Sampling with/without replacement, tree diagrams, and Venn diagrams

~45 min · WST01 · 3.4

WST01 · 3.4 · 45 min

A tree diagram for sampling without replacement is not one bag drawn from twice — it is two different bags, drawn from once each. The first draw changes what is left, so every branch after it describes a smaller, different sample space than the one before it. The single most-repeated way to lose marks on this topic, confirmed directly from a real question, is writing the first bag's fraction onto the second bag's branch, as though nothing had been removed at all.

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Sampling with and without replacement — what changes between draws

Spec 3.4 groups three ideas under one heading — "sum and product laws" — because tree diagrams and Venn diagrams are just two different pictures of the same two rules: multiply probabilities along a sequence of events, add probabilities across separate ways an outcome can happen. Sampling with or without replacement decides what the numbers in that multiplication actually are, which is why it sits inside the same spec point rather than off on its own.

With replacement, whatever is drawn goes back before the next draw, so the bag looks identical every time. Draw after draw is independent of every draw before it — the conditional-probability shortcut from spec 3.3, $P(B \mid A) = P(B)$ , applies exactly, and every branch leaving a given colour keeps the same fractions the whole way down the tree.

Without replacement, whatever is drawn stays out, so the bag is smaller and differently composed for the next draw — one fewer of whatever colour was just removed, one fewer overall. The branches leaving "first was black" and the branches leaving "first was white" describe two different bags, and they need two different sets of fractions. This is not a rule to memorise; it is a direct consequence of what "without replacement" physically means, and it is worth re-deriving from the picture every time rather than pattern-matching to a remembered fraction.

One habit catches almost every error in this section before it is even written down: at any single node of a tree, the branches leaving it have to add to 1, because between them they cover every possibility from that point. If a pair of branches you have just written down does not sum to 1, something about that node is wrong — usually a leftover number from before the first draw.

Mechanism

Why branches multiply and paths add — a tree diagram is the product and sum laws, drawn

Rearranging the conditional-probability formula from spec 3.2, $P(A \cap B) = P(A)P(B \mid A)$ , already says exactly what a tree diagram draws: the probability of following two branches in sequence — first $A$ , then $B$ — is the first branch's probability multiplied by the second branch's probability, where the second is conditioned on the first having happened. That is the product law, and a tree diagram does not add anything new to it; it is a picture of applying it once per draw. For a two-draw tree, the probability of reaching any single end-point is the product of every fraction on the path that leads there — three draws multiply three fractions, four multiply four, and so on, which is exactly why forgetting to update even one of them (the without-replacement trap above) corrupts the whole path, not just the branch where the mistake happened.

The sum law answers a different question: not "what is the probability of this one path", but "what is the probability of this event, which several different paths could produce". Two different complete paths through a tree — say, black-then-white and white-then-black — cannot both happen on the same pair of draws, so they are mutually exclusive by construction, whatever the actual probabilities turn out to be. That is what licenses $P(\text{event}) = P(\text{path 1}) + P(\text{path 2}) + \ldots$ with no overlap term to subtract: add the probability of every complete path that counts as a "success", and nothing is double-counted, because no single pair of draws can walk two different paths at once. This is also the logic behind the branches-sum-to-1 rule above: the full set of paths through a tree is mutually exclusive and exhaustive, so summing every end-point's probability has to return exactly 1.

Real WST01 mark schemes reward exactly this structure. The general marking guidance printed on every one of the 14 series' mark schemes reviewed for this course defines an 'M' mark as one "given for a correct method or an attempt at a correct method" — here, setting up the right multiplication or the right sum — and defines an 'A' mark as a dependent accuracy mark, awarded only once the method mark behind it has been earned. A note specific to Statistics papers adds a further rule: "For method marks, we generally allow or condone a slip or transcription error if these are seen in an expression. We do not, however, condone or allow these errors in accuracy marks." In practice, that means the structure of a tree-diagram calculation — which branches were multiplied, which paths were summed — is worth setting out visibly even under time pressure, because that structure is where the more forgiving mark lives.

Worked, in full

The bag of counters: is the second draw the same colour as the first?

