 3.1/3.2, underlying both pictures). Every probability calculation in this topic starts from a sample space $\Omega$ of equally-weighted or otherwise-specified outcomes, with $P(\Omega)=1$. A tree diagram and a Venn diagram are two different ways of drawing the same partition of $\Omega$ into mutually exclusive pieces whose probabilities sum to 1 — a tree diagram partitions by sequence of events (branches), a Venn diagram partitions by combination of events (regions).

The product law (tree diagrams). The multiplication law printed on the WST01 formula sheet, $P(A\cap B)=P(A)P(B\mid A)$, is the rule a tree diagram draws: the probability of following branch $A$ then branch $B$ is the first branch's probability multiplied by the second branch's probability, conditioned on the first having happened. Extending along $n$ branches multiplies $n$ conditional probabilities, one per branch — this is why a single omitted or un-updated branch corrupts the entire path's probability, not just that one branch's own value.

The sum law (both diagrams, in opposite directions). On a tree diagram, distinct complete paths are mutually exclusive by construction (two different sequences of draws cannot both occur on the same trial), so the probability of an event realised by several different paths is the sum of those paths' probabilities, with nothing to subtract. On a Venn diagram, the same sum law runs the other way: the diagram is built by cutting $\Omega$ into mutually exclusive regions in the first place (event-combination, not event-sequence), so any union of interest is read off by adding the regions that make it up — which is exactly why the two-event addition law below has to hold, rather than being a rule bolted on separately.

What must be memorised versus looked up, specifically for 3.4 (verified against the formula booklet, Issue 2, January 2021, pp. 13–17). The general product law $P(A\cap B)=P(A)P(B\mid A)$ and the two-event addition law $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ are both printed in the booklet — students look them up. The independence special case $P(A\cap B)=P(A)P(B)$ is explicitly listed in the spec's own notation-and-formulae box as one students must know from memory, not one the booklet supplies. For 3.4 specifically, this means the tree-diagram product rule itself never needs memorising (it's the same booklet formula applied branch by branch), but recognising when a with-replacement tree collapses to the independence special case — and therefore needs no conditioning at all — is a recall skill, not a look-up.

#2. Tree-diagram sub-cases, each with a short micro-example

#2.1 With replacement — draws are genuinely independent trials, not just "the same numbers again"

When an item is replaced before the next draw, the bag's composition is identical at every stage, so $P(B\mid A)=P(B)$ by the definition of independence (spec 3.3) — every branch leaving a given outcome carries the same fraction as the branch before it, for the entire depth of the tree, not merely "by coincidence" but because the conditioning event genuinely doesn't change anything about the sample space.

Micro-example. A spinner has 4 equal sectors, 3 red and 1 blue, spun twice with the outcome recorded each time (nothing physically removed, so this behaves exactly like with-replacement sampling). $P(\text{both red}) = \frac34\times\frac34=\frac{9}{16}$ — the same fraction, $\frac34$, on both spins, because nothing about the spinner has changed between them.

#2.2 Without replacement, one container — the rebuilt branch

Removing an item without replacing it changes both the numerator (the count of whatever was just removed) and the denominator (the total remaining) for every subsequent draw — the mechanism the live lesson's disc micro-example and its "bag of counters" worked-chain already drill at length against the real Jan 2023 Q2 finding that students carried a first-bag fraction "onto the branches following the 1st was blue" and the real Jun 2022 Q3(d) finding of a fixed-denominator error, $(62/88)^4$, across four genuine without-replacement draws.

Fresh micro-example, illustrating the sum law on top of the rebuilt branch (not reproduced in the live lesson's own chains). A box holds 7 tiles, 4 red and 3 white; two are drawn without replacement. $P(\text{red then white}) = \frac47\times\frac36=\frac{2}{7}$; $P(\text{white then red}) = \frac37\times\frac46=\frac27$ — two different paths, but identical totals, because the same two tiles are removed either way, just in the opposite order. $P(\text{exactly one red}) = \frac27+\frac27=\frac47$ by the sum law, since the two orders are mutually exclusive ways of getting one of each colour.

