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How do you describe the probabilities of all outcomes at once?

Probability distributions, the discrete uniform distribution, the binomial distribution, and expected values.

A focused answer to AQA GCSE Statistics on probability distributions, covering what a probability distribution is, the discrete uniform distribution, the binomial distribution, and calculating expected values.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
What a probability distribution is
The discrete uniform distribution
The binomial distribution
Expected value

What this dot point is asking

AQA wants you to understand what a probability distribution is, work with the discrete uniform distribution, recognise and use the binomial distribution, and calculate expected values. At GCSE Statistics level the binomial is treated through tree-diagram reasoning rather than the full formula, so the focus is on conditions and short multiplications.

What a probability distribution is

A distribution can be shown as a table, a bar chart or a formula. The "sums to $1$ " property is the workhorse of exam questions: it lets you find a missing probability by subtracting the known ones from $1$ , as in the spinner question above. It also lets you check your own table for arithmetic slips before reading off any probability.

The discrete uniform distribution

Uniform distributions describe fair mechanisms: a fair coin ( $n = 2$ ), a fair die ( $n = 6$ ), or drawing one named card position at random. Recognising uniformity lets you replace counting with a single fraction $\frac{1}{n}$ , and the expected value of a uniform distribution is just the ordinary mean of the outcomes. A discrete uniform distribution can be shown as a bar chart in which every bar is the same height, which is the visual signature of "all outcomes equally likely". This is also the model you implicitly assume when you say a die or spinner is fair: any departure from equal probabilities (a biased die) means the distribution is no longer uniform, and you would then need experimental probabilities instead of the theoretical $\frac{1}{n}$ .

The binomial distribution

The four binomial conditions, often summarised as a checklist, are: a fixed number of trials $n$ ; each trial has only two outcomes; the trials are independent; and the probability of success $p$ is constant. For example, the number of heads in $10$ tosses of a biased coin is binomial, but the number of red counters drawn without replacement is not, because $p$ changes each draw. At this level you find probabilities of "all successes" or "all failures" by multiplying along a tree: the chance of $n$ successes is $p^n$ , and of $n$ failures is $(1 - p)^n$ .

Expected value

Expected value of a game

A game pays $\pounds 0$ , $\pounds 2$ or $\pounds 5$ with probabilities $0.5$ , $0.3$ and $0.2$ respectively. We want to find the expected payout per game, which tells us whether the game is worth playing.

Step 1: Confirm the probabilities sum to 1

Before calculating, check the distribution is valid. A complete probability distribution must always sum to $1$ :

0.5 + 0.3 + 0.2 = 1 \checkmark

Step 2: Multiply each outcome by its probability

The expected value formula $E(X) = \sum x\,P(x)$ weights each payout by how likely it is. More probable outcomes count more in the average:

0 \times 0.5 = 0, \quad 2 \times 0.3 = 0.6, \quad 5 \times 0.2 = 1.0

Step 3: Sum the products

Adding the weighted payouts gives the long-run average outcome per game:

E(X) = 0 + 0.6 + 1.0 = \pounds 1.60

Step 4: Interpret in context

If the game costs $\pounds 2$ to play, the player loses $40$ p per game on average, so over many games it is not worth playing. The expected value is a long-run average, not the guaranteed result of any single game.

Final answer: the expected payout is $\pounds 1.60$ per game.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20214 marksA biased four-sided spinner has scores

1, 2, 3, 4

with probabilities

0.4, 0.3, 0.2

and

p

. (a) Find

p

. (b) Calculate the expected score for one spin.

Show worked answer →

(a) Probabilities sum to $1$ : $0.4 + 0.3 + 0.2 + p = 1$ , so $p = 1 - 0.9 = 0.1$ .

(b) $E(X) = \sum x\,P(x) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 0.4 + 0.6 + 0.6 + 0.4 = 2.0$ .

Markers reward using "probabilities sum to $1$ " to find $p$ , then multiplying each score by its probability and summing for the expected value.

AQA 20193 marksA coin is biased so that

P(\text{head}) = 0.6

. It is tossed

3

times. The number of heads follows a binomial distribution. Calculate the probability of getting exactly

3

heads, and state the two conditions that make this binomial.

Show worked answer →

For $3$ heads in $3$ independent tosses with constant $p = 0.6$ : $P(3 \text{ heads}) = 0.6^3 = 0.216$ .

Binomial conditions: a fixed number of independent trials, and each trial has two outcomes with the same probability of success.

Markers reward $0.6^3 = 0.216$ (multiplying along the all-heads branch) and naming the fixed-independent-trials and constant- $p$ conditions.

Related dot points

Sources & how we know this

AQA GCSE Statistics (8382) specification — AQA (2017)