01
Draw the tree and label only what is known before any draw happens: a bag of 8 counters, 5 green and 3 blue. The first pair of branches is $P(G_1) = \frac{5}{8}$ and $P(B_1) = \frac{3}{8}$ — and check the habit from above immediately: $\frac{5}{8} + \frac{3}{8} = 1$ . ✓
Earns: B1 — first-draw probabilities correctly stated from the given composition of the bag.
02
Before labelling the second pair of branches, rebuild the bag twice — once for each first-draw outcome — because "without replacement" means each one really is a different bag now. Following $G_1$ : 4 green and 3 blue remain, 7 counters in total. Following $B_1$ : 5 green and 2 blue remain, 7 counters in total. Writing this out explicitly, before touching a single fraction, is what stops the second-draw branches from silently copying the first.
Earns: Nothing yet — this stage sets up the M1 below. There is no separate mark on a real mark scheme for stating the reduced composition on its own, only for what it produces.
03
Label the second-draw branches from each of those two rebuilt bags, not from the original one. Following $G_1$ : $P(G_2 \mid G_1) = \frac{4}{7}$ , $P(B_2 \mid G_1) = \frac{3}{7}$ . Following $B_1$ : $P(G_2 \mid B_1) = \frac{5}{7}$ , $P(B_2 \mid B_1) = \frac{2}{7}$ . Both pairs sum to 1 again — $\frac{4}{7}+\frac{3}{7}=1$ and $\frac{5}{7}+\frac{2}{7}=1$ — which is exactly the check that catches the real, documented error of writing $\frac{5}{8}$ or $\frac{3}{8}$ here instead: those fractions belong to the original 8-counter bag, and neither pair built from them would sum to 1 against a 7-counter bag.
Earns: M1 — attempts the second-draw conditional probabilities from the reduced, correctly-composed bag for at least one branch.
04
"Same colour" happens down exactly two of the four complete paths — green-then-green, or blue-then-blue — and those two paths cannot both happen on the same pair of draws, so the product law gives each path and the sum law will add them: $P(GG) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ and $P(BB) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$ .
Earns: A1 — both individual path probabilities correct, built on the correctly-composed branches from stage 3.
05
Add the two paths: $P(\text{same colour}) = \frac{5}{14} + \frac{3}{28} = \frac{10}{28} + \frac{3}{28} = \frac{13}{28}$ . A free check costs one more line: the complementary event, "different colours", covers the other two paths, $P(GB) + P(BG) = \frac{5}{8}\times\frac{3}{7} + \frac{3}{8}\times\frac{5}{7} = \frac{15}{56}+\frac{15}{56} = \frac{15}{28}$ , and $\frac{13}{28} + \frac{15}{28} = 1$ exactly, as it has to.
Earns: A1 — correct final probability, $\frac{13}{28}$ .
06
A natural follow-up, mirroring the shape of a real WST01 tree-diagram question exactly: given that the two counters drawn WERE the same colour, find $P(\text{both green})$ . This is $P(GG \mid \text{same colour})$ — restrict attention to the "same colour" outcome (the denominator, already found in stage 5) and ask what share of it is the $GG$ path specifically (the numerator, one of the two individual path probabilities already found in stage 4): $P(GG \mid \text{same colour}) = \frac{P(GG)}{P(\text{same colour})} = \frac{5/14}{13/28} = \frac{5}{14}\times\frac{28}{13} = \frac{10}{13}$ . Nothing here needed calculating fresh — both numbers were already sitting on the page from stages 4 and 5. This is the same conditional-probability formula from the Venn-diagram teach block below, $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$ , applied here to a tree diagram instead of a Venn diagram: the "event" is the single path $GG$ , and the "condition" is the two-path union "same colour" found by the sum law one line above.
Earns: A1ft-style — the correct numerator (one of the two already-found path probabilities) over the already-found part-(c)-style denominator; ft their tree diagram.

Source — Examiner report, Jan 2023

"common errors seen were to repeat the 4/8 (given in the question) following the 1st was green onto the branches following the 1st was blue"

Complete it yourself

Complete the chain — a box of 9 pens, 6 blue and 3 red, two taken without replacement: what is $P(\text{both blue})$ ?

01
Before any pen is taken, the box holds 6 blue and 3 red — 9 pens in total, so $P(B_1) = \frac{6}{9} = \frac{2}{3}$ .
02
Rebuild the box for what happens after one blue pen is removed, before writing any second-draw fraction: 5 blue and 3 red remain, 8 pens in total. (If the first pen taken had been red instead, the rebuilt box would look different again — 6 blue and 2 red, 8 total — but that branch is not needed for "both blue".)

