HookThe number that put Sally Clark in prison
In November 1999 a Manchester solicitor named Sally Clark was convicted of murdering her two baby sons, who had died suddenly fourteen months apart in 1996 and 1998. A large part of the case rested on a single number. A paediatrician, Professor Sir Roy Meadow, told the jury that the chance of two cot deaths in one family like hers was '1 in 73 million' — a figure he reached by taking a roughly 1-in-8,500 risk for a single cot death and simply squaring it.
The number was wrong, and the mathematics of why it was wrong is the whole of this section. Squaring assumes the two deaths were independent — that the first tells you nothing about the chance of the second — when shared genes and environment plainly link siblings. And the way the figure was then used confused two different conditional probabilities, a mistake statisticians call the prosecutor's fallacy. The Royal Statistical Society took the rare step of writing to the Lord Chancellor in 2001, and Sally Clark's conviction was quashed in 2003. Probability is the mathematics of the uncertain but measurable, on a fixed scale from 0 to 1, and this section builds from that scale up to exactly the ideas the court got wrong: theoretical probability from equally likely outcomes, the rule that probabilities sum to 1, relative and expected frequency, sample spaces, Venn and tree diagrams, independent versus dependent events, and conditional probability.
ModelThe probability scale and theoretical probability
Every probability lives on a scale from 0 to 1: 0 means impossible, 1 means certain, \(\tfrac{1}{2}\) an even chance, and nothing is ever below 0 or above 1 — a value like \(1.3\) or \(-0.2\) is a guaranteed error. Probabilities may be written as fractions, decimals or percentages, but you should not give them as odds ('1 in 6') unless the question asks. When every outcome is equally likely — a fair coin, a fair die, a well-shuffled pack — the theoretical probability of an event is the number of favourable outcomes divided by the total number of outcomes.
The load-bearing word is fair. \(P(\text{even on a fair die})=\tfrac{3}{6}=\tfrac{1}{2}\) only because the six faces are genuinely equally likely; on a loaded die the formula collapses. Judging fairness — pointing to equally likely outcomes, or their absence — is itself examined, and it is precisely the question a court should have asked before treating two family tragedies as if they were two rolls of the same fair die.
MechanismAdding to 1 — mutually exclusive and exhaustive events
Two events are mutually exclusive if they cannot both happen at once, like rolling a \(2\) and rolling a \(5\) on one throw; for these you find 'one or the other' by adding the probabilities. A set of outcomes is exhaustive if it covers every possibility, and an exhaustive set of probabilities always adds to exactly \(1\) — the mathematical way of saying that something must happen.
The most useful consequence is the complement rule \(P(\text{not }A)=1-P(A)\): if the chance of rain is \(0.3\), the chance of no rain is \(0.7\). This one line is the fastest route through 'at least one' problems, which are almost always quicker as \(1\) minus the probability of 'none'. The warning is that adding only works when the events genuinely cannot overlap — you cannot add \(P(\text{even})\) and \(P(\text{prime})\) on a die, because \(2\) is both and would be counted twice, which is exactly what the next block on Venn diagrams is built to prevent.
A spinner lands on red, blue or yellow, with \(P(\text{red})=0.4\) and \(P(\text{blue})=0.25\). Find \(P(\text{yellow})\) and \(P(\text{not red})\). The three colours are exhaustive, so they add to \(1\): \(P(\text{yellow})=1-0.4-0.25=0.35\). And \(P(\text{not red})=1-0.4=0.6\), which you can check against \(P(\text{blue})+P(\text{yellow})=0.25+0.35=0.6\). The two routes agreeing is your proof the arithmetic holds.
DataRelative frequency, expected frequency and the long run
When outcomes are not equally likely — a drawing pin, a biased spinner, a team's home form — you estimate probability from data. The relative frequency of an outcome is the number of times it happened divided by the number of trials, and it is your best estimate of the true probability. Run the idea the other way and you get expected frequency: probability \(\times\) number of trials predicts how many successes to expect. This is the step students most often miss, handing back a probability where the question asked for a count.
Why trust an experiment at all? The law of large numbers: as the number of trials grows, the relative frequency of an unbiased experiment settles towards the theoretical probability. Ten coin tosses might give seven heads and look rigged; ten thousand will hug \(0.5\). Small samples wobble, large ones converge — which is why a probability estimated from a big sample is trusted over one from a handful of trials, and why bias should never be judged from a short run. Frequency trees and two-way tables are the tools for recording these counts stage by stage before the probabilities are read off.
A biased spinner is spun \(200\) times and lands on 'win' \(46\) times. Estimate the probability of a win, then predict the number of wins in \(1{,}500\) spins. The relative frequency is \(\tfrac{46}{200}=0.23\), the best estimate of \(P(\text{win})\). Expected wins in \(1{,}500\) spins is \(0.23\times 1{,}500=345\). Note the final answer is a count, \(345\), not the probability \(0.23\) — handing back the probability here is the classic dropped mark.
ModelListing outcomes — sample spaces, grids and Venn diagrams
To count outcomes without missing or repeating any, you organise them. A sample space is a complete list of every outcome; for combined experiments — two dice, two spinners — a grid with one experiment along the top and the other down the side guarantees you catch them all. Two ordinary dice give a \(6\times 6\) grid of \(36\) equally likely outcomes, and any combined probability is just the count of wanted cells over \(36\).
