Conditional probability, and independence vs. mutually exclusive

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.

Sample space, mutually exclusive events, and where the conditional probability formula actually comes from

The sample space $\Omega$ is the set of every possible outcome of an experiment, and an event is a subset of it — 'the die shows an even number' is the subset $\{2, 4, 6\}$ of $\Omega = \{1,2,3,4,5,6\}$ . Two events $A$ and $B$ are mutually exclusive (spec 3.2) when they share no outcomes at all: $A \cap B = \emptyset$ , and therefore $P(A \cap B) = 0$ . This is a statement about the *sets themselves* — whether the two subsets of $\Omega$ overlap — decided entirely by what the events mean, before any probability is attached to them.

One immediate consequence, and a genuinely useful shortcut: for mutually exclusive events, the general addition law $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ simplifies to $P(A \cup B) = P(A) + P(B)$ , because the overlap term is already zero. The general form is worth knowing precisely because it doesn't assume mutual exclusivity — it's the version that's always true, with the $-P(A \cap B)$ term correcting for double-counting the overlap when there is one. The complement rule $P(A') = 1 - P(A)$ is the other basic tool here: $A$ and $A'$ partition the whole sample space, so between them they always account for probability 1.

Now for the memorise-vs-look-up split, checked directly against the real WST01 formula booklet (WST01-verified-facts.md §2a): $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ is printed on the formula sheet, along with the general multiplication law $P(A \cap B) = P(A)P(B \mid A)$ . The complement rule $P(A') = 1 - P(A)$ is not — the spec's own notation box states outright that formulae 'expected to know' are not repeated in the booklet, and this is one of them. It costs nothing to memorise (it's one line) but it is genuinely not there to fall back on in the exam.

Conditional probability — $P(A \mid B)$ , 'the probability of $A$ , given that $B$ has happened' — is not a separate rule to memorise so much as a rearrangement of the multiplication law that already IS on the formula sheet. That law states $P(A \cap B) = P(A)P(B \mid A)$ ; since $A \cap B$ is the same event as $B \cap A$ , the identical law also reads $P(A \cap B) = P(B)P(A \mid B)$ . Divide both sides by $P(B)$ (which must be strictly positive — conditioning on an impossible event is undefined) and the conditional probability formula falls straight out: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ . The intuition matches the algebra exactly: conditioning on $B$ shrinks the sample space down to $B$ itself, and $P(A \mid B)$ asks what proportion of that shrunk space is also in $A$ — which is exactly $\frac{P(A \cap B)}{P(B)}$ , the overlap as a fraction of the new, smaller whole.

Independence — a different kind of question entirely, and one you have to calculate

Independence (spec 3.3) asks something conditional probability makes precise: does knowing $B$ happened change the probability of $A$ ? If it doesn't — if $P(A \mid B) = P(A)$ exactly — then $A$ and $B$ are independent. Two other forms of the same statement follow immediately and are equally valid starting points: $P(B \mid A) = P(B)$ (the same question asked the other way round), and, substituting $P(A \mid B) = P(A)$ into the conditional probability formula derived above, $P(A) = \frac{P(A \cap B)}{P(B)}$ , which rearranges to $P(A \cap B) = P(A) \times P(B)$ . All three are checked directly against the real formula booklet in WST01-verified-facts.md §2a as things the spec expects memorised, not looked up — the booklet carries only the general multiplication law $P(A \cap B) = P(A)P(B \mid A)$ , not its independent-events special case.

$P(A \cap B) = P(A) \times P(B)$ is the version worth defaulting to on the exam, because it's the only one of the three that needs no conditional probability computed first — plug in the two marginal probabilities, multiply, and compare to the given (or computed) $P(A \cap B)$ .

The dice example above is worth re-reading as a warning rather than a curiosity. $A$ = 'first die is 6' and $B$ = 'the total is 7' turned out independent — $P(A \cap B) = \frac{1}{36} = P(A) \times P(B)$ — despite $B$ being built directly out of $A$ 's own die. Nothing about how the events were defined made that obvious in advance; only running the actual calculation settled it. This is precisely the discipline WST01's own examiner reports repeatedly find missing: the general comment for one series states plainly that 'candidates often assume independence when an appropriate conditional probability should be used instead' (Jun 2022, general) — independence is being treated as a default assumption about a situation, when it is really a specific numerical claim that has to be earned by a calculation every single time.

