Probability: enumerate sets and combinations of sets systematically, using tables, grids, Venn diagrams and tree diagrams

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Overview

Probability questions become manageable when every possibility has a clear place. This topic teaches you to enumerate, meaning list completely, the outcomes of choices or events without missing or repeating any. You will use systematic lists, sample-space grids, two-way tables, Venn diagrams and tree diagrams. These are different tools for organising information, not different rules of probability.

In an examination, candidates may be asked to list all possible codes, complete a table of dice outcomes, sort people into overlapping sets, or draw branches for successive events. A correct diagram makes your reasoning visible and provides a reliable starting point for calculating probabilities. The key question is always: have I represented every allowed outcome exactly once?

This work connects to fractions, percentages, counting, algebra and interpreting data. It also prepares you for combined-event and conditional probability questions. Your aim is not merely to remember a picture, but to select the right representation and explain why it is complete. Read the restrictions first, organise the information, and check the total before calculating. That routine prevents many avoidable errors under time pressure.

Specification and tier: Pearson Edexcel GCSE Mathematics 1MA1, primarily probability statement P6, with applications to P7 and P8. The core enumeration skills in this guide are relevant to both Foundation and Higher. P9 conditional probability is Higher-only. 1

Marking note: All worked and practice questions here are original Edexcel-style teaching questions. Their exact allocations are illustrative rubrics, not reproduced Pearson mark schemes. Official marks awarded depend on the published scheme for the particular question.

Key Concepts

Concept 1: A complete, systematic list

An outcome is one possible result. A sample space is the set of all possible outcomes. An event is a selection of outcomes meeting a condition. For a die, the sample space is {1, 2, 3, 4, 5, 6}; the event 'even' is {2, 4, 6}.

To list two choices systematically, hold the first choice fixed while the second cycles through every allowed value. Then change the first choice and repeat. With tops R and B and trousers K, G and W, write RK, RG, RW, BK, BG, BW. The visible pattern helps candidates check completeness.

Why this works: each block exhausts the possibilities associated with one first choice. Different blocks have different first choices, so they cannot duplicate one another. There are 2 × 3 = 6 outfits when every pairing is allowed.

Restrictions change the list. For two-digit codes using 1, 2 and 3 without repetition, the complete list is 12, 13, 21, 23, 31, 32. There are 3 first choices and 2 remaining second choices. Do not include 11, 22 or 33.

Check whether order matters. The code 12 differs from 21 because the positions have different roles. A team consisting of two named students is normally unchanged when their names are reversed. For an unordered team from A, B and C, list AB, AC and BC, not six ordered pairs. You do not need a factorial formula for these small enumeration tasks.

Cover and recall: hide the code list. Rebuild it, then explain why it contains six rather than nine outcomes.

Concept 2: Sample-space grids preserve the pairs

Use a sample-space grid when one outcome from each of two choices is combined. Label the rows with one choice and the columns with the other. Every cell records the outcome determined by its row and column. A coin and a six-sided die produce a 2-by-6 grid.

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The twelve outcomes are H1 through H6 and T1 through T6. 'Heads and an even number' selects H2, H4 and H6. For a fair coin and fair die acting independently, all twelve pairs are equally likely, so P(heads and even) = 3/12 = 1/4.

An outcome grid may contain pairs, sums or products according to the question. State what each cell represents. For two distinguishable fair dice, there are 36 ordered pairs. If the cells show sums, several cells contain 7, but only one contains 2. The sums are not equally likely merely because the pairs are.

Examiner focus: candidates who list the eleven totals 2 to 12 and use 11 as the denominator have not represented the equally likely outcomes. The correct denominator is 36. A total of 7 has six successful cells, giving 6/36 = 1/6.

What if? If the coin is biased, the grid still lists possible pairs, but counting successful cells is no longer sufficient unless the pairs have equal probabilities.

Concept 3: Two-way frequency tables count a population

A two-way frequency table classifies people or objects by two properties. Unlike an outcome grid, each cell contains a frequency, not a single trial outcome. Totals at the ends of rows and columns are sums of the relevant cells.