Worked, in full

Two containers, not one: transferring counters from P into Q before drawing again

01
Every tree diagram so far has reduced a single bag across its own successive draws. Real without-replacement questions are often built differently: two separate containers, with counters physically moved from one into the other partway through. Container P holds 6 red and 4 yellow counters (10 total); Container Q is a separate, already-existing stock of 5 red and 3 yellow counters (8 total) — not touched until something is actually placed into it. Label only the first pair of branches, leaving Container P: $P(R_1) = \frac{6}{10} = \frac{3}{5}$ , $P(Y_1) = \frac{4}{10} = \frac{2}{5}$ . Check: $\frac{3}{5} + \frac{2}{5} = 1$ . ✓
Earns: B1 — first-draw probabilities correctly stated from the given composition of Container P.
02
Two counters are drawn one after another from Container P, without replacement, and each is placed into Container Q as it is drawn — Container Q stays exactly as given until a counter physically lands in it. The second draw still comes out of P, so P still has to be rebuilt twice for it, exactly as with a single bag: following $R_1$ , P has 5 red and 4 yellow left (9 total); following $Y_1$ , P has 6 red and 3 yellow left (9 total).
Earns: Nothing yet — sets up the M1 below, the same role stage 2 plays in the single-bag chain above.
03
Label the second-draw branches from each rebuilt version of P, not the original 10-counter one: following $R_1$ , $P(R_2 \mid R_1) = \frac{5}{9}$ , $P(Y_2 \mid R_1) = \frac{4}{9}$ ; following $Y_1$ , $P(R_2 \mid Y_1) = \frac{6}{9} = \frac{2}{3}$ , $P(Y_2 \mid Y_1) = \frac{3}{9} = \frac{1}{3}$ . Both pairs sum to 1 again.
Earns: M1 — attempts the second-draw conditional probabilities from the reduced, correctly-composed version of P for at least one branch.
04
Multiply along each of the four complete paths through Container P to get the probability that each specific pair of colours is the one transferred: $P(RR) = \frac{3}{5} \times \frac{5}{9} = \frac{1}{3}$ , $P(RY) = \frac{3}{5} \times \frac{4}{9} = \frac{4}{15}$ , $P(YR) = \frac{2}{5} \times \frac{6}{9} = \frac{4}{15}$ , $P(YY) = \frac{2}{5} \times \frac{3}{9} = \frac{2}{15}$ . These four paths are mutually exclusive and exhaust every possibility for the two counters leaving P, so they have to sum to 1: $\frac{1}{3} + \frac{4}{15} + \frac{4}{15} + \frac{2}{15} = \frac{5}{15} + \frac{4}{15} + \frac{4}{15} + \frac{2}{15} = 1$ . ✓
Earns: A1 — all four two-counter path probabilities correct, built on the correctly-composed branches from stage 3.
05
Before writing a single third-draw fraction, rebuild Container Q's composition for each of those four paths — the step a single-bag tree never needs, because the third draw here comes out of a genuinely different container from the one the first two draws came out of. Container Q's new composition = its original stock (5 red, 3 yellow) + whichever two counters were just transferred into it: following $RR$ , Q gains two reds → 7 red, 3 yellow, 10 total; following $RY$ , Q gains one red and one yellow → 6 red, 4 yellow, 10 total; following $YR$ , Q also gains one red and one yellow → 6 red, 4 yellow, 10 total — identical to the line above, because Q's composition depends only on how many of each colour arrived, not on the order they arrived in; following $YY$ , Q gains two yellows → 5 red, 5 yellow, 10 total.
Earns: Nothing yet — this is the rebuild step itself, the direct analogue of stage 2 above but for a second container instead of a smaller version of the first one. There is no separate mark for stating it, only for what it produces below — but skipping it is exactly what the real Jan 2023 Q2 mark scheme is checking for when it awards its two 3rd-branch marks separately for the two differently-rebuilt compositions.
06
Label the third-draw branch — drawn from Q, not from P — for each of the four rebuilt compositions: $P(R_3 \mid RR) = \frac{7}{10}$ ; $P(R_3 \mid RY) = P(R_3 \mid YR) = \frac{6}{10} = \frac{3}{5}$ , the identical fraction in both places, exactly because stage 5 showed those two paths rebuild Q identically; $P(R_3 \mid YY) = \frac{5}{10} = \frac{1}{2}$ .
Earns: B1 — third-draw probability correct following the RR-transfer composition. B1 — third-draw probabilities correct and IDENTICAL in both correct places following the RY- and YR-transfer compositions (the real Jan 2023 Q2 mark scheme awards its equivalent line only when both are present and equal, not just one of them).
07
Add every complete three-draw path that ends in a red third counter — the sum law, because the four paths from Container P are still mutually exclusive: $P(R_3) = P(RR)\times\frac{7}{10} + P(RY)\times\frac{3}{5} + P(YR)\times\frac{3}{5} + P(YY)\times\frac{1}{2}$ . Since $P(RY)$ and $P(YR)$ carry the identical third-draw fraction, they can be grouped before multiplying: $(P(RY)+P(YR))\times\frac{3}{5} = \frac{8}{15}\times\frac{3}{5} = \frac{8}{25}$ . So $P(R_3) = \frac{1}{3}\times\frac{7}{10} + \frac{8}{25} + \frac{2}{15}\times\frac{1}{2} = \frac{7}{30} + \frac{8}{25} + \frac{1}{15}$ . Over a common denominator of 150: $\frac{35}{150} + \frac{48}{150} + \frac{10}{150} = \frac{93}{150} = \frac{31}{50}$ .
Earns: A1 — correct final probability, $\frac{31}{50}$ .