Mechanism

Why mutually exclusive events (with nonzero probability) can never be independent — and why 'they overlap' isn't the whole story either

Run the two definitions against each other directly. If $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$ — that's what mutually exclusive means. If $A$ and $B$ are independent, $P(A \cap B) = P(A) \times P(B)$ — that's what independent means. If both held at once, $P(A) \times P(B)$ would have to equal $0$ . But a product of two numbers is zero only if at least one of them is zero — and if $P(A) > 0$ and $P(B) > 0$ , that's impossible. So for any two events that both have positive probability, mutually exclusive and independent are not just different — they are mutually incompatible: proving one is true is a proof the other is false. This is the precise version of the pattern an examiner report names directly, calling it chronic rather than occasional: 'there was the usual confusion between events being independent and events being mutually exclusive highlighted by statements such as "they are not independent as they overlap"' (Oct 2021, Q1(b)) — the word 'usual' there is Pearson's own, describing a recurring pattern rather than a one-off. But read that quoted student statement carefully, because the mechanism above actually cuts partway in the student's favour: NOT overlapping (being mutually exclusive) genuinely does rule out independence, whenever both probabilities are positive — that half of the reasoning is sound. Where it breaks is the unstated assumption running the other way: that overlapping is enough, on its own, to conclude independence. It isn't. Overlapping is necessary for independence but nowhere near sufficient, and the dice example shows exactly how a small change in the numbers exposes the gap: let $B'$ = 'the two dice sum to 8' instead of 7. $A$ = 'first die is 6' and $B'$ certainly overlap — 6-then-2 is a valid outcome in both. $P(B') = \frac{5}{36}$ (five ways to make 8: 2+6, 3+5, 4+4, 5+3, 6+2), and $P(A \cap B') = \frac{1}{36}$ (only 6-then-2). Check independence: $P(A) \times P(B') = \frac{1}{6} \times \frac{5}{36} = \frac{5}{216}$ , but $P(A \cap B') = \frac{1}{36} = \frac{6}{216}$ . $\frac{6}{216} \neq \frac{5}{216}$ — not independent, despite genuinely overlapping. So the full picture has three regions, not two: mutually exclusive events (with both probabilities positive) are never independent; independent events are never mutually exclusive; but 'overlaps and isn't mutually exclusive' is a large middle ground that still needs the actual product-rule check, because it contains both independent pairs (first-die-6 and sum-is-7) and dependent ones (first-die-6 and sum-is-8) side by side, indistinguishable without calculating.

Worked, in full

A bag holds 5 red and 3 blue counters. Three are drawn, one at a time, without replacement. Given that the second counter drawn is blue, find the probability the draw sequence was exactly red, blue, red.

01
Name the events precisely before touching any numbers. Let $C$ = 'the second counter drawn is blue' (the given, conditioning event) and $D$ = 'the sequence drawn, in order, is red, blue, red'. The question wants $P(D \mid C) = \dfrac{P(D \cap C)}{P(C)}$ . Notice that $D$ 's own second draw IS blue — so $D$ automatically satisfies $C$ , meaning $D$ is entirely contained inside $C$ , and $P(D \cap C) = P(D)$ . That observation doesn't skip any working; it just tells you which two quantities to go and find: $P(D)$ and $P(C)$ separately.
Earns: M1 — correct conditional probability structure with $D$ and $C$ correctly identified, and the containment $D \subseteq C$ correctly justified rather than assumed. Writing $P(D)/P(C)$ with no justification for dropping the $\cap$ would not earn this: the simplification has to be shown to be valid, not just used.
02
Find $P(D)$ using a tree diagram, one branch per draw, remembering the bag shrinks with each draw and is not replaced. $P(\text{1st red}) = \frac{5}{8}$ . Given the first was red, 4 red and 3 blue remain (7 total): $P(\text{2nd blue} \mid \text{1st red}) = \frac{3}{7}$ . Given red-then-blue, 4 red and 2 blue remain (6 total): $P(\text{3rd red} \mid \text{1st red, 2nd blue}) = \frac{4}{6}$ . All three branches multiply: $P(D) = \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} = \frac{60}{336} = \frac{5}{28}$ .
Earns: M1 — multiplies all THREE branch probabilities along the one path that gives red-blue-red. This is the exact step a real WST01 examiner report flags as the most common failure on a structurally identical three-branch conditional-probability question: 'the most common error was using a product of 2 probabilities rather than 3 in the numerator' (Oct 2021, Q4(c)) — forgetting one branch's contribution, not making an arithmetic slip within a branch.
03
Find $P(C) = P(\text{2nd draw is blue})$ — and notice this needs BOTH possible colours for the first draw, since the second draw's probability depends on what the first draw removed. First red then second blue: $\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$ . First blue then second blue: $\frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$ . Sum the two disjoint cases: $P(C) = \frac{15}{56} + \frac{6}{56} = \frac{21}{56} = \frac{3}{8}$ .
Earns: M1 — sums both first-draw cases that lead to a blue second draw, rather than computing only one of them. A1 — correct value $\frac{3}{8}$ , which is worth checking against a genuinely useful symmetry fact: for sampling without replacement, the marginal probability that the $k$ -th draw is any given colour equals the ORIGINAL proportion of that colour in the bag — here $\frac{3}{8}$ , identical to $P(\text{1st is blue})$ . That symmetry doesn't replace the casework (a mark scheme would still want the working shown), but it's a fast way to sanity-check the answer once it's found.
04
Combine the two results: $P(D \mid C) = \dfrac{P(D)}{P(C)} = \dfrac{5/28}{3/8} = \dfrac{5}{28} \times \dfrac{8}{3} = \dfrac{40}{84} = \dfrac{10}{21}$ .
Earns: A1 — correct final value $\frac{10}{21}$ (or an equivalent form, e.g. awrt $0.476$ ), correctly simplified.
05
Sanity-check before moving on. $\frac{10}{21} \approx 0.476$ is a valid probability (between 0 and 1) — a good first filter, since dividing by the wrong denominator or dropping a branch factor tends to produce values outside $[0,1]$ or suspiciously close to 1. It's also larger than the unconditional $P(D) = \frac{5}{28} \approx 0.179$ , which makes sense: knowing the second counter WAS blue removes all the sequences where it wasn't, concentrating the probability onto the sequences — like red-blue-red — that remain possible.
Earns: Nothing on the mark scheme — the working above has already earned every available mark. It's included because this exact check (is the conditional probability bigger than the unconditional one, and does that direction make sense?) is what catches a correct-looking fraction that's actually wrong, before the paper is handed in.