When events can overlap, a Venn diagram keeps the accounts straight: a box for everything, overlapping circles for the events, and the overlap — the intersection — filled in first so nothing is double-counted. When events happen in sequence, a tree diagram branches stage by stage. Each tool is bookkeeping that makes the final probability fall out, and each earns method marks even when the last number slips, which is why a fully drawn diagram always beats a bare answer.
Two spinners, one numbered \(1,2,3\) and the other \(1,2,3,4\), are spun and the numbers multiplied. Find \(P(\text{product}>6)\). The grid holds \(3\times 4=12\) equally likely outcomes; the products are \(1,2,3,4\) / \(2,4,6,8\) / \(3,6,9,12\). Those greater than \(6\) are \(8\), \(9\) and \(12\) — three cells — so \(P(\text{product}>6)=\tfrac{3}{12}=\tfrac{1}{4}\). Listing the grid in full is far safer than trying to picture the products in your head.
CaseCombined events — independent, dependent and the Clark error
For events in sequence, the tree diagram runs on two rules: multiply along the branches for 'this and then that', and add between branches for separate paths that both count. The decisive distinction is independence. Events are independent if one does not affect the other — two spins, or drawing a card and replacing it — so the probabilities stay fixed. They are dependent if the first changes the second, the classic case being drawing objects without replacement: remove a red counter and there are now fewer counters and fewer reds, so the second fraction shifts.
This is the exact point the Sally Clark case turned on. Multiplying the risk of one cot death by itself to reach '1 in 73 million' assumed the two deaths were independent; they were not, because shared genes and environment link siblings, so the chance of a second death given a first is far higher than the headline suggests. Whenever you multiply probabilities you are assuming independence, and saying so is exactly what AQA means by 'write down any assumptions you have made'.
A bag holds \(5\) red and \(3\) blue counters. Two are drawn without replacement. Find \(P(\text{both red})\) and \(P(\text{different colours})\). The first draw is \(P(\text{red})=\tfrac{5}{8}\); after removing a red, \(7\) remain with \(4\) red, so the second is \(\tfrac{4}{7}\). Multiply along the branch: \(P(\text{both red})=\tfrac{5}{8}\times\tfrac{4}{7}=\tfrac{20}{56}=\tfrac{5}{14}\). Likewise \(P(\text{both blue})=\tfrac{3}{8}\times\tfrac{2}{7}=\tfrac{6}{56}=\tfrac{3}{28}\), so \(P(\text{same colour})=\tfrac{20}{56}+\tfrac{6}{56}=\tfrac{26}{56}=\tfrac{13}{28}\) and \(P(\text{different})=1-\tfrac{13}{28}=\tfrac{15}{28}\). The second fraction being \(\tfrac{4}{7}\), not \(\tfrac{5}{8}\), is the whole meaning of 'without replacement'.
MechanismConditional probability — and the prosecutor's fallacy
The Higher headline is conditional probability — the chance of one event given that another has already happened, written \(P(A\mid B)\). The word 'given' does one thing: it shrinks the world you are counting in down to just the cases where \(B\) is true, so the denominator becomes the size of \(B\), not the whole sample. On a two-way table you read straight along the relevant row or column; on a tree diagram, the without-replacement second branch is a conditional probability; on a Venn diagram you divide the overlap by the total of the 'given' circle.
This is where Sally Clark's second error lived — the prosecutor's fallacy, the confusion of \(P(A\mid B)\) with \(P(B\mid A)\). The probability of the evidence given innocence is not the same as the probability of innocence given the evidence, yet the trial treated a tiny \(P(\text{two deaths}\mid\text{innocent})\) as if it were the probability of innocence itself. Keeping the order of the condition straight — knowing which event is the 'given' and which is being asked — is the single most important idea in Higher probability.
Of \(120\) patients, \(40\) have a condition and \(80\) do not. A rapid test is positive for \(36\) of the \(40\) with the condition, and for \(12\) of the \(80\) without it. A patient tests positive; find the probability they actually have the condition. The positives total \(36+12=48\). Restricting to that column, \(P(\text{condition}\mid\text{positive})=\tfrac{36}{48}=\tfrac{3}{4}\). Contrast the reverse, \(P(\text{positive}\mid\text{condition})=\tfrac{36}{40}=0.9\): the two conditionals are different numbers from the same table, and mistaking one for the other is the prosecutor's fallacy in miniature.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Probability runs across all three papers and leans on the calculator for the fraction arithmetic, but the marks are for reasoning, not button-pressing. Give probabilities as fractions, decimals or percentages — never '1 in 6' odds unless asked — and if you are told to compare probabilities, convert them all to the same form first. Any value below 0 or above 1 is automatically wrong, so glance at your final answer.
The reliable dropped marks are avoidable. Expected frequency is a count: multiply the probability by the number of trials and give a whole-ish number (345 wins, not 0.23). On tree diagrams, multiply along the branches for 'and', add between branches for 'or', and check each split's probabilities sum to \(1\). On without-replacement questions, change the second fraction's numerator and denominator to match what was removed — writing the \(\tfrac{4}{7}\) explicitly earns the method mark — and state the independence assumption whenever you multiply.
Conditional probability is the Higher discriminator. Read 'given that' as an instruction to restrict yourself to a subset, so the denominator is that subset's total, not the grand total. Never swap \(P(A\mid B)\) for \(P(B\mid A)\); on a two-way table, decide which is the fixed row or column before you divide. For a relative-frequency estimate use the largest sample available, since bigger samples give better estimates, and show your sample space, Venn diagram or frequency tree in full — a completed diagram carries method marks even when the final probability slips.