Source — Mark scheme, Jan 2023

"B1 for 8/13 & 5/13 in the correct place on the 3rd branches; B1 for 7/13 & 6/13 in both correct places on the 3rd branches"

Venn diagrams: reading regions, and the rule about what "blank" means

A Venn diagram for two events $A$ and $B$ splits the whole sample space into exactly four regions: inside $A$ only, inside $B$ only, inside both (the overlap, $A \cap B$ ), and inside neither — outside both circles but still inside the rectangle that represents everything possible. Every probability or frequency the question gives has to land in exactly one of those four regions, and the four regions between them always account for the whole sample space — the Venn-diagram version of the same "branches sum to 1" check from the tree-diagram section above.

Reading $P(A \cap B)$ off the diagram is direct: it is whatever value sits in the overlap. Reading a conditional probability is almost as direct once the picture is understood correctly — $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ restricts attention to the $B$ circle only (the denominator) and asks what fraction of it also lies inside $A$ (the numerator, the overlap). The diagram makes the restriction visible: the whole rectangle is no longer in view, only one circle within it.

The region outside both circles is real and has to carry a value, even when that value turns out to be zero — leaving it blank does not mean zero, it means unfinished. If three of the four regions are given or calculated, the fourth is found the same way the last branch of a tree diagram is: everything has to sum to 1 (or to the total, if the diagram is drawn in frequencies rather than probabilities), so it is whatever the other three haven't accounted for.

Two events are mutually exclusive when their circles do not overlap at all — $P(A \cap B) = 0$ — which is a different question from whether they are independent. Independence compares $P(A \cap B)$ against $P(A) \times P(B)$ ; mutual exclusivity is a statement about the diagram's shape. Treating "the circles touch" as evidence about independence confuses the two: circles touching is only ever evidence about $P(A \cap B)$ itself, and that value still has to be checked against $P(A) \times P(B)$ before independence can be claimed.

Marked, line by line

A survey of 50 students in a sports club asked whether each plays badminton (event $A$ ) or netball (event $B$ ). 15 play badminton only, 15 play netball only, 10 play both, and the rest play neither. (a) Find $P(A \cap B)$ . (2) (b) Find $P(A \mid B)$ . (2) (c) Determine, with a reason, whether $A$ and $B$ are independent. (3) — VERIDIAN-original question. The scenario, and the 15/15/10 split it is built from, are original; the mark-scheme conventions attached to each line (M for method, A for accuracy dependent on M, B for an independent mark, and the 'oe' convention allowing an equivalent fraction, decimal or percentage — verified live on a real WST01 Venn-diagram-and-tree-diagram question, Oct 2022 MS, where a mark point is written '= 40/117 oe') are drawn from WST01-verified-facts.md §4, not invented.

7 marks available

(a) — 2 marks

01
50 students in total: 15 badminton only, 15 netball only, 10 both — so the "both" region gives $P(A \cap B)$ directly, with no calculation needed. Check the fourth region against the total anyway: $15+15+10 = 40$ , leaving $10$ students who play neither, and $15+15+10+10 = 50$ — the whole club accounted for, the same "everything has to add up" rule a tree diagram's branches obey.
Independent mark for correctly identifying that the intersection region is read straight off the diagram. Does not depend on any other part of the question.
B1
02
$P(A \cap B) = \frac{10}{50} = \frac{1}{5}$
oe — accept an equivalent fraction, decimal (0.2) or percentage (20%). The 'oe' convention is verified in live use on a real WST01 Venn-and-tree question, allowing exactly this kind of equivalent-form answer.
A1

(b) — 2 marks

101
$P(B) = \frac{15 + 10}{50} = \frac{25}{50} = \frac{1}{2}$
Method mark for finding P(B) from the diagram as (B only) + (both), over the total — the quantity a conditional probability restricts attention to.
M1
102
$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/5}{1/2} = \frac{2}{5}$
oe. Dependent on the method mark above: the conditional-probability formula applied correctly to the two values just found.
A1