Source — Examiner report, Oct 2021

"the most common error was using a product of 2 probabilities rather than 3 in the numerator"

Complete it yourself

Complete the chain — a bag holds 6 red and 4 blue counters. Three are drawn without replacement. Given the second draw is red, find P(the sequence was blue, red, blue).

01
Let $C$ = 'the second counter drawn is red' and $D$ = 'the sequence drawn, in order, is blue, red, blue'. $D$ 's own second draw is red, so $D \subseteq C$ , and $P(D \cap C) = P(D)$ . The target is $P(D \mid C) = \dfrac{P(D)}{P(C)}$ .
02
That splits the problem into two separate calculations: find $P(D)$ (a single three-branch path through the tree — blue, then red, then blue) and $P(C)$ (summing both first-draw cases that lead to a red second draw), then divide.

Worked, in full

A Venn diagram of three events $X$ , $Y$ , $Z$ has two unknown regions, $x = P(X \cap Y \cap Z)$ and $y = P(X \cap Y \text{ only})$ . Given that $P(Z \mid Y) = P(Z)$ , find $x$ and $y$ .

01
A three-event Venn diagram has eight regions in total; six are named here (four given numbers, two unknowns) and the remaining two — ' $Y \cap Z$ only' and ' $Z$ only' — are both empty. The six named regions: ' $X$ only' $= 0.05$ , ' $X \cap Z$ only' $= 0.08$ , ' $X \cap Y \cap Z$ ' $= x$ , ' $X \cap Y$ only' $= y$ , ' $Y$ only' $= 0.25$ , 'outside all three' $= 0.37$ . This is the companion algebraic-equation skill to the same real anchor question already named in the trap-taxonomy above for its part (b) (Oct 2021, Q1) — this worked chain mirrors part (c) of that same question, on VERIDIAN-original regions rather than Pearson's own. Since the probabilities of all eight regions must sum to $1$ : $0.05 + 0.08 + x + y + 0.25 + 0.37 = 1$ , so $x + y = 0.25$ — a relationship between the two unknowns that comes from total probability alone, not from any condition given about the events. It isn't enough on its own to find either unknown, but stage 4 below will need it. First, find $P(Y)$ the way any part (a) would: sum every region inside $Y$ . $P(Y) = 0.25 + y + x + 0 = 0.25 + (x + y) = 0.25 + 0.25 = 0.5$ — found from the sum alone, without knowing $x$ and $y$ individually.
Earns: Groundwork, not itself a scored line for the equation below — the same 'read the diagram, use total probability = 1' step that would earn the marks for a real part (a) 'find P(Y)' question, exactly as the anchor's own part (a) finds P(B) = 0.5 from p + q = 0.2 before part (c) needs that same relationship again.
02
The question's condition is $P(Z \mid Y) = P(Z)$ . Read both sides off the diagram using the conditional probability formula, rather than assuming what the condition means. $Y \cap Z$ consists only of the region ' $X \cap Y \cap Z$ ' (since ' $Y \cap Z$ only' is empty), so $P(Y \cap Z) = x$ and $P(Z \mid Y) = \dfrac{P(Y \cap Z)}{P(Y)} = \dfrac{x}{0.5}$ . $Z$ itself consists of ' $X \cap Z$ only' and ' $X \cap Y \cap Z$ ' (since ' $Z$ only' is also empty), so $P(Z) = 0.08 + x$ . Substituting both sides into the given condition: $\dfrac{x}{0.5} = 0.08 + x$ — an equation in $x$ , with $x$ appearing on both sides.
Earns: M1 — a correct equation in $x$ based on the given statement, with the unknown correctly appearing on both sides after both $P(Y \cap Z)$ and $P(Z)$ are read off the diagram and the conditional probability formula is applied, not asserted. The real anchor mark scheme rewards exactly this shape of equation, on its own regions and its own unknown: "p / 0.5 = p + 0.06 (o.e.)" (WST01 Oct 2021 Q1(c) mark scheme) — the identical move, condition-into-equation, applied here to a different diagram.
03
Solve for $x$ : $\dfrac{x}{0.5} = 0.08 + x \Rightarrow x = 0.5(0.08 + x) = 0.04 + 0.5x \Rightarrow x - 0.5x = 0.04 \Rightarrow 0.5x = 0.04 \Rightarrow x = 0.08$ .