For 30 students, suppose 18 are in A, 14 are in B, and 8 are in both. With columns B, not B and total, the A row is 8, 10, 18. The not-A row is 6, 6, 12. The totals row is 14, 16, 30.

Build the table using subtraction: A but not B is 18 − 8 = 10; B but not A is 14 − 8 = 6. The remaining frequency is 30 − (8 + 10 + 6) = 6. Check both directions. Each row sum must match its row total, each column sum its column total, and the grand total must agree.

When a student is chosen uniformly at random from the whole class, P(A and B) = 8/30. The denominator is the size of the selection pool, not the largest row total. This is the same information you will place in a Venn diagram next.

Concept 4: Venn diagrams separate whole sets from regions

The rectangle represents the universal set, meaning everything under consideration. A circle represents a set of items with a stated property. The overlap is the intersection, written A ∩ B. Each person or item belongs to exactly one disjoint region.

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Always fill a two-set count diagram from the middle outwards. Put 8 in the overlap, then 18 − 8 = 10 in A only, and 14 − 8 = 6 in B only. The union contains 10 + 8 + 6 = 24, so neither is 30 − 24 = 6.

A set total is not an 'only' value. Writing 18 in A only would incorrectly imply that A contains 18 + 8 = 26 members. Visible subtractions show the intended method, although the particular examination scheme determines whether method credit is available.

Read the words precisely: A ∪ B means A or B or both; A ∩ B means both. 'Exactly one' includes A only and B only but excludes the overlap. 'At least one' means the union. 'Neither' means outside the union. A′ means not in A, so it includes B only and neither.

For this diagram, n(A ∩ B) = 8, n(A ∪ B) = 24, n(A′) = 12 and the number in exactly one set is 16. A prime means complement, while n means number of elements. A ∩ B′ describes A only.

Some questions require individual elements rather than frequencies. If the universal set is {1, 2, 3, 4, 5, 6}, A is the even numbers and B is numbers greater than 3, put {4, 6} in the overlap, {2} in A only, {5} in B only, and {1, 3} outside. Do not replace elements with their count unless requested.

Why subtract the overlap once? Adding the two whole-set totals counts every overlap member twice. Subtracting one copy leaves each union member counted once.

Concept 5: Trees show complete sequences

Choose a tree when events happen in stages. Draw one first-stage branch for each possible result. From every first-stage endpoint, draw all allowed second-stage branches. Read one complete path from the start to a final endpoint as one combined outcome.

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Two coin tosses produce HH, HT, TH and TT. HT and TH remain different sequences, even though both contain exactly one head. Forgetting the second-stage branches after T would miss half the outcomes.

A tree can enumerate outcomes without probabilities. Add branch probabilities when required. For independent fair tosses, each branch is 1/2 and each full path has probability (1/2) × (1/2) = 1/4. Exactly one head uses HT or TH, so add their probabilities: 1/4 + 1/4 = 1/2.

The branching count is a check, not a substitute for drawing the requested tree. With two possible results at each of three unrestricted stages, there are 2 × 2 × 2 = 8 paths. If a rule forbids a continuation, omit that continuation and count the surviving endpoints.

Branch check: probabilities on the alternative branches leaving the same point add to 1. Do not add probabilities from unrelated parts of the tree. Independent events have unchanged later-stage probabilities. Dependent events may have different later-stage probabilities, for example when a counter is removed without replacement. Use the remaining counts, not the original bag total.

Mathematical Relationships

The supplied formulae below refer to Pearson's 2026 exam aids. Check the aid for your own examination series if taking a later exam. Here, N is the universal-set count, n(A) is the number in A, and P(A) is the probability of A. 2 3

Must memorise: multiplication principle. Total complete outcomes = m × k for m first choices and k available second choices for each first choice. For more stages, multiply the successive numbers of choices. If the branch counts vary, count each branch separately and add.