201
$P(A) = \frac{15+10}{50} = \frac{1}{2}$ , so $P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ .
Method mark for computing P(A)×P(B) — the actual comparison the independence test needs, not a guess read off the picture.
M1
202
Compare with $P(A \cap B) = \frac{1}{5}$ from part (a): $\frac{1}{4} \neq \frac{1}{5}$ .
Correct comparison, both values stated. This is the line real Venn-diagram conditional-probability questions are documented to lose most often — the comparison has to be made explicitly, not just implied.
A1
203
Since $P(A)P(B) \neq P(A \cap B)$ , badminton and netball are NOT independent.
Correct conclusion, drawn from the comparison in the line above rather than asserted. cao — no credit for the right conclusion reached by the wrong route (e.g. "not independent because the circles overlap" — see the common wrong path).
A1

In your own words

In one sentence: why does the region outside both circles on a Venn diagram still need a value written in it, even when that value turns out to be zero?

Named traps

original-fraction-reused-after-removal: Confirmed directly on a real without-replacement counters question: "common errors seen were to repeat the 4/8 (given in the question) following the 1st was green onto the branches following the 1st was blue" (Jan 2023, Q2). The mechanism is not carelessness with arithmetic — the fraction $\frac{4}{8}$ is not wrong in itself, it is simply the wrong bag's fraction, carried onto branches that describe a bag with one fewer counter in it. The fix is procedural, not conceptual: rebuild the composition of what remains before writing a single second-draw fraction, every time.
second-draw-branches-swapped: The research bank's own account of the same Jan 2023, Q2 finding (its documented summary of the examiner report, not a direct quotation) is that a reversed-order error was also seen on the second-branch probabilities, alongside the reused-fraction error above. No specific wrong values survive in the research pass, so none are reproduced here. The mechanism is nonetheless real and worth teaching: the bag can be correctly reduced and the right pair of fractions correctly derived, and still end up written onto the wrong colour's branch, as if the reducing had been done right and then mis-filed. A tree diagram rebuilt with the right numbers can still be wrong if those numbers are not checked against which branch they actually belong to.
denominator-kept-constant-across-repeated-draws: Confirmed on a real without-replacement counting question spanning four draws: "the most common response was scoring 1 mark for the special case using (62/88)⁴" (Jun 2022, Q3(d)) — raising a single fraction to the fourth power, which keeps the denominator fixed at 88 the whole way through, exactly as though every counter were replaced before the next draw. Without replacement, a bag that loses one counter per draw needs a different denominator — and often a different numerator — at every single stage; four genuine draws without replacement need four genuinely different fractions multiplied together, never the same one raised to a power.
venn-blank-region-assumed-zero: Confirmed directly on a real Venn-diagram question: "some candidates left out the 0 in the outside region of N and should be reminded that blank spaces are not assumed to be 0s in Venn diagrams" (Jun 2022, Q4(c)). Even when the correct value for a region genuinely is zero, it has to be written there — an empty space on the diagram reads as unfinished working, not as an answer.
independence-assumed-instead-of-tested: A real examiner report's own general, paper-wide comment states plainly: "Candidates often assume independence when an appropriate conditional probability should be used instead" — named as a pattern across the whole paper, not tied to one question. The same series' report names the concrete version of it on a Venn-diagram question: "common errors were to assume independence" (Jun 2022, Q4(b)). Independence is a conclusion a calculation reaches — comparing $P(A \cap B)$ with $P(A) \times P(B)$ — never a shortcut taken on the way to one; see the common wrong path attached to the marked Venn-diagram solution above for exactly what this costs.
conditioning-denominator-folded-into-numerator: Confirmed directly on the real examiner report for a tree-diagram conditional-probability question — the part (d) that follows on from a "P(same colour)"-style part (c), exactly the shape of stage 6 above (Jan 2023, Q2(d)): the two most common wrong answers BOTH used the correct conditioning denominator, carried over from part (c) as $\frac{61}{234}$ — the denominator was never the problem. Each wrong answer instead folded an extra, spurious copy of that same $\frac{61}{234}$ into the numerator as well, either added in, $\left(\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13} + \frac{61}{234}\right) \div \frac{61}{234}$ , or multiplied in, $\left(\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13} \times \frac{61}{234}\right) \div \frac{61}{234}$ , instead of leaving the numerator as the single clean path probability it should have been, $\frac{5}{9}\times\frac{4}{8}\times\frac{8}{13} = \frac{20}{117}$ . The fix is procedural, the same one stage 6 above models: write $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$ out with its numerator and denominator settled as two separate, already-known numbers first, and substitute only once — a denominator that has just been used once should not reappear a second time inside the line above it.