Earns: A1 — correct value $x = 0.08$ , the first of the two unknowns, found by the identical rearrangement move the real anchor mark scheme itself credits on its own numbers: "2p = p + 0.06, so p = 0.06" (WST01 Oct 2021 Q1(c) mark scheme).
04
The condition on $Y$ and $Z$ is now used up — it found $x$ , but says nothing more about $y$ . Finding $y$ needs the OTHER relationship instead: $x + y = 0.25$ , from stage 1, which came from total probability equalling $1$ , not from the conditional-probability condition at all. Substituting $x = 0.08$ : $y = 0.25 - 0.08 = 0.17$ .
Earns: A1 — correct value $y = 0.17$ , earned by a genuinely different move from the one that found $x$ : falling back on a second, independent linear relationship between the two unknowns, rather than trying to re-apply the conditional-probability formula a second time. The real anchor mark scheme awards its second accuracy mark the identical way: "use of p + q = 0.2 gives q = 0.14" (WST01 Oct 2021 Q1(c) mark scheme). A solution that finds the first unknown correctly and then stalls, guesses, or repeats the conditional-probability calculation for the second one loses this mark even though the harder step is already done.
05
Check both values are valid probabilities ( $0 \le x, y \le 1$ — both are) and check the original condition genuinely holds: $P(Z) = 0.08 + 0.08 = 0.16$ , and $P(Z \mid Y) = x / 0.5 = 0.08 / 0.5 = 0.16$ — equal, exactly as the condition required.
Earns: Nothing on the mark scheme — both marks for this part are already earned by stage 4. Included because checking the condition actually holds, using the two values just found, is the fastest way to catch a slip in either $x$ or $y$ before it's handed in.

Source — Mark scheme, Oct 2021

"p / 0.5 = p + 0.06 (o.e.)"

Marked, line by line

In a survey of 200 university applicants, 90 applied to study Mathematics (event $M$ ) and 70 applied to study Physics (event $P$ ); 30 applied to study both subjects. (a) Find $P(M \cup P)$ . (2) (b) Find $P(M \mid P)$ , giving your answer as a fraction in its simplest form. (2) (c) Determine, showing your working, whether $M$ and $P$ are independent. (3) (d) Determine, showing your working, whether $M$ and $P$ are mutually exclusive. (1) (e) Using your answers to parts (c) and (d), explain why $M$ and $P$ could never have turned out to be both mutually exclusive and independent. (2) — VERIDIAN-original question, built to combine spec items 3.2 and 3.3 in a single multi-part item the way real WST01 questions combine sub-skills across a Section 3 question. Not a reproduction of any past-paper question; the per-line mark allocations are modelled on the verified WST01 mark-scheme conventions in WST01-verified-facts.md §4 (M/A/B mark meanings, dependency rules, 'oe'), not transcribed from a real scheme, which does not exist for an original question.

10 marks available

(a) — 2 marks

01
$P(M) = \dfrac{90}{200} = 0.45$ , $P(P) = \dfrac{70}{200} = 0.35$ , $P(M \cap P) = \dfrac{30}{200} = 0.15$ . Using the addition law: $P(M \cup P) = P(M) + P(P) - P(M \cap P) = 0.45 + 0.35 - 0.15$ .
Method mark for the addition law with all three probabilities correctly substituted. This formula is printed on the WST01 formula sheet (verified against the real booklet), so the mark is for correct substitution, not for recalling the formula from memory.
M1
02
$P(M \cup P) = 0.65$
Accuracy mark, correct answer only. Accept oe: 13/20 or 65%. A candidate who instead added P(M) and P(P) without subtracting the overlap would reach 0.80, double-counting the 30 applicants who applied to both.
A1

(b) — 2 marks

101
$P(M \mid P) = \dfrac{P(M \cap P)}{P(P)} = \dfrac{0.15}{0.35}$
Method mark for the correct conditional probability structure — intersection over the CONDITIONING event's own probability, P(P), not P(M) and not the unconditioned P(M∩P) alone.
M1
102
$P(M \mid P) = \dfrac{15}{35} = \dfrac{3}{7}$
Accuracy mark, correct answer only, in simplest form as the question requires. Accept oe: awrt 0.429 or 42.9%.
A1

