HookThe number that jailed Sally Clark
In 1999 a Cheshire solicitor named Sally Clark was convicted of murdering her two baby sons, who had died suddenly thirteen months apart. The prosecution's case leaned on a single number. The paediatrician Sir Roy Meadow told the jury that the chance of two cot deaths in one affluent, non-smoking family was about 1 in 73 million — a figure he reached by taking the chance of one such death, roughly 1 in 8,500, and squaring it. That squaring is a probability calculation, and it is wrong. Squaring 1 in 8,500 only gives 1 in 73 million if the two deaths are independent — if the first death tells you nothing about the chance of the second. Cot deaths in the same family are not independent: shared genes, shared environment and shared risk factors link them. The Royal Statistical Society wrote to the Lord Chancellor in 2001 to say exactly this, and Sally Clark's conviction was quashed in 2003.
Every idea in this section is a defence against that kind of error. You will learn when two events are mutually exclusive and when they are independent — two words that sound similar and mean almost opposite things — and precisely when the multiplication that convicted Sally Clark is allowed. You will learn conditional probability, the mathematics of how one event shifts the odds of another, using tree diagrams, Venn diagrams and two-way tables. And you will learn to model with probability: to state your assumptions out loud, ask whether they are realistic, and work out how the answer would move if they are not. The whole of S3 is the discipline of knowing which probabilities you are allowed to multiply, and which you are not.
ModelMutually exclusive vs independent — two words, opposite meanings
Two events are mutually exclusive if they cannot both happen: rolling an even number and rolling an odd number on one die. For mutually exclusive events \(P(A\cap B)=0\), and the probability that one or the other happens is just the sum, \(P(A\cup B)=P(A)+P(B)\). In general, when events can overlap, you must subtract the double-counted middle: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). This is the addition rule, and forgetting the final term is the single most common slip on Venn-diagram questions.
Two events are independent if one happening does not change the probability of the other: the second toss of a coin does not care how the first landed. For independent events, and only for independent events, you may multiply — the multiplication rule, \(P(A\cap B)=P(A)\,P(B)\). It also helps to know the complement: \(P(A')=1-P(A)\), which turns an awkward 'at least one' into an easy 'one minus none'.
Here is the distinction the examiner hammers, and the one the Sally Clark jury never heard: mutually exclusive and independent are almost opposites. If \(A\) and \(B\) both have non-zero probability and are mutually exclusive, they cannot be independent — because if \(A\) has happened, \(B\) now has probability zero, so \(A\) has changed \(B\)'s odds enormously. Independence links to the wider course too: the binomial distribution in S4 is built on the assumption of independent trials, and continuous distributions assign probability to intervals rather than to single points, but the multiplication rule for independence is the engine underneath both.
Roll one fair six-sided die. Let \(A\) be 'even' \((\{2,4,6\})\) and \(B\) be 'a multiple of 3' \((\{3,6\})\). Then \(P(A)=\tfrac{3}{6}=\tfrac12\) and \(P(B)=\tfrac{2}{6}=\tfrac13\). The overlap is \(A\cap B=\{6\}\), so \(P(A\cap B)=\tfrac16\).
Are they independent? Check the multiplication rule: \(P(A)\,P(B)=\tfrac12\times\tfrac13=\tfrac16\), which equals \(P(A\cap B)\) — so yes, \(A\) and \(B\) are independent even though they overlap. Now the addition rule: \(P(A\cup B)=\tfrac12+\tfrac13-\tfrac16=\tfrac{3}{6}+\tfrac{2}{6}-\tfrac{1}{6}=\tfrac{4}{6}=\tfrac23\). Contrast this with \(A\) and \(C=\)'odd' \((\{1,3,5\})\): these are mutually exclusive, \(P(A\cap C)=0\), but they are the opposite of independent — \(P(A\mid C)=0\neq P(A)=\tfrac12\). One die shows both ideas at once.
MechanismConditional probability — how one event rewrites the odds of another
Conditional probability is written \(P(A\mid B)\) and read 'the probability of \(A\) given \(B\)'. It is defined by \[P(A\mid B)=\frac{P(A\cap B)}{P(B)},\] which simply says: restrict your attention to the world where \(B\) has happened, then ask what fraction of that world also has \(A\). Rearranged, it gives the general multiplication rule \(P(A\cap B)=P(B)\,P(A\mid B)\) — the honest version of 'multiply the probabilities', valid whether or not the events are independent. Independence is now just the special case where the condition changes nothing: \(P(A\mid B)=P(A)\).
Three tools compute the same thing. On a tree diagram you multiply probabilities along a branch and add across branches that reach the same outcome; the second layer of branches carries conditional probabilities. On a two-way table you find the relevant row or column total and read the conditional probability as a fraction of it. On a Venn diagram you shade \(B\) and ask what proportion of it is also \(A\). The trap that convicted Sally Clark lives here: \(P(A\mid B)\) is generally not equal to \(P(B\mid A)\). The probability of the evidence given innocence is a completely different number from the probability of innocence given the evidence — confusing the two is the 'prosecutor's fallacy'.
A rapid test for a condition that is present in 1% of the population is 99% sensitive (it catches 99% of true cases) but has a 5% false-positive rate. A randomly chosen person tests positive — what is the probability they actually have the condition?