Retrieval — with feedback on every choice

Question 1

3 marks

A bag contains 7 counters: 4 orange and 3 purple. Two counters are drawn at random, one after another, without replacement. What is $P(\text{first orange, then purple})$ ?

Question 2

3 marks

On a Venn diagram, $P(C) = 0.6$ , $P(D) = 0.5$ and $P(C \cap D) = 0.2$ . What is $P(C \mid D)$ ?

Question 3

2 marks

A Venn diagram for a class of 60 students and events $A$ (studies French) and $B$ (studies Spanish) shows 18 students in $A$ only, 22 in $B$ only, and 12 in both. How many students are outside both circles, and how should that be shown on the diagram?

Question 4

3 marks

A bag contains 5 counters: 2 red and 3 green. Two counters are drawn at random, WITH replacement. What is $P(\text{both red})$ ?

Question 5

3 marks

A spinner is divided into 5 equal sections: 3 blue and 2 yellow. It is spun twice. What is $P(\text{exactly one yellow})$ ?

Question 6

3 marks

A bag contains 8 counters: 5 green and 3 blue. Three counters are drawn at random, one after another, without replacement. What is $P(\text{all three the same colour})$ ?

Reference — not a study method, a lookup

Product law along a path: multiply the branches. P(A∩B) = P(A) × P(B|A).
Sum law across paths: add complete, mutually exclusive paths. Branches from one node always sum to 1.
Without replacement: total AND the count just removed both fall, every single draw. With replacement: nothing changes between draws.
Venn diagram: 4 regions (A only, B only, both, neither) sum to 1 (or to the total). Never leave a region blank — write 0 if it is 0.
P(A|B) = P(A∩B) / P(B). Overlap ⇒ not mutually exclusive. Independence is P(A)P(B) = P(A∩B) — a separate check, never assumed.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation attributed to an examiner report or mark scheme in this lesson — the reused-fraction wording (Jan 2023, Q2), the fixed-denominator error (Jun 2022, Q3(d)), the Venn-diagram findings (Jun 2022, Q4(b) and Q4(c)), the 'oe' mark-scheme convention (Oct 2022 MS), the two-container 3rd-branch mark-scheme note (Jan 2023 MS, Q2(a)), and the folded-numerator conditional-probability error (Jan 2023 examiner report, Q2(d)) — was independently verified against the primary Pearson document and is reproduced verbatim, with its series and question number stated alongside it. One further trap-taxonomy finding (the second-draw-branches-swapped item, also Jan 2023, Q2) is present in the research bank only as the facts-bank author's own documented summary of the examiner report rather than a directly quoted sentence, and is presented that way here rather than dressed as a verbatim quotation. Every actual QUESTION in this lesson — all three prequestions, the counters worked chain (including its stage-6 conditional-probability extension), the two-container transfer worked chain, the pens chain-drill, the badminton/netball marked solution, and every MCQ — is VERIDIAN-original: none of them reproduces a real Pearson question, and no full original numeric scenario for this topic was available in the research bank to reproduce even if that were the goal. What is real, in every case, is the specific wrong-answer mechanism the original numbers are built to demonstrate. The mark-scheme conventions attached to these original questions (M/A/B, cao, oe) are modelled on WST01's own verified general marking guidance, not transcribed from a real mark scheme for a question that does not exist.

Question 13 marks

A bag contains 7 counters: 4 orange and 3 purple. Two counters are drawn at random, one after another, without replacement. What is $P(\text{first orange, then purple})$ ?

$\frac{2}{7}$
Correct. $P(O_1) = \frac{4}{7}$ , and once an orange counter is gone, 3 purple remain out of 6 total: $P(P_2 \mid O_1) = \frac{3}{6}$ . Multiplying along the path: $\frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7}$ .
B $\frac{12}{49}$
This is $\frac{4}{7} \times \frac{3}{7}$ — the second fraction's denominator was left at 7, as though the first counter had been put back. This is exactly the mechanism a real examiner report documents on a without-replacement question spanning several draws: keeping the denominator fixed instead of letting it shrink by one at every draw.
C $\frac{4}{21}$
This computes $\frac{4}{7} \times \frac{2}{6}$ , reducing the purple count from 3 to 2 as though a purple counter had been the one removed. It was an orange counter that came out first — the purple count stays at 3, and it is the total and the orange count that fall.
D $\frac{1}{2}$
This is only the second branch's own probability, $P(P_2 \mid O_1)$ , on its own. A path through a tree diagram is a sequence of two events happening one after the other, and the product law says its probability is the product of both branches, not just the last one.