(c) — 3 marks

201
Independence requires $P(M \cap P) = P(M) \times P(P)$ . Compute the right-hand side: $P(M) \times P(P) = 0.45 \times 0.35$ .
Method mark for stating and attempting the correct independence test. A candidate who instead asserts independence or dependence without this calculation cannot access any mark in this part — there is no partial credit for a conclusion with no test behind it.
M1
202
$P(M) \times P(P) = 0.1575$
Accuracy mark for the correct product, dependent on the method mark above.
A1
203
$0.1575 \neq 0.15 = P(M \cap P)$ , so $M$ and $P$ are NOT independent.
Accuracy mark, correct answer only, for the explicit numerical comparison AND the correctly stated conclusion — both parts are needed; stating '0.1575' with no comparison to the known 0.15, or a bare 'not independent' with no supporting number, would not earn this on its own.
A1

(d) — 1 mark

301
$P(M \cap P) = 0.15 \neq 0$ , so $M$ and $P$ are NOT mutually exclusive.
Independent accuracy mark — no separate method needed, since P(M∩P) was already established in part (a). This is the kind of mark a B code exists for: a correct, standalone conclusion read straight from a value already found.
B1

(e) — 2 marks

401
If $M$ and $P$ HAD been mutually exclusive, $P(M \cap P)$ would have to be $0$ ; if they had ALSO been independent, $P(M \cap P)$ would have to equal $P(M) \times P(P) = 0.1575 \neq 0$ . Both cannot be true at once, since $P(M)$ and $P(P)$ are both nonzero.
Independent mark for correctly stating the general reason — a product of two positive probabilities can never equal zero — not just repeating the two separate answers from (c) and (d).
B1
402
In fact $M$ and $P$ turned out to be neither: not independent (part (c)) and not mutually exclusive (part (d)) — which is the far more common outcome for two real, overlapping events than either extreme.
Independent mark for correctly relating the general reason back to the specific numbers in this question, rather than leaving the explanation purely abstract. This part sits in the AO3/AO4 territory the spec itself weights unusually heavily for S1 (15–20 of 75 marks on AO3 alone, per WST01-verified-facts.md §2b) — a worded conclusion drawn from a calculation, not the calculation itself.
B1

In your own words

In one sentence: why can two events with positive probability never be both mutually exclusive and independent at the same time?

Named traps

independence-and-mutual-exclusivity-conflated: The chronic, named error across the whole archive: "there was the usual confusion between events being independent and events being mutually exclusive highlighted by statements such as 'they are not independent as they overlap'" (Oct 2021, Q1(b)). Pearson's own word "usual" marks this as a recurring pattern, not a single script's slip. The mechanism block above shows exactly where the reasoning goes wrong: NOT overlapping does rule out independence (when both probabilities are positive) — but overlapping on its own proves nothing about independence, which needs the actual P(A∩B) = P(A)P(B) check.
conditional-probability-as-raw-ratio-of-marginals: Verified directly, quoted from a real conditional-probability question: "we occasionally still saw P(0.35∩0.4)/0.4 or 0.35/0.4" (Jan 2021, Q1(c)) — dividing two given probabilities directly, as though P(A|B) meant P(A) ÷ P(B) rather than P(A∩B) ÷ P(B). A fast self-check: if the "conditional probability" you've computed by dividing two marginals exceeds 1, you've made exactly this error — a real probability never can.
independence-assumed-instead-of-tested: "Candidates often assume independence when an appropriate conditional probability should be used instead" (Jun 2022, general examiner comment — stated as a paper-wide diagnosis, not tied to one question). A second, more specific instance on a Venn-diagram conditional-probability question: "common errors were to assume independence or to do 1 − 1/15 before dividing by 3/8" (Jun 2022, Q4(b)) — independence substituted in as a shortcut for the actual conditioning calculation the question required.
tree-diagram-numerator-missing-a-branch: "the most common error was using a product of 2 probabilities rather than 3 in the numerator" (Oct 2021, Q4(c)), on a three-branch tree-diagram conditional probability question. The worked-chain example above is built around this exact trap: forgetting one branch of a multi-stage path produces a wrong but perfectly plausible-looking fraction, with no obvious sign anything went wrong.
extra-probability-folded-into-the-numerator: A second, independently confirmed instance of the numerator/denominator confusion (Jan 2023, Q2(d)) — corrected 2026-08-28 against the real examiner-report PDF, since an earlier reading of this citation had the mechanism backwards: the conditioning denominator itself (61/234, carried from an earlier part) was correct in both wrong answers candidates gave. The actual slip was in the numerator — an extra, spurious copy of that same 61/234 got folded in, either added or multiplied: (5/9 × 4/8 × 8/13 + 61/234)/(61/234) and (5/9 × 4/8 × 8/13 × 61/234)/(61/234), instead of the correct numerator 5/9 × 4/8 × 8/13 = 20/117. The lesson here: once a probability from an earlier part is sitting on the page, it's tempting to reuse it a second time inside a new calculation where it doesn't belong — write P(A|B) = P(A∩B)/P(B) out in full first, work out P(A∩B) as its own clean fraction, and only then substitute, so there's a formula on the page to check the substitution against.

