In simple terms
A friendly intro before the formal notes — no formulas yet.
The Long-Run Average
A discrete random variable attaches a probability to each countable numerical outcome of a random event, like the score on a die. The expected value is the average result you would get if you repeated the event a very large number of times — a weighted average, not necessarily an outcome you could ever see on a single trial.
Picture a charity raffle where each ticket costs 1 unit of money. One ticket in 500 wins a prize of 100; the rest win nothing. Your winnings from a single ticket are random: almost always 0, occasionally 100. The expected value is what those winnings average out to over millions of imaginary plays: . You can never actually win 0.20 on one ticket — but that number is the honest long-run value of the chance, and it is exactly what tells you the ticket price of 1 makes the raffle a poor bet.
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Identify the discrete random variable and list every possible numerical value it can take.
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Write the probability distribution as a table: each value paired with its probability . Check the probabilities are each between 0 and 1 and sum to exactly 1.
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If a probability is unknown, use to form an equation and solve it.
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Compute by multiplying each outcome by its probability and adding, then interpret the result in context (for a game, a profit of 0 means fair).
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Key formulas
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Full topic notes
Formal explanation with the rigour you need for the exam.
Discrete random variables and probability distributions
A random variable is a variable whose value is a numerical outcome of a random experiment. We write it with a capital letter such as , and use the matching lowercase letter for a particular value it can take. A random variable is discrete when the values it can take form a finite or countably infinite list of separate numbers — the kind of thing you count. The score on a die, the number of sixes in five rolls, and the number of faulty bulbs in a box are all discrete. A continuous variable, by contrast, could be any value in an interval, like a length or a time; those are not the subject of this topic.
A probability distribution for a discrete random variable pairs each possible value of with its probability. It is almost always displayed as a table, with the values across the top and the probabilities beneath. Whatever else a table shows, two conditions must hold for it to be a genuine probability distribution.
Each probability is valid: for every value .
The probabilities are exhaustive: they cover every outcome and sum to exactly one, .
These two facts are the engine of the whole topic — the sum-to-one rule in particular is what lets you find an unknown probability or constant.
Expected value: the long-run mean
The expected value of a discrete random variable is the number you would get by averaging the variable over a very large number of repetitions of the experiment. It is a weighted average of the possible values, where each value is weighted by how likely it is. Because it is the mean of the distribution it is also written .
In words: multiply every outcome by its probability , then add all those products together. Two features of trip students up. First, it is a long-run average, so it need not equal any value the variable can actually take — the expected score on a fair die is , which no single roll can produce. Second, it is not the same as the most likely outcome (the mode); a lopsided distribution can have its mean well away from its most probable value.
Fair games and applications
Expected value is the backbone of decisions under uncertainty — in insurance, finance and gambling. A game is fair when a player's expected gain is zero: over the long run they neither win nor lose. If the variable you are working with is the player's profit (winnings minus stake), fairness means . If instead you are working with the gross winnings, the equivalent condition is that the expected winnings equal the stake, . A positive expected profit favours the player; a negative one favours the house, which is why commercial games are always tilted so the operator's expected profit is positive.
Common mistakes examiners penalise
Forgetting the sum-to-one rule — a distribution with an unknown must be pinned down by first; skipping it loses the opening method mark and every mark that depends on the missing value.
Confusing with — the expected value weights each outcome by its probability, ; it is not the sum of the probabilities (which is always 1) nor the plain average of the outcomes.
Expecting to be a possible outcome — it is a long-run mean and may fall between the values, like for a fair die; do not 'round it to a real outcome'.
Confusing the expected value with the mode — the most likely single outcome is the mode, not the mean; they coincide only by chance.
Mishandling the stake in a fair game — decide whether your variable is profit or gross winnings, then use or accordingly; mixing the two shifts the answer by exactly the stake.
Giving a rounded decimal when an exact form is available — a probability such as or an expected value such as should be left exact unless the question asks for a decimal; premature rounding can lose the accuracy mark.
Writing only a final number with no working — in an analytic scheme the method marks live in the lines above the answer; without them a wrong final value can cost most of the marks.
Model answer — marked the way our engine marks it
In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A), where an A-mark is dependent on the corresponding M-mark being earned. Follow-through (FT) means a correct value carried on from an earlier slip still earns the later marks, and ISW ('ignore subsequent working') means once you have written a correct answer, extra scribble does not lose it. The examples above already carry the [M]/[A] tags line by line; the one below shows the full protocol on a classic 5-mark distribution question, and then spells out exactly how each mark is awarded.
Where this leads
The expected value is the first of the summary numbers that describe a distribution; the variance and standard deviation, which measure spread, are built on the very same weighted-sum idea. The habit you have practised here — tabulate the outcomes, enforce , then weight and sum — reappears directly in the binomial distribution, where the probabilities come from a formula rather than a table, and again in continuous distributions such as the normal, where the sum becomes an integral. Master the routine now and each of those topics becomes a variation on a method you already own.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
A biased four-sided die is rolled and the score is the random variable . Its probability distribution is:
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
(a) Find the value of . (b) Hence find . [4]
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(a) Use the sum-to-one rule. The probabilities must total 1. [M1: set the sum equal to 1] . [A1: ]
The random variable has the probability distribution below.
| 0 | 1 | 2 | 3 | |
|---|---|---|---|---|
Find . [3]
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Apply the expected-value formula. [M1: use ] [M1: correct products substituted] . [A1]
A player pays 5 to play a game and spins a wheel with four sectors labelled 1, 2, 3, 4 whose probabilities are , , and . The winnings are , , and for the four sectors respectively. (a) Find , the expected winnings, in terms of . (b) Given the game is fair, find . [5]
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(a) Write the winnings distribution and apply the formula.
A discrete random variable has for . Find the value of , and hence find . [5]
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Model answer — full working.
How it all connects
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Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
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Quick check
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Revision flashcards
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Discrete random variable
A variable whose value is a numerical outcome of a random experiment, taking a finite or countably infinite set of distinct values — things you count, such as the score on a die or the number of defective items.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
- ✓
Each probability is valid: for every value .
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The probabilities are exhaustive: they cover every outcome and sum to exactly one, .
- ✓
These two facts are the engine of the whole topic — the sum-to-one rule in particular is what lets you find an unknown probability or constant.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Get a Paper 2 calculation marked: find $k$ and $E(X)$ for a distribution with full working
Get a Paper 2 calculation marked: find and for a distribution with full working
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Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do Get a Paper 2 calculation marked: find $k$ and $E(X)$ for a distribution with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.