HookERNIE has been drawing random numbers since 1957
Since June 1957 the winners of Britain's Premium Bonds have been chosen by a machine called ERNIE — the Electronic Random Number Indicator Equipment — first switched on at Lytham St Annes in Lancashire. The original ERNIE used the electronic noise of neon tubes to generate genuinely unpredictable numbers; today's version uses quantum effects in silicon, and it draws millions of prizes every month from over 100 billion eligible £1 bonds. For any single £1 bond the chance of winning a prize in a given month is currently about 1 in 22,000 — small, fixed, and utterly indifferent to whether that bond won last month or has never won in forty years. The machine has no memory.
That last point is where most probability marks are won and lost, and it runs through this whole section. Probability is the mathematics of the uncertain but measurable: how likely an outcome is, on a fixed scale from 0 (impossible) to 1 (certain). You will learn to calculate a theoretical probability from equally likely outcomes, to use the fact that all the probabilities of a situation add to 1, to estimate probabilities from experiments and understand why big experiments are more trustworthy than small ones, and to handle combined events with sample spaces, Venn diagrams and tree diagrams. The arithmetic is mostly fractions; the difficulty is in the reasoning, and in resisting the very human urge to believe a result is somehow 'due'.
ModelThe probability scale and theoretical probability
Every probability sits on a scale from 0 to 1: 0 means impossible, 1 means certain, ½ means an even chance. Nothing is ever negative, and nothing is ever more than 1 — an answer of 1.2 or −0.3 is a signal you have made an error. You can write a probability as a fraction, a decimal or a percentage, but 'evens' as an everyday phrase becomes ½ or 0.5 or 50%.
When every outcome is equally likely — a fair coin, a fair die, a well-shuffled pack — the theoretical probability of an event is simply the number of outcomes you want divided by the total number of outcomes. The chance of rolling an even number on a fair die is 3 (the twos, fours and sixes) out of 6, which is ½. The word doing the heavy lifting is 'fair': the formula only works when the outcomes really are equally likely, which is exactly what makes ERNIE fair and a loaded die not.
This is also where the language marks live. AQA will ask you to place events on a 0-to-1 line, or to judge them 'unlikely', 'even chance', 'likely' and so on, and to justify a claim of fairness by pointing to equally likely outcomes.
A bag holds 3 red, 4 blue and 5 green counters — 12 in total. Find the probability of drawing a blue counter. There are 4 blue outcomes out of 12 equally likely draws, so P(blue) = 4/12 = 1/3. The probability of green is 5/12, and of red 3/12 = 1/4. As a check, the three add to 4/12 + 5/12 + 3/12 = 12/12 = 1 — every possible outcome is covered, so the total must be 1.
MechanismAdding to 1 — mutually exclusive and exhaustive events
Two events are mutually exclusive if they cannot both happen at once — rolling a 2 and rolling a 5 on a single throw, for example. For mutually exclusive events you find the probability of one or the other by adding their probabilities. A set of outcomes is exhaustive if it covers every possibility, and the probabilities of an exhaustive set always add to exactly 1 — that is just the statement that something must happen.
The most useful consequence is the 'not' rule: since an event either happens or it does not, and those two cover everything, P(not A) = 1 − P(A). If the chance of rain tomorrow is 0.3, the chance of no rain is 1 − 0.3 = 0.7. This single line saves enormous effort — working out the probability of 'at least one' something is almost always quicker as 1 minus the probability of 'none'.
The warning is that adding only works when events genuinely cannot overlap. You cannot add P(even) and P(prime) on a die, because 2 is both even and prime and you would count it twice — a mistake the next block on Venn diagrams is built to prevent.
A spinner lands on red, blue or yellow. P(red) = 0.35 and P(blue) = 0.2. Find P(yellow) and P(not red). The three colours are exhaustive, so they add to 1: P(yellow) = 1 − 0.35 − 0.2 = 0.45. And P(not red) = 1 − P(red) = 1 − 0.35 = 0.65 — which you can check equals P(blue) + P(yellow) = 0.2 + 0.45 = 0.65. The two routes agreeing confirms the arithmetic.
DataRelative frequency, expected frequency and the long run
Not everything is a fair die. When you cannot assume equal outcomes — is this drawing pin biased? does this football team win at home? — you run an experiment and use the results. The relative frequency (or experimental probability) of an outcome is the number of times it happened divided by the number of trials. Throw a pin 200 times, see it land point-up 130 times, and your best estimate of P(point-up) is 130/200 = 0.65.
Run the machine the other way and you get expected frequency: if you know a probability and are about to do many trials, the expected number of successes is probability × number of trials. This is the step students most often skip — they give the probability where the question asked for a count of how many times.
Why trust an experiment at all? Because of the law of large numbers: as the number of trials grows, the relative frequency settles down towards the true theoretical probability. Ten coin tosses might give seven heads (0.7), which looks biased; ten thousand tosses will sit very close to 0.5. Small samples wobble; large unbiased samples converge. This is precisely why ERNIE's monthly draw, across billions of bonds, distributes prizes in almost exactly the proportions the odds predict, even though any one bond's fate is pure chance.
A fair six-sided die is rolled 300 times. How many sixes are expected? The theoretical probability of a six is 1/6, so the expected frequency is 1/6 × 300 = 50. You would not bet on getting exactly 50 — you might get 44 or 57 — but 50 is the long-run expectation, and over 30,000 rolls the proportion would hug 1/6 far more tightly than over 300. Note the answer is a count, 50, not the probability 1/6: giving the fraction here is the classic dropped mark.