Beyond the spec

WST01's own spec content list (WST01-verified-facts.md §1) never names 'Bayes' theorem' anywhere in items 3.1–3.4 — but the real formula booklet does print the full ratio-of-products form of it, in the S1 section, right next to the general multiplication and addition laws (WST01-verified-facts.md §2a). That's worth knowing precisely because it means recognising the shape of a Bayes'-theorem question is useful exam technique even though the name itself is never examined — the tool is sitting on the sheet whether or not you know what to call it.

The conditional probability formula reverses cleanly. Start from the two ways of writing $P(A \cap B)$ from the multiplication law: $P(A \cap B) = P(B)P(A \mid B) = P(A)P(B \mid A)$ . Set the middle and right expressions equal — $P(B)P(A \mid B) = P(A)P(B \mid A)$ — and divide through by $P(B)$ : $P(A \mid B) = \dfrac{P(A)P(B \mid A)}{P(B)}$ . This is Bayes' theorem: it lets you find $P(A \mid B)$ from $P(B \mid A)$ — the *reverse* conditional — which matters whenever the reverse direction is the one you can actually measure or are actually given. A concrete shape this takes on a WST01-style question: a factory's three machines produce known proportions of its output, each machine has a different known defect rate — that's $P(\text{machine})$ and $P(\text{defective} \mid \text{machine})$ , both easy to state — but the question asks the reverse, $P(\text{a specific machine} \mid \text{item is defective})$ , which needs Bayes' theorem to get at directly. The formula booklet's own version of this (verified in WST01-verified-facts.md §2a) is written as a ratio of two products rather than the two-line derivation above, but it is the identical statement — this derivation just shows where that printed ratio actually comes from, the same discipline the mechanism block above applies to the independence rules.

Retrieval — with feedback on every choice

Question 1

2 marks

Events $C$ and $D$ satisfy $P(C) = 0.6$ , $P(D) = 0.5$ , and $P(C \cap D) = 0.3$ . Are $C$ and $D$ independent, mutually exclusive, both, or neither?

Question 2

3 marks

In a class, $P(\text{studies Chemistry}) = 0.5$ , $P(\text{studies Biology}) = 0.6$ , and $P(\text{studies both}) = 0.2$ . Find $P(\text{Biology} \mid \text{NOT Chemistry})$ .

Question 3

2 marks

Events $E$ and $F$ satisfy $P(E) = 0.2$ , $P(F) = 0.3$ , $P(E \cap F) = 0.06$ . A student writes: 'E and F are independent, because their probabilities are both small, so they can't really affect each other.' Is the student's final answer correct, and is the reasoning acceptable?

Question 4

3 marks

Two fair six-sided dice are rolled. $A$ = 'the first die shows a 6' and $B$ = 'the two dice sum to 7' — the same pair of events already shown earlier in this lesson to be independent. Which of the following events $D$ is mutually exclusive with $A$ , but still has $P(B \cap D) > 0$ ?

Reference — not a study method, a lookup

P(A|B) = P(A∩B)/P(B), requires P(B) > 0. Conditioning restricts the sample space to B.
Independent (memorise, not on the sheet): P(A|B)=P(A), P(B|A)=P(B), P(A∩B)=P(A)P(B) — any one implies the others.
Mutually exclusive: A∩B=∅, so P(A∩B)=0 and P(A∪B)=P(A)+P(B).
If P(A)>0 and P(B)>0: mutually exclusive and independent can never both hold. Overlap is necessary for independence but never sufficient — always run the product-rule check.
On the formula sheet: P(A∪B)=P(A)+P(B)−P(A∩B), and P(A∩B)=P(A)P(B|A). Not on it: P(A′)=1−P(A) and all three independence rules.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation attributed to a WST01 examiner report or mark scheme in this lesson (Jan 2021, Oct 2021, Jun 2022) was checked against WST01-verified-facts.md, itself independently verified against the real Pearson PDFs — there is no prior AI-authored WST01 material in this repo to have carried an error over from. Every scenario used to teach with — the 200-applicant Mathematics/Physics table, both counter-drawing bags, the dice examples, the three-event Venn diagram in the second worked-chain, and every prequestion and MCQ — is VERIDIAN-original, checked with exact fraction (or, for the Venn diagram's two linked unknowns, exact algebraic) arithmetic before being written in (including the deliberately wrong answers, so the wrong values shown are the actual numbers those specific errors produce). The exceptions are the malformed expressions quoted directly from Jan 2021 Q1(c) — 'P(0.35∩0.4)/0.4' and '0.35/0.4' — and the short mark-scheme fragment quoted from Oct 2021 Q1(c) in the Venn-diagram worked-chain ('p / 0.5 = p + 0.06 (o.e.)', '2p = p + 0.06, so p = 0.06', 'use of p + q = 0.2 gives q = 0.14'), all of which are Pearson's own numbers from real questions this research pass did not have full access to reconstruct as complete questions; they are presented only as quoted evidence for the mark-scheme criterion they support, never reconstructed into a fabricated full question of their own.