#2.3 Without replacement, across two containers — why transfer order never matters, in general

The live lesson's own seven-stage worked-chain (Container P transferring counters into Container Q) demonstrates, on one specific pair of numbers, that moving a red-then-blue counter into a container rebuilds it identically to moving blue-then-red. That is not a coincidence of those particular numbers: a container's composition after a transfer depends only on the multiset of items moved in, not the sequence they arrived in, because forming the new count is addition, and addition is commutative. If a container starts with $r$ red and $b$ blue, and one red and one blue counter are transferred into it from elsewhere, its new composition is $(r+1)$ red and $(b+1)$ blue regardless of which one physically arrived first — which is exactly why the real anchor's own (a) mark scheme rewards "8/13 & 5/13" and "7/13 & 6/13 in BOTH correct places" for the two transfer orders: both orders are being credited for reconstructing the same destination container, because they provably must.

Minimal illustration of the counting fact itself (not a probability chain — the lesson's own chain already covers the full tree). A container starts with 4 red and 2 blue counters (6 total). One red and one blue counter are transferred into it from elsewhere, in either order. Composition after the transfer is $4+1=5$ red and $2+1=3$ blue (8 total) — the same 5-red-3-blue result whichever counter physically landed first, because the final counts are just $4+1$ and $2+1$ either way.

#2.4 Reading a conditional probability directly off a tree diagram's own path-sums

Spec 3.2's conditional-probability formula, $P(A\mid B)=P(A\cap B)/P(B)$, applies to a tree diagram by treating $B$'s probability as the sum of every path-end that counts as $B$, and $A\cap B$'s probability as the sum of only those $B$-paths that also satisfy $A$ — the live lesson demonstrates this once, in stage 6 of its "bag of counters" chain, and the real Jan 2023 Q2(d) finding (a spurious extra copy of the $61/234$ denominator folded into the numerator, rather than the correct clean path-product) is the confirmed, quoted trap this exact skill guards against.

Fresh micro-example, same method, different numbers. A bag has 6 counters, 4 red and 2 blue; two are drawn without replacement. $P(RR)=\frac46\times\frac35=\frac25$; $P(BB)=\frac26\times\frac15=\frac{1}{15}$; $P(\text{same colour})=\frac25+\frac{1}{15}=\frac{7}{15}$. Given the two counters were the same colour, $P(\text{both red}) = P(RR)/P(\text{same}) = \frac{2/5}{7/15}=\frac67$ — the numerator is exactly one of the two already-found path probabilities, the denominator is their sum, and nothing else is added in, which is precisely the discipline the real Jan 2023 Q2(d) error abandoned.

#3. Venn-diagram sub-cases, each with a short micro-example

#3.1 The two-event addition law, derived from disjoint regions rather than just quoted

A Venn diagram for two events splits $\Omega$ into four mutually exclusive regions: $A$-only, $B$-only, $A\cap B$, and neither. Since $A$ is the union of its own two disjoint pieces, $P(A)=P(A\text{-only})+P(A\cap B)$, and likewise $P(B)=P(B\text{-only})+P(A\cap B)$. Adding these two equations counts the $A\cap B$ region twice: $P(A)+P(B)=P(A\text{-only})+P(B\text{-only})+2P(A\cap B)$. But $P(A\cup B)$ — the region-sum of $A$-only, $B$-only and $A\cap B$ — counts that same intersection only once, so $P(A)+P(B)=P(A\cup B)+P(A\cap B)$, which rearranges to the booklet's own formula, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. This is why the formula subtracts the intersection: it is undoing the double-count that adding the two marginal probabilities directly created.

Micro-example, checked against the region sum directly. $P(A)=0.7$, $P(B)=0.5$, $P(A\cap B)=0.3$. Regions: $A$-only $=0.7-0.3=0.4$; $B$-only $=0.5-0.3=0.2$; both $=0.3$; neither $=1-(0.4+0.2+0.3)=0.1$ — a genuine, checkable partition (all four regions sum to exactly 1). The addition law gives $P(A\cup B)=0.7+0.5-0.3=0.9$, which matches summing the three regions inside the union directly, $0.4+0.2+0.3=0.9$ — the formula and the direct region count agree, as the derivation above guarantees they always will.