MechanismRecording outcomes — frequency trees and two-way tables
Before you can calculate, you often have to organise the data, and AQA has two favourite tools. A two-way table sorts a group by two features at once — say gender across the top and whether they wear glasses down the side — with the row and column totals around the edge. The skill is filling in a missing cell by using the totals: every row and every column must add up correctly, so one gap can always be found by subtraction.
A frequency tree tracks a group as it splits stage by stage. Start with the total, branch it by the first feature, then branch each of those by the second, and the numbers on the final branches must add back up to the starting total. Unlike a probability tree (which carries fractions), a frequency tree carries actual counts of people or things.
Both are just bookkeeping, but they are worth several marks because they make the probabilities fall out at the end: once the table or tree is complete, a probability is just the right final number over the total.
80 students each study French or Spanish. 45 study French, and of those 45 French students 20 are boys. Altogether there are 38 boys. Build the two-way table. French students: 45, so Spanish students: 80 − 45 = 35. French boys: 20, so French girls: 45 − 20 = 25. Total boys: 38, so Spanish boys: 38 − 20 = 18, and Spanish girls: 35 − 18 = 17. Now any probability drops out: P(a randomly chosen student is a Spanish girl) = 17/80. The whole method is subtraction using the totals.
ModelSample spaces — listing every possibility
A sample space (or possibility space) is a complete list of every outcome of an experiment. For a single die it is just 1, 2, 3, 4, 5, 6. For combined experiments — two dice, a coin and a spinner — the neatest tool is a grid, with one experiment's outcomes along the top and the other's down the side. The grid guarantees you miss nothing and repeat nothing, which is the same 'be systematic' discipline that runs through all of counting.
Rolling two ordinary dice produces a 6-by-6 grid of 36 equally likely outcomes. If you fill each cell with the total of the two dice, patterns leap out: a total of 7 appears in six cells, a total of 2 (double one) in just one. Every combined-event probability then comes straight off the grid — count the cells you want, divide by 36. This is why 7 is the most likely total with two dice and 2 or 12 the least, a fact every board-game player half-knows and every exam tests.
Two fair dice are rolled and their scores added. Find the probability the total is 7, and the probability it is 10 or more. In the 6-by-6 grid of 36 outcomes, a total of 7 comes from 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 — six cells — so P(7) = 6/36 = 1/6. A total of 10 or more means 10, 11 or 12: that is 3 cells (for 10), 2 (for 11) and 1 (for 12), so 6 cells, giving P(10+) = 6/36 = 1/6 as well. Counting cells on the grid is safer than trying to list combinations in your head.
MechanismVenn diagrams — where events overlap
When events can happen together, adding their probabilities double-counts the overlap, and a Venn diagram is how you keep the accounts straight. Draw a box for everything (the total, or 'universal set'), and overlapping circles inside it for each event. The intersection — the overlap — holds the things in both events; the union is everything in either; and anything outside the circles is in neither.
The method is to fill the overlap first, then work outwards, subtracting the overlap from each circle's total so nothing is counted twice. Once every region holds a number, any probability is the count in the region you want over the grand total. Venn diagrams are the visual answer to the earlier warning about 2 being both even and prime: put it in the overlap, once, and the double-counting problem disappears.
In a class of 30, 18 play football, 12 play tennis, and 7 play both. Complete the Venn diagram. The overlap is 7. Football only: 18 − 7 = 11. Tennis only: 12 − 7 = 5. Neither: 30 − 11 − 7 − 5 = 7. Now P(a student plays only tennis) = 5/30 = 1/6, and P(plays neither) = 7/30. Filling the overlap first is what stops you from writing 18 and 12 in the circles and overshooting the class size.
CaseCombined events — independent, dependent and tree diagrams
For events happening in sequence — two draws, two spins, rain then a bus being late — the tool is a tree diagram, and two rules run it. To find the probability of one outcome and another, you multiply along the branches. To combine separate paths that each satisfy the question, you add the branch totals. At each split, the branch probabilities must add to 1.
The crucial distinction is independent versus dependent. Events are independent if one does not affect the other — two spins of a spinner, or drawing a card and replacing it — so the probabilities stay fixed. Events are dependent if the first changes the second, the classic case being drawing objects without replacement: take a red counter out and there are now fewer counters, and fewer reds, so the second probability shifts. Spotting 'without replacement' and adjusting the second fraction is the single most tested idea in the topic, and the assumption underneath every calculation — that draws are random and the stated probabilities hold — is what AQA means by 'state your assumptions'.
A bag holds 4 red and 6 blue counters. Two are drawn without replacement. Find P(both red). First draw: P(red) = 4/10. After removing a red, 9 counters remain and 3 are red, so the second draw: P(red) = 3/9. Multiply along the branch: 4/10 × 3/9 = 12/90 = 2/15. Because it is without replacement, the second fraction is 3/9, not 4/10 — the bag has changed. Had the counter been replaced, the draws would be independent and the answer 4/10 × 4/10 = 16/100 = 4/25.
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 papers for the fraction arithmetic, but the marks are for reasoning, not button-pressing. Give probabilities as fractions, decimals or percentages — never as '1 in 6' odds unless the question asks for odds — and if you are told to compare probabilities, convert them all to the same form first. Any answer below 0 or above 1 is automatically wrong, so glance at your final value.
The most reliable dropped marks are avoidable. Expected frequency is a count: multiply the probability by the number of trials and give a whole-ish number (50 sixes, not 1/6). 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 has been removed — writing the 3/9 explicitly earns the method mark.
AQA loves two phrasings. 'Explain whether this is fair' wants you to compare the outcomes and point to whether they are equally likely. 'Estimate the probability' from a table wants the relative frequency, so use the largest total available, because bigger samples give better estimates. Show your sample space, Venn diagram or frequency tree in full: a completed diagram carries method marks even when the final probability slips, and an unsupported answer risks scoring one mark where the working would have scored three.