Question 12 marks

Events $C$ and $D$ satisfy $P(C) = 0.6$ , $P(D) = 0.5$ , and $P(C \cap D) = 0.3$ . Are $C$ and $D$ independent, mutually exclusive, both, or neither?

Independent, and therefore not mutually exclusive — $P(C) \times P(D) = 0.6 \times 0.5 = 0.3 = P(C \cap D)$
Correct. The product-rule test passes exactly, so $C$ and $D$ are independent — and since $P(C \cap D) = 0.3 \neq 0$ , they are not mutually exclusive, consistent with the mechanism proved earlier: independent events with positive probabilities can never be mutually exclusive.
BMutually exclusive, since $C$ and $D$ are clearly two separate, distinct events
'Distinct events' is not the same as 'mutually exclusive' — mutually exclusive specifically means they share no outcomes, i.e. $P(C \cap D) = 0$ . Here $P(C \cap D) = 0.3$ , given directly in the question, so they plainly do overlap.
CBoth independent and mutually exclusive
Impossible whenever both probabilities are positive, by the same argument proved in the mechanism block: mutually exclusive would force $P(C \cap D) = 0$ , but it's given as $0.3$ . Only one of the two properties can hold here, and the calculation shows it's independence.
DNeither — $0.3$ doesn't obviously relate to $0.6$ and $0.5$
$0.3$ is exactly $0.6 \times 0.5$ — the relationship is there, it just needs the multiplication carried out rather than eyeballed. This is the whole reason the independence test exists: the connection between the numbers is rarely visible without actually computing it.

Traps tested: Distinct events assumed mutually exclusive · Mutually exclusive and independent treated as compatible · Independence check not attempted

Question 23 marks

In a class, $P(\text{studies Chemistry}) = 0.5$ , $P(\text{studies Biology}) = 0.6$ , and $P(\text{studies both}) = 0.2$ . Find $P(\text{Biology} \mid \text{NOT Chemistry})$ .

$0.8$ — from $P(B \cap C') = P(B) - P(B \cap C) = 0.6 - 0.2 = 0.4$ and $P(C') = 1 - 0.5 = 0.5$ , so $P(B \mid C') = 0.4 \div 0.5$
Correct. $P(B \cap C')$ is the students who study Biology but NOT Chemistry — everyone who studies Biology, minus the ones who also study Chemistry — and $P(C')$ is the complement rule applied to the conditioning event. Both intermediate steps have to be found before the conditional probability formula can be applied at all.
B $0.6$ — since 'not Chemistry' shouldn't change anything about Biology
This assumes the answer rather than testing it — exactly the habit this lesson is built to break. $P(\text{Biology}) = 0.6$ unconditionally, but conditioning on NOT studying Chemistry does change the figure here, to $0.8$ : among the non-Chemistry students specifically, Biology is more common than in the class as a whole, because the overlap between the two subjects has been removed from the comparison.
C $0.4$ — using $P(B \cap C')$ alone, without dividing by $P(C')$
$0.4$ is a correct intermediate value — but it's $P(B \cap C')$ , the probability of studying Biology and not Chemistry out of the WHOLE class, not the conditional probability restricted to non-Chemistry students. The division by $P(C') = 0.5$ is the step that's missing.
D $\dfrac{1}{3}$ — from $P(C \cap B) \div P(B) = 0.2 \div 0.6$
This computes $P(\text{Chemistry} \mid \text{Biology})$ — the wrong pair of events conditioned the wrong way round, not $P(\text{Biology} \mid \text{NOT Chemistry})$ . Both the event being conditioned ON and which event is complemented need to match the question exactly.

Traps tested: Conditioning assumed to have no effect · Conditioning step omitted · Conditioning event reversed

Question 32 marks

Events $E$ and $F$ satisfy $P(E) = 0.2$ , $P(F) = 0.3$ , $P(E \cap F) = 0.06$ . A student writes: 'E and F are independent, because their probabilities are both small, so they can't really affect each other.' Is the student's final answer correct, and is the reasoning acceptable?

