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Probability fairness relative frequency and models - Mathematics Year 7

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Introduction

Does counting two colour names make each colour equally likely? State the possible outcomes, check their probabilities and distinguish a theoretical model from what happens in a short experiment.

You will calculate six simple probabilities and evaluate a fairness claim using explicitly fictional trial counts. Then compare mutually exclusive alternatives with two successive events. Begin with the chance of an even result on a fair die numbered 1 to 6.

Define events and check six simple chances

A probability model assigns numbers between 0 and 1 to possible events. Zero means impossible under the model; one means certain under it. The sample space lists the possible elementary outcomes. For a fair six-sided die numbered 1 to 6, we assume each face has probability 1/6. The chance of an even result is $P (e v e n) = \frac{3}{6} = \frac{1}{2}$ , because 2, 4 and 6 are the three favourable faces.

The favourable-outcomes-over-total rule requires equally likely elementary outcomes. If a bag contains three red counters and one blue, randomly selecting one counter does not make the two colour names equally likely: there are four equally selectable counters but three belong to red. Counting category labels instead of elementary outcomes would answer a different question.

Six original probability calculations.

Event and stated model	Probability	Reason
Even on a fair 1–6 die	1/2	Three of six equally likely faces
Above 4 on the same die	1/3	Two faces: 5 and 6
Result 7 on the same die	0	No such face in this model
A result from 1 to 6 on that die	1	Every possible face is included
Not red if red has probability 0.3	0.7	Complement of the red event
Other if red 0.3 and blue 0.45 are disjoint and exhaustive with other	0.25	Remaining probability after red and blue

The complement of red is every outcome outside red, so their probabilities sum to 1. For the last row, we explicitly partition all outcomes into red, blue and other, with no overlap. Therefore $P (o t h e r) = 1 - 0.30 - 0.45 = 0.25$ . If categories overlap or fail to cover all possible outcomes, subtracting their displayed values in this way may not represent the named event.

A four-sector model. Imagine numbering the four equal sectors in a copy of the source 1, 2, 3, 4 and assume a fair central spinner. Each sector has probability 1/4. If three sectors are assigned red and one blue, red has probability 3/4 and blue 1/4. The unmodified orange source does not already show those labels or colours; draw the model and state the fair-mechanism assumption yourself.

Probability has no physical unit like centimetres. It may be expressed equivalently as a fraction, decimal or percentage; one half is 0.5 or 50%, not 50 on the 0–1 scale. Keep the event and model attached to each number so a calculation remains interpretable.

Distinguish alternatives successive stages and models

Alternatives need a rule about overlap. Events are mutually exclusive when they cannot both occur on the same trial. On one die roll, “result 1” and “result 2” are disjoint, so the probability of either is 1/6+1/6=1/3. This addition is justified by the lack of shared outcomes, not simply by seeing the word “or”.

Even and a multiple of 3 are not mutually exclusive: 6 belongs to both. Their union is{2, 3, 4, 6}, so $P (e v e n o r m u l t i p l e o f 3) = \frac{4}{6} = \frac{2}{3}$ . Adding 1/2 and 1/3 without removing the shared 6 would give 5/6 and count that face twice. Listing outcomes is especially reliable for a small finite sample space; set reasoning explains the overlap.

Two draws with replacement. An original bag model contains 3 red and 2 blue counters, each equally selectable. Replace and remix after the first draw. With independent random draws, the chance of red twice is $P (R R) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$ . A two-stage outcome tree or 25 equally likely ordered counter pairs provides a second representation. Nine pairs are red-red.

Without replacement, the second red probability changes after a first red: only 2 red remain among 4 counters. Then $P (R R) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ . This multiplication uses the conditional second chance; the two draws are not independent. Replacement is an assumption that changes the problem, not a decorative phrase that can be ignored.

An unshaded ten-by-ten grid. — Use a copy to represent100equally weighted model outcomes or100recorded trials,but declare which interpretation you mean. The source has no pre-coloured probabilities or actual trial results already filled in.

For a model with probabilities 0.30, 0.45 and 0.25, the grid can represent expected shares 30, 45 and 25 out of 100. Those are model-based quantities, not a claim that every real run of 100 must contain exactly those counts. A grid of observed frequencies and a theoretical share diagram can look alike while describing different evidence.

Use Corbