Probability language, scales and theoretical models
Describe random events precisely, use a probability scale and calculate probabilities from stated equally likely outcomes or angular weights.
Exam scope and assessment notes
The first five lessons are shared Foundation work inherited by Higher; the final four are additional Higher trees, simple conditioning and applications. State the actual random mechanism, equally likely elementary outcomes, complete sample spaces and event membership. Distinguish independence from replacement and mutual exclusivity, observed relative frequency from theoretical probability, and expected frequency from a guaranteed count. Retain every ordered without-replacement history and its remaining contents. Conditional probabilities use the conditioning population; an empty condition is undefined. Inverse object counts are nonnegative integers in the original positive-denominator model. Written reasoning and assumptions still need human review; no marks or mastery are awarded. Specialist and formal-assessment approval remain pending. This is Issue2 linear4MA1; November2026/June2027 and regional-centre eligibility differ from modular awards. Literal Book1 HandlingData1/2/3 and Hub tags are navigation only; no Book2 tag or probability chapter/Hub body is verified. Keep map228–242/proof229–243/next244, map303/proof304–328/next329 and map398–428/proof397/summary426/next428 separate. Missing explicit Hub tags A/B/F/J stay missing. The common elementary two-coin model is not claimed as a novel model. Statistical-measures and diagram prerequisites do not themselves establish probability coverage.
Before you begin
- fractions and ratios
- angles and percentages
- ordering numbers from0to1
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A uniform draw is made from physical cards carrying repeated labels. What is equally likely in this stated model?
What you will learn
- Explain outcomes, events, randomness and equal likelihood; random does not imply equally likely labels.
- Use the scale from 0 to 1 and equivalent fractions/decimals/percentages, including certain/impossible events.
- Calculate model probabilities from equally likely physical outcomes or given sector angle weights; name the assumptions rather than infer fairness from appearance.
A bag has 5 red, 4 blue and 3 green tokens. Each physical token is equally likely to be selected once. Find the three colour probabilities, the probability of purple and the probability of any listed colour. Place the three colour probabilities on a 0–1 scale and identify a certain and an impossible event.
Enter the original physical identities, ordered outcomes, selected populations, weights, observed counts and actual geometry yourself. No space, history, support, denominator, rate, count, unknown proposal or probability is filled. A recorded representation is an attempt, not a correctness result.
Original givens and my literal probability work
Original supplied information
labels
- red
- blue
- green
counts
- 5
- 4
- 3
selection
one uniformly chosen physical tokenThe original physical labels, given weights and stated mechanism are supplied. No completed sample space, tree, event selection, total, relative frequency, expected count, unknown count or requested probability is filled. Counts determine masses; drawn circle areas and tree lengths do not.
Every planar panel shares this uniform camera. Experiment rate positions are categorical layout positions, not a fabricated cumulative trial history. The prediction panel labels its separate trials/expected-events units. Pan, zoom and turn affect the view only.
Empty own drawing plane; no probability representation is supplied
Blue supplied diagrams are schematic; amber records are actual learner work; purple is dirty retained work; dashed work is unready. Native clipping does not clamp, snap or repair points. Exact keyboard tokens and numeric-only free inverse coordinates are separate: a finite inverse click remains actual raw work even when exact-token bounds later reject it. The finite tool uses coordinates±1,000,000, up to40 tree nodes/81 terminals/4 branches,100 grid cells or weights. These are tool limits. No numerical agreement, written-text presence, drawn tree, sample space or reference proves reasoning, grants marks/mastery, or supplies teacher/specialist/formal-assessment or book-body approval.
Place my probability-scale markers
No actual representation is recorded.
The question’s event names are red; blue; green; purple; anyListed. Enter an event row and its actual favourable/conditioning populations yourself. The original replacement or independence mechanism remains the supplied question; the interface never generates remaining counts or conditional weights.
My quantities, conclusions and explanation
Use the original physical integer-count domain; inadmissible roots remain explicit only where requested.
Keep the original event support, probability/conditioning assumptions and requested format. No denominator or answer is inserted.
Keep the original event support, probability/conditioning assumptions and requested format. No denominator or answer is inserted.