The answer is correct ( $0.2 \times 0.3 = 0.06 = P(E \cap F)$ , so they ARE independent) — but the reasoning given would not earn the method mark, because 'small probabilities' was never actually tested
Correct on both counts. The independence test happens to pass here, so the conclusion is right — but a mark scheme rewards the calculation, not a correct guess about it. Swap in different small-but-not-multiplying-to-match probabilities (say $P(E)=0.2$ , $P(F)=0.3$ , $P(E \cap F)=0.1$ ) and the same reasoning ('both small, so independent') would produce the wrong answer, because it was never actually connected to the numbers.
BThe answer is correct and the reasoning is acceptable, since it correctly identifies that small probabilities tend to indicate independence
There is no general rule connecting the SIZE of two probabilities to whether they're independent — a large $P(E)$ and small $P(F)$ can just as easily be independent, or a small $P(E)$ and small $P(F)$ can just as easily be dependent. The only valid test is the specific equality $P(E \cap F) = P(E) \times P(F)$ , checked with these exact numbers.
CThe answer is wrong — $E$ and $F$ are not independent, because $0.06$ is a small number and small intersections usually mean dependent events
$0.06$ being 'small' says nothing on its own; it has to be compared to $P(E) \times P(F) = 0.06$ specifically, and it matches exactly. The size of $P(E \cap F)$ in isolation, without that comparison, cannot answer an independence question.
DThere isn't enough information to check independence without knowing more about what $E$ and $F$ actually represent
The independence test needs only the three numbers given — $P(E)$ , $P(F)$ , and $P(E \cap F)$ — and nothing about what the events physically represent. That's precisely what makes the test objective rather than a matter of judgement: two events can be independent (or not) regardless of how intuitively 'related' they seem.

Traps tested: Probability magnitude substituted for independence test · Intersection magnitude substituted for independence test · Independence treated as unanswerable without context

Question 43 marks

Two fair six-sided dice are rolled. $A$ = 'the first die shows a 6' and $B$ = 'the two dice sum to 7' — the same pair of events already shown earlier in this lesson to be independent. Which of the following events $D$ is mutually exclusive with $A$ , but still has $P(B \cap D) > 0$ ?

$D$ = 'the first die shows a 1 or a 2' — $A$ needs a first die of 6, so $A$ and $D$ share no outcomes at all, and $D$ still overlaps $B$ : the outcomes $(1,6)$ and $(2,5)$ both sum to 7 and both have a first die of 1 or 2
Correct. Both conditions have to hold at once, and this is the only choice here that actually satisfies both rather than just one. Mutually exclusive with $A$ means sharing NO outcome with 'first die is 6' — true here, since $D$ only allows a first die of 1 or 2 — while $P(B \cap D) > 0$ means at least one outcome has to land in both $B$ and $D$ , which $(1,6)$ and $(2,5)$ do: $P(B \cap D) = \frac{2}{36}$ .
B $D$ = 'the first die shows an even number (2, 4, or 6)' — this clearly overlaps $B$ , since $(2,5)$ , $(4,3)$ , and $(6,1)$ all sum to 7 with an even first die
$D$ does overlap $B$ — but that is only half of what the question asks. $D$ also includes 'first die is 6', the exact outcome $A$ is built from, so $A$ and $D$ share the outcome $(6,1)$ and are NOT mutually exclusive. Checking only the $P(B \cap D) > 0$ half of the question, and stopping there, misses the other, equally necessary condition.
C $D$ = 'the two dice sum to 2' (the single outcome $(1,1)$ ) — the first die is 1, not 6, so $D$ is mutually exclusive with $A$
$D$ genuinely is mutually exclusive with $A$ — that half is correctly checked. But $D$ contains only the outcome $(1,1)$ , which sums to 2, not 7, so $B$ and $D$ share no outcome and $P(B \cap D) = 0$ — the second condition fails. An event can satisfy 'mutually exclusive with $A$ ' perfectly and still be the wrong answer, because that was never the whole question.
D $D = B$ itself, i.e. $D$ = 'the two dice sum to 7' — then $P(B \cap D) = P(B) > 0$ automatically, since $B \cap B = B$
The $P(B \cap D) > 0$ half is satisfied trivially this way — but $A$ and $B$ are NOT mutually exclusive (they were shown to be independent earlier in this lesson, and $A \cap B$ contains the outcome $(6,1)$ ), so setting $D = B$ makes $D$ share that same outcome with $A$ . Reusing $B$ guarantees the overlap condition by construction but breaks the mutual-exclusivity condition by the very same stroke — $D$ has to be a genuinely new event, not a shortcut built directly from $A$ or $B$ .

Traps tested: Mutual exclusivity condition not checked · Overlap with b condition not checked · Constructed event reuses b and breaks mutual exclusivity

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: Oct 2021 · Q4(c) — cited directly in this lesson
Mark scheme: Oct 2021 · Q1(c) — cited directly in this lesson

Pearson's official past-papers portal

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