Must memorise: equally likely probability. P(E) = number of outcomes in E / total number of outcomes. Use only when the underlying outcomes are equally likely.

Must memorise: two-set counts. n(A only) = n(A) − n(A ∩ B); n(B only) = n(B) − n(A ∩ B); n(A ∪ B) = n(A) + n(B) − n(A ∩ B); n(neither) = N − n(A ∪ B). Also, n(A′) = N − n(A), and n(exactly one) = n(A) + n(B) − 2n(A ∩ B).

Must memorise: complements and totals. P(A′) = 1 − P(A). Probabilities of a complete set of mutually exclusive outcomes add to 1. Every probability lies between 0 and 1 inclusive.

Must memorise: tree rules. Multiply the stage probabilities along a path. Add probabilities of distinct successful paths. For independent A and B, P(A and B) = P(A)P(B). For disjoint events, P(A or B) = P(A) + P(B).

Given on formula sheet, both tiers, 2026: P(A or B) = P(A) + P(B) − P(A and B). This inclusive-or rule subtracts the overlap once. 2 3

Given on formula sheet, Higher, 2026: P(A and B) = P(A given B)P(B). 3

Must memorise, Higher-only formal extension: P(A | B) = P(A ∩ B)/P(B), provided P(B) > 0. The symbol | means 'given'. In count form this is n(A ∩ B)/n(B). This rearranges the supplied Higher relationship and explains why the denominator changes when selection is restricted to B. It is beyond core P6 enumeration.

Practical Applications and Data Skills

Meal choices, clothing combinations and short access codes make systematic enumeration memorable. Surveys about clubs or subjects use tables and Venn diagrams because a person can have both properties. Successive quality checks or repeated coin tosses are naturally represented by trees. In every case, define the allowed outcomes before counting.

There is no required practical for this GCSE Mathematics topic. You can test the ideas by tossing two coins repeatedly, but observed frequencies are estimates and need not equal theoretical probabilities in a small sample.

No coordinate graph is required. The relevant data skills are reading row and column headings, checking marginal totals, matching regions to wording and tracing tree paths in order. Circle areas and branch lengths do not indicate probability unless the question explicitly says so. Use the written counts or labels.

Probabilities have no units. Keep an exact fraction where possible. To convert a probability to a percentage, multiply by 100: 1/4 = 0.25 = 25%. Never write 0.25% for one quarter. If counts represent people, use whole numbers and check that no region is negative.

Exam Routine and Active Revision

Use read, represent, calculate, check. Read the command word and restrictions. Represent every outcome or disjoint region. Calculate from the correct event and selection pool. Check completeness, totals and whether the answer is plausible.

Each Edexcel paper has 80 marks and lasts 90 minutes. Paper 1 is non-calculator; Papers 2 and 3 allow a calculator. 1 Allow about 1.1 minutes per mark, including checking, as a flexible pacing guide. A six-mark question should normally receive about seven minutes before you decide whether to move on.

Cover and recall: reconstruct the four Venn regions from the totals 30, 18, 14 and 8. Then say which regions form the union, intersection and complement of A. Explain your reason before checking the diagram.

After a short delay, redraw the coin tree without looking and identify the exactly-one-head paths. Revisit missed ideas the next day and again later in the week. Correcting a recalled mistake is more useful than repeatedly reading a solution that already looks familiar.

Ten-Minute Tutor Podcast

Listen once with the diagrams visible, then replay the recall quiz with the guide covered. The episode follows a one-minute introduction, five minutes of core concepts, two minutes of exam tips, a one-minute recall quiz, and a one-minute summary.

Source Notes

In Pearson's June 2023 Foundation Paper 1, Question 24 carries 5 marks: B3 for the fully correct Venn diagram, then M1 and A1 for the probability of B′, with answer 7/10. Question 31 carries 2 marks: M1 for 0.2 × 0.3, then A1 for 0.06. These are verified official allocations, unlike this guide's illustrative rubrics. B denotes independent accuracy credit, M method credit and A accuracy credit dependent on method. 4 5