#3.2 The blank-region convention

A confirmed, specific presentation trap, verified directly (Jun 2022 Q4(c) examiner report): candidates left the "neither" region of a two-event diagram blank, and the report states plainly that "blank spaces are not assumed to be 0s in Venn diagrams." The 0.1 "neither" region in §3.1's own micro-example is exactly the kind of value this convention requires writing in explicitly, even when it is small, or genuinely zero, rather than leaving the space empty.

#3.3 Independence and mutual exclusivity, read off Venn-diagram values

Testing whether two events shown on a Venn diagram are independent means comparing the diagram's own $P(A\cap B)$ value against the computed product $P(A)\times P(B)$ — never inferring the answer from whether the regions visually overlap, which is a distinct, geometric question about mutual exclusivity, not independence. This exact by-eye substitution is a confirmed, named misconception (Oct 2021 Q1(b) examiner report quotes a student line, "they are not independent as they overlap"), and Jun 2022's own general commentary states a paper-wide version of the same failure: candidates "often assume independence when an appropriate conditional probability should be used instead." Both the underlying proof (mutually exclusive events with positive probability can never be independent) and a full worked derivation of this specific pairing are already covered in depth in this course's sibling document, codex-deep-dive-conditional-probability-independence-vs-mutually-exclusive.md, and are not re-derived here — they are cross-referenced because a genuinely complete account of this topic's Venn-diagram sub-case has to name the test at all, even without repeating its proof.

#3.4 Three-event Venn diagrams: real diagrams are not a fixed 8-region template

The spec's own guidance for 3.4 names "Venn diagrams" without restricting the count of events, and this course's own real-anchor record confirms three-event diagrams have actually been set, twice, each with a genuinely different region layout — a structural point about diagram construction worth stating explicitly, even though both anchors are filed in this repo's curriculum under adjacent spec points (3.2 and 3.3 respectively) rather than under this lesson.

• Jan 2021 Q1: events $A$, $B$, $C$ with six named regions — $A$-only $=0.15$, $A\cap B=0.20$, $B$-only $=0.17$, $B\cap C=0.23$, $C$-only $=0.12$, none $=0.13$ — independently confirmed to sum to exactly 1. This diagram has no drawn $A\cap C$ region at all: $A$ and $C$ are mutually exclusive on this specific figure, which is why the mark scheme's own alternative route to part (b), "$0.35+0.35$," works — it is adding two genuinely disjoint marginal probabilities, not a coincidence of arithmetic.
• Oct 2021 Q1: events $A$, $B$, $C$ with regions $0.04$ ($A$-only), $0.06$ ($A\cap C$-only), an unknown $p$ (the full three-way overlap $A\cap B\cap C$), an unknown $q$ ($A\cap B$-only), $0.3$ ($B$-only), and $0.4$ (none) — solved via $p+q=0.2$ (from the total-probability-equals-1 constraint) and the given condition $P(C\mid B)=P(C)$, yielding $p=0.06$, $q=0.14$. This diagram has a genuine three-way overlap region but no drawn $B\cap C$-only region.

The general lesson for diagram-reading: a real three-event Venn diagram on this paper is drawn to fit the specific scenario's events, not stamped from a generic eight-region template with every pairwise and triple overlap always present — a student has to read which regions the specific diagram actually shows, rather than assume a fixed shape and hunt for a missing box to fill in.

Micro-example, VERIDIAN-original, illustrating region bookkeeping with one unknown. A three-event Venn diagram for a sports club's members records: swims only $=0.30$, runs only $=0.15$, cycles only $=w$ (unknown), swims-and-runs only $=0.10$, runs-and-cycles only $=0.05$, none of the three $=0.10$ (six named regions, no drawn swims-and-cycles overlap, mirroring the Jan 2021 Q1 diagram's own "one pair entirely absent" shape). Since all regions sum to 1: $0.30+0.15+w+0.10+0.05+0.10=1$, so $w=0.30$. $P(\text{cycles})$ is then read off directly as the sum of every region touching "cycles" that the diagram actually draws — here, cycles-only ($0.30$) plus runs-and-cycles-only ($0.05$), since no swims-and-cycles region exists on this diagram: $P(\text{cycles})=0.35$.