Imagine 10,000 people. About \(1\%\) — that is 100 people — have the condition; 9,900 do not. Of the 100 who have it, the test finds \(99\%\), giving 99 true positives. Of the 9,900 who do not, \(5\%\) test positive anyway, giving \(0.05\times 9900 = 495\) false positives. So the total positives are \(99+495=594\), of whom only 99 are genuine. Therefore \[P(\text{condition}\mid\text{positive})=\frac{99}{594}=0.167,\] about 17%. Notice the two conditionals: \(P(\text{positive}\mid\text{condition})=0.99\), but \(P(\text{condition}\mid\text{positive})=0.167\). They are wildly different, because the condition is rare — exactly the base-rate reasoning the Sally Clark jury was denied.
MechanismTesting independence — and where it quietly breaks
To decide whether two events really are independent, you do not guess from the story — you check the arithmetic. Either confirm \(P(A\cap B)=P(A)\,P(B)\), or equivalently confirm \(P(A\mid B)=P(A)\). If the two sides disagree, the events are associated, and the sign of the gap tells you the direction: \(P(A\cap B)\) below \(P(A)\,P(B)\) means they tend to avoid each other, above means they tend to occur together.
The place independence quietly dies is sampling without replacement. Draw one card from a pack and the probability the second is a heart depends on what the first card was — the trials are conditional, not independent, so the branches of your tree carry different probabilities on the second layer. Draw with replacement and independence is restored. Whenever a question says 'without replacement', 'chosen at random from those remaining', or describes a small closed population, expect conditional probabilities and reach for the tree diagram rather than a blind multiplication.
In a survey of 500 commuters, let \(A\) be 'owns a car' with \(P(A)=0.6\), and \(B\) be 'commutes by train' with \(P(B)=0.3\). The survey finds \(P(A\cap B)=0.15\). Are car ownership and train commuting independent?
Test the multiplication rule: \(P(A)\,P(B)=0.6\times 0.3=0.18\). But the observed \(P(A\cap B)=0.15\), which is smaller, so the events are not independent — car owners are a little less likely to commute by train, as you might expect. Confirm with the conditional: \(P(A\mid B)=\dfrac{0.15}{0.3}=0.5\), whereas \(P(A)=0.6\). Because \(0.5\neq 0.6\), knowing someone takes the train genuinely lowers the probability that they own a car. The mismatch is the whole answer — never assume independence just because the two facts feel unrelated.
CaseModelling with probability — say your assumptions out loud
A probability model is a deliberate simplification: you assume outcomes are equally likely, or trials are independent, or a probability stays constant, so that the maths becomes tractable. The A-Level skill the spec singles out is not just computing within a model but critiquing it — naming each assumption, judging whether it is realistic, and predicting which way the answer would move if you replaced it with something truer to life. Is a die really fair? Are newborns really equally likely to be boys or girls (in fact about 51% are boys)? Are successive calls to a helpline really independent, or does one incident trigger a cluster?
The Sally Clark case is the definitive worked example of a bad assumption. The 1-in-73-million figure assumed the two infant deaths were independent. A more realistic model treats the second death as conditional on the first, because a family that has suffered one cot death may carry a shared genetic or environmental risk factor that makes a second more likely. Replacing independence with a realistic conditional probability does not tweak the answer at the second decimal place — it moves it by orders of magnitude, which is why getting the assumption right mattered more than any single calculation. When you meet a modelling question, the marks are in the sentence that says which assumption you made and how a better one would change the conclusion.
Suppose the chance of a single cot death in a comparable family is about \(\tfrac{1}{8500}\). Under the independence assumption the jury heard, two deaths have probability \[\left(\tfrac{1}{8500}\right)^2=\tfrac{1}{72\,250\,000}\approx\text{1 in 73 million}.\] Now relax that assumption. Suppose a shared risk factor, once the first death has occurred, raises the chance of a second to around \(\tfrac{1}{100}\). The realistic joint probability becomes \[P(\text{first})\times P(\text{second}\mid\text{first})=\tfrac{1}{8500}\times\tfrac{1}{100}=\tfrac{1}{850\,000},\] roughly 1 in 850,000 — about 85 times more likely than the figure the jury was given. One swapped assumption, an 85-fold shift in the answer: that is why modelling questions ask you to defend, not just compute.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Define your events with letters before you compute — writing '\(A=\) owns a car, \(B=\) takes the train' costs one line and stops you muddling which conditional you want. Whenever you multiply probabilities, state the justification: either 'independent, so \(P(A\cap B)=P(A)P(B)\)' or 'using \(P(A\cap B)=P(B)P(A\mid B)\)'. Examiners award method marks for the rule, not just the number.
On tree diagrams, multiply along branches and add across them, and check every set of branches from a node sums to 1. On two-way tables, a conditional probability is a fraction of a row or column total, never of the grand total — misreading the denominator is the classic dropped mark. For 'modelling' questions, name the assumption explicitly (independence, equally likely, constant probability), say whether it is realistic, and predict the direction the answer moves under a more realistic one; that sentence is what separates a full-mark response from a bare calculation. And never write \(P(A\mid B)\) when you mean \(P(B\mid A)\) — the prosecutor's fallacy is examined precisely because it is so tempting.