Keep the original event support, probability/conditioning assumptions and requested format. No denominator or answer is inserted.
Keep the original event support, probability/conditioning assumptions and requested format. No denominator or answer is inserted.
Keep the original event support, probability/conditioning assumptions and requested format. No denominator or answer is inserted.
A nonblank explanation is an attempt for human review, not approval of reasoning.
These original finite cases are self-checks, not automatic marking, mastery, teacher/specialist/formal-assessment or unavailable book-body approval. Equally likely physical outcomes, independence, replacement and comparable experimental conditions must be justified from each original question. An empty conditioning population is undefined, not probability zero. Expected counts are model predictions, not guaranteed observations. No statistics or modular credit is inferred.
State the outcome and the mechanism
An outcome is one possible result of the random process. An event is the collection of outcomes that meets a stated condition. Describe what is selected, how it is selected and what the event asks for before writing a fraction. Random selection means the result is uncertain in the model; it does not automatically give every displayed label the same probability.
Equally likely elementary outcomes must come from the stated mechanism. Uniform selection among physical objects gives every object the same chance, even when several objects share a label. A fair die supplies equally likely faces. A spinner with a uniform pointer angle instead weights a label by its angular share. A neat sketch, a short label list or the word random alone does not establish equal likelihood.
Measure an event within the whole model
For a finite collection of equally likely elementary outcomes, divide the number belonging to the event by the total number of possible elementary outcomes. Keep repeated physical objects distinct while counting. Only then combine objects that share a requested label. Check that the event count cannot be negative or exceed the whole population; a resulting probability must lie between zero and one.
For a uniform angular pointer, a sector probability is its angle divided by a full turn. Add angular shares if one event contains several sectors. The ratio of two sector probabilities is the ratio of their angles, provided their denominator is the same full turn. These are consequences of the declared angular model, not a rule that any physical spinner must behave perfectly.
Place and interpret the probability
A probability scale has zero for an impossible event and one for a certain event within the stated model. Intermediate values express uncertainty. Convert a fraction to a decimal by division, or to a percentage by multiplying by one hundred, without changing its position on the scale. Label the endpoints and place your own event markers according to their numerical values.
Order probabilities by value rather than by the size of their denominators or labels. Equal probabilities share a position; a larger probability lies farther towards one. A probability is not a promise about the next result, and a model value does not require a matching proportion in every short run. State the fairness, uniformity or angular assumption that supports the calculation.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue2 · Foundation printed24–25/physical30–31; Higher printed39/physical45; inheritance printed27/physical33
- Pearson2026/2027 entry manual · centre regions physical3; Mathematics A physical25
- Pearson second-edition modular book/Hub map · literal navigation only · physical4,6,7
Pause and explain
A bag has 2 silver and 5 black counters, each physical counter equally likely. What is P(black)?
Worked example
A bag has 5 red, 4 blue and 3 green tokens. Each physical token is equally likely to be selected once. Find the three colour probabilities, the probability of purple and the probability of any listed colour. Place the three colour probabilities on a 0–1 scale and identify a certain and an impossible event.
Show the worked solution
- Identify the equally likely physical outcomes or the stated uniform-angle mechanism. Mark the requested event and its whole reference space before forming a probability; draw and label the requested 0–1 scale yourself.
- There are 12 equally likely tokens. The colour event counts are 5, 4 and 3, so the probabilities are 5/12, 4/12=1/3 and 3/12=1/4.
- Purple is absent, giving probability 0. Every token has a listed colour, so that event has probability 1.
- On the scale, 1/4<1/3<5/12; all lie strictly between 0 and 1. A random draw does not make the three colours equally likely.
Answer P(red)=5/12, P(blue)=1/3, P(green)=1/4. P(purple)=0 and P(any listed colour)=1. On the labelled scale the distinct colour markers are green 1/4 < blue 1/3 < red 5/12. Purple is impossible and a listed colour is certain in this model.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A fair eight-sided die is labelled 1 to 8. Find the probabilities of an even score, a score greater than 6, a score greater than 8 and a score at most 8. Give the first two as a fraction, decimal and percentage. Place the resulting distinct event probabilities on a labelled 0–1 scale, including the endpoints.