Traps tested: Denominator kept constant across repeated draws · Wrong colours count reduced · Second branch alone not multiplied

Question 23 marks

On a Venn diagram, $P(C) = 0.6$ , $P(D) = 0.5$ and $P(C \cap D) = 0.2$ . What is $P(C \mid D)$ ?

$0.4$
Correct. $P(C \mid D) = \frac{P(C \cap D)}{P(D)} = \frac{0.2}{0.5} = 0.4$ — restrict attention to the $D$ circle (denominator 0.5) and ask what share of it is also inside $C$ (numerator 0.2).
B $0.6$
This assumes $C$ and $D$ are independent and uses $P(C \mid D) = P(C)$ without checking — and here the assumption is wrong: $P(C) \times P(D) = 0.6 \times 0.5 = 0.3$ , which does not equal $P(C \cap D) = 0.2$ , so they are not independent. A real examiner report names exactly this shortcut as a pattern across the whole paper, not a one-off: candidates assume independence "when an appropriate conditional probability should be used instead."
C $\frac{1}{3}$
This divides the other way round, $\frac{P(C \cap D)}{P(C)} = \frac{0.2}{0.6}$ , which gives $P(D \mid C)$ instead of $P(C \mid D)$ . The event named after the vertical bar is the one to divide by.
D $0.7$
This adds $P(C \cap D) + P(D) = 0.2 + 0.5$ . A conditional probability is a ratio — what fraction of $D$ 's probability also lies in $C$ — not a sum of the two given values.

Traps tested: Independence assumed instead of tested · Conditioned on the wrong event · Added instead of divided

Question 32 marks

8, written explicitly in the region outside both circles.
Correct. $18 + 22 + 12 = 52$ , leaving $60 - 52 = 8$ students who study neither — and that 8 has to be written on the diagram itself, in the region outside both circles, not left blank.
B0, so nothing needs to be written there.
The arithmetic gives 8, not 0 — three named regions here only account for 52 of the 60 students. More generally, even a genuinely empty region has to carry a written 0: a real examiner report specifically records candidates losing marks for leaving it blank, stating plainly that blank spaces are not assumed to be 0s in Venn diagrams.
C8, but it doesn't need to be shown since it can be worked out from the other three.
The value 8 is correct, but the same examiner-report finding applies regardless of whether it CAN be worked out — the diagram is marked on what it shows, not on what a reader could reconstruct from it.
D42, since only the overlap should be subtracted from the total.
This is $60 - 18$ , subtracting only the $A$ -only region and leaving out both the $B$ -only region and the overlap. All three named regions — $A$ only, $B$ only, and both — have to come off the total, not just one of them.

Traps tested: Venn blank region assumed zero · Region value not required on diagram · Only one region subtracted

Question 43 marks

A bag contains 5 counters: 2 red and 3 green. Two counters are drawn at random, WITH replacement. What is $P(\text{both red})$ ?

$\frac{4}{25}$
Correct. With replacement, the bag is identical for both draws: $P(R_1) = \frac{2}{5}$ and $P(R_2) = \frac{2}{5}$ again, so $P(\text{both red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$ .
B $\frac{1}{10}$
This is $\frac{2}{5} \times \frac{1}{4}$ — the without-replacement answer, reducing both the red count and the total for the second draw. The question states the counter is replaced, so the second draw sees exactly the same bag as the first.
C $\frac{2}{5}$
This is only the probability of the first counter being red. "Both red" is two events happening in sequence, and the product law needs both branches multiplied, not just the first one stated on its own.
D $\frac{1}{5}$
This is $\frac{2}{5} \times \frac{2}{4}$ — reducing the total for the second draw as though a counter had been removed, while the question says it was replaced. With replacement, nothing about the bag changes between draws, so the second fraction should be identical to the first, not partly adjusted.

Traps tested: With replacement treated as without · Second draw not considered · Denominator reduced numerator not

Question 53 marks

A spinner is divided into 5 equal sections: 3 blue and 2 yellow. It is spun twice. What is $P(\text{exactly one yellow})$ ?

$\frac{12}{25}$
Correct. Two different paths give exactly one yellow — blue-then-yellow and yellow-then-blue — and a spinner resets itself every spin, so both spins use the same fractions, $P(\text{blue})=\frac{3}{5}$ and $P(\text{yellow})=\frac{2}{5}$ . Each path: $\frac{3}{5}\times\frac{2}{5} = \frac{6}{25}$ . Both paths count as "exactly one yellow" and cannot both happen on the same two spins, so the sum law adds them: $\frac{6}{25}+\frac{6}{25}=\frac{12}{25}$ .
B $\frac{6}{25}$
This is only one of the two paths that give exactly one yellow — blue-then-yellow, say — leaving out yellow-then-blue. Both orders count as "exactly one yellow", and the sum law needs both added, not just one.
C $\frac{4}{25}$
This is $P(\text{both yellow}) = \frac{2}{5}\times\frac{2}{5}$ , a different event from "exactly one yellow". Both yellow is one single path; exactly one yellow is two.
D $\frac{3}{5}$
This is just $P(\text{blue})$ on a single spin, with no second spin considered at all. The question asks about two spins together.