#4. Synthesis — every real exam angle this topic has actually been tested from

Grounded entries only, one row per distinct angle, all traced to WST01-verified-facts.md:

Honest frequency statement, no invented likelihoods. Sampling/tree-diagram/Venn-diagram construction (3.4 as its own named skill, distinct from the adjacent conditional-probability and independence content) is directly evidenced in 2 of the 6 examiner-report series fully reviewed for this facts bank — Jun 2022 and Jan 2023. Venn diagrams also appear as the vehicle for adjacent 3.2/3.3 questions in a further 2 of the 6 (Jan 2021, Oct 2021) — real diagram-construction material, but testing a different named spec point through the same picture. Mark-scheme-level confirmation of tree-diagram/sampling mechanics (without an accompanying examiner report) additionally exists in at least 14 series' worth of question papers reviewed for this facts bank, per its own §7 ranking of this pairing as having "multiple clean, fully-quoted worked contexts" — but that is a mark-scheme count, not an examiner-report count, and the two should not be conflated.

#5. Open flags — where the real evidence currently runs out

• 8 of the 14 mark-scheme series that exist for this unit have no confirmed, independently-verified examiner report read yet (Jun 2021, Jan 2022, Oct 2022, Jun 2023, Oct 2023, Jan 2024, Oct 2024, Jan 2025 — per the facts bank's §8/§9). The one spot-check done outside the original five plus Jun 2019 surfaced real gaps in two other topics (outliers/box plots; discrete random variables) and reported nothing new for this topic specifically — one series out of eight unread, a lower bound, not a closed inventory.
• §2.3's "transfer order never matters" claim is a general counting argument this document derives, not a Pearson-stated rule. It is a straightforward consequence of addition being commutative, and it correctly explains the live lesson's own numeric example, but it should be taught as a reason, not quoted as if the mark scheme states it in those words.
• §3.4's "real diagrams aren't a fixed template" observation rests on exactly two confirmed anchors. Two independent, differently-shaped real diagrams is a real pattern worth teaching, but it is not a claim that every future three-event Venn question will omit some pairwise region — only that assuming a full generic template is not safe, on the evidence actually reviewed.
• No standalone past-paper question testing a three-outcome-per-draw tree diagram (as opposed to two-outcome branches) was found in the material reviewed for this document — every real tree-diagram anchor cited above uses binary branches at each stage (a counter is one of two colours). This document does not claim a non-binary branch case is untested on this paper generally, only that none was found in the specific sources read for it.

#Sources

research/veridian/WST01-verified-facts.md — §1 (spec text, formula-booklet memorise/look-up split), §2a (formula booklet contents), §3 (source archive and its honest denominators), §4 (mark-scheme conventions, ft/oe), §5 Section 3 (all examiner-report quotes cited above), §7 (topic-richness ranking), §8's dedicated audit of sampling-tree-diagrams-venn-diagrams.ts (the Jan 2023 Q2 anchor, all three closed gaps), and the elementary-probability-and-conditional-notation.ts / conditional-probability-independence-vs-mutually-exclusive.ts audits (Jan 2021 Q1 and Oct 2021 Q1 three-event Venn anchors). content/veridian/curriculum/wst01/sampling-tree-diagrams-venn-diagrams.ts (live lesson, read in full). content/veridian/technique-hub/wst01/codex-formula-and-method-bank.md, forge-exam-technique-guide.md, signal-frequency-and-boundaries.md, codex-deep-dive-conditional-probability-independence-vs-mutually-exclusive.md (cross-referenced for §3.3). content/veridian/question-bank/wst01.ts (checked, no overlapping item found).