Give me a hint
Identify the elementary objects or angular shares before counting labels; then place your own numerical results on the scale.
Compare my reasoning
- Identify the equally likely physical outcomes or the stated uniform-angle mechanism. Mark the requested event and its whole reference space before forming a probability; draw and label the requested 0–1 scale yourself.
- The eight scores are equally likely. Four are even, giving 4/8=1/2=0.5=50%.
- Only 7 and 8 exceed 6, giving 2/8=1/4=0.25=25%.
- No score exceeds 8, so that probability is 0. A score at most 8 is certain, giving 1; neither random result is promised on one trial.
P(even)=1/2=0.5=50%; P(score>6)=1/4=0.25=25%; P(score>8)=0; P(score≤8)=1. Place the event markers at 0,1/4,1/2,1 on a labelled 0–1 scale.
Look for these in your work
- List favourable elementary scores.
- Convert each requested probability exactly between fraction, decimal and percentage forms.
- Keep impossible/certain at the scale endpoints.
A spinner has sectors A 45°, B 90°, C 135° and D 90°. Its pointer angle is uniform over a full turn. Find each label probability and compare A with C. Explain why counting four labels as four equally likely outcomes would be wrong. Place all four event probabilities on a labelled 0–1 scale, showing equal markers at the same position.
Give me a hint
Identify the elementary objects or angular shares before counting labels; then place your own numerical results on the scale.
Compare my reasoning
- Identify the equally likely physical outcomes or the stated uniform-angle mechanism. Mark the requested event and its whole reference space before forming a probability; draw and label the requested 0–1 scale yourself.
- A full turn is 360°, so weight each event by its angular share: 45/360, 90/360, 135/360 and 90/360.
- The probabilities simplify to 1/8, 1/4, 3/8 and 1/4. Sector C is three times the angle and probability of A.
- The labels have unequal angular shares. Uniform angle does not mean uniform labels; a sketch that merely looks balanced is not evidence of fairness.
P(A)=1/8, P(B)=1/4, P(C)=3/8 and P(D)=1/4. C has three times A’s probability. Scale order: A 1/8 < B=D 1/4 < C 3/8, between endpoints 0 and 1. The pointer is uniform in angle; the four labels do not have equal angular shares.
Look for these in your work
- Use the supplied angular model.
- Check probabilities total 1.
- Explain why label counting alone fails.
Six cards carry labels 1, 1, 1, 2, 2, 5 and one physical card is chosen uniformly. Find the probability of each possible label and order them on a probability scale. Does random selection make 1, 2 and 5 equally likely?
Give me a hint
Identify the elementary objects or angular shares before counting labels; then place your own numerical results on the scale.
Compare my reasoning
- Identify the equally likely physical outcomes or the stated uniform-angle mechanism. Mark the requested event and its whole reference space before forming a probability; draw and label the requested 0–1 scale yourself.
- Each of the six physical cards is equally likely, but three carry 1, two carry 2 and one carries 5.
- The label probabilities are 3/6=1/2, 2/6=1/3 and 1/6. In increasing order they are 5, 2, 1.
- Randomness describes the selection mechanism; equal probabilities apply to physical cards here, not to the three distinct labels.
P(1)=1/2, P(2)=1/3, P(5)=1/6. Scale order: label 5 at 1/6 < label 2 at 1/3 < label 1 at 1/2. The physical cards are equally likely; their three distinct labels are not.
Look for these in your work
- Distinguish physical outcomes from labels.
- Count repeated physical cards within each requested label event.
- Explain randomness versus equal likelihood.
Common mistakes
- Treating random labels as equally likely without checking their physical weights.
- Dividing by the number of distinct labels when repeated objects are equally likely.
- Placing fractions on a scale by denominator alone.
What can you explain now?
Why can a random choice produce unequal probabilities for different labels?
Compare with the explanation
The labels may correspond to different numbers or weights of elementary outcomes.
Equal likelihood concerns the specified elementary mechanism; combining repeated objects or unequal angular shares can give different label probabilities.
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Make it stick
Explain an outcome, an event, the equally likely mechanism and the zero-to-one position before computing a new model probability.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.