Traps tested: Only one order counted · Wrong event computed · Single spin probability given

Question 63 marks

A bag contains 8 counters: 5 green and 3 blue. Three counters are drawn at random, one after another, without replacement. What is $P(\text{all three the same colour})$ ?

$\frac{11}{56}$
Correct. Two different paths give "all three the same colour" — green-green-green, or blue-blue-blue — and each path is a product of three sequentially-conditioned fractions, not two: $P(GGG) = \frac{5}{8}\times\frac{4}{7}\times\frac{3}{6} = \frac{5}{28}$ , and $P(BBB) = \frac{3}{8}\times\frac{2}{7}\times\frac{1}{6} = \frac{1}{56}$ . The two paths cannot both happen on the same three draws, so the sum law adds them: $\frac{5}{28}+\frac{1}{56} = \frac{10}{56}+\frac{1}{56} = \frac{11}{56}$ . This is exactly the extend-the-product-law-then-sum-two-paths skill a real WST01 tree-diagram question needs (Jan 2023, Q2(c)'s mark scheme rewards precisely a three-term product along each of two paths, added together) — just worked here on one bag instead of two.
B $\frac{13}{28}$
This stops the product law one draw too early — it is $P(GG)+P(BB)$ from only the first two draws, $\frac{5}{8}\times\frac{4}{7} + \frac{3}{8}\times\frac{2}{7} = \frac{5}{14}+\frac{3}{28} = \frac{13}{28}$ , and never accounts for the third draw at all. "All three the same colour" needs three fractions multiplied along each path, not two — the product law extends as far as the sequence of draws actually goes, not just to however many terms feel natural to stop at.
C $\frac{33}{256}$
This keeps the denominator fixed at 8 for all three draws — $\frac{5}{8}\times\frac{4}{8}\times\frac{3}{8} + \frac{3}{8}\times\frac{2}{8}\times\frac{1}{8} = \frac{15}{128}+\frac{3}{256} = \frac{33}{256}$ — exactly the mechanism a real examiner report documents on a without-replacement question spanning several draws: treating the bag as though every counter gets replaced before the next draw, instead of a genuinely smaller bag at every single stage.
D $\frac{5}{28}$
This is $P(GGG)$ alone, correctly extended to three terms — but it stops after the green path and never adds the blue path, $P(BBB) = \frac{1}{56}$ . "All three the same colour" can happen two different ways, and the sum law needs both paths added, not just the more likely one.

Traps tested: Product law stopped at two terms · Denominator kept constant across repeated draws · Second colour path not summed

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2023 · Q2 — cited directly in this lesson
Mark scheme: Jan 2023 · Q2(a) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WST01.

Statistics 1 · progress saved in this browser · sign in to sync across devices

Up next

Correlation coefficient and regression — the calculation mechanics

This is the thinnest evidence base of any WST01 lesson in this batch, and it's worth saying plainly rather than dressing it up. Where the regression-interpretation lesson draws on five separate examiner reports, spec item 4.3 — the product moment correlation coefficient itself — is anchored by exactly two sources: one examiner report and one mark scheme. What those two sources hand over is precise and genuinely useful all the same: a real, evidenced trap on the r-calculation itself — an omitted square root in the denominator — a well-documented Pearson accuracy discipline for 'show that' questions this lesson extends to r's own square-root division, and a clean, quotable coding-invariance property Pearson credits outright. This lesson assumes the Sxx/Sxy machinery already taught elsewhere and adds exactly what that machinery didn't need — a third summary sum, and the coefficient it unlocks.

40 min

Sampling with/without replacement, tree diagrams, and Venn diagrams

Sampling with and without replacement — what changes between draws

Why branches multiply and paths add — a tree diagram is the product and sum laws, drawn

The bag of counters: is the second draw the same colour as the first?

Complete the chain — a box of 9 pens, 6 blue and 3 red, two taken without replacement: what is P(both blue)P(\text{both blue})P(both blue)?

Two containers, not one: transferring counters from P into Q before drawing again

Venn diagrams: reading regions, and the rule about what "blank" means

Deep dive

Practice this for real

Complete the chain — a box of 9 pens, 6 blue and 3 red, two taken without replacement: what is $P(\text{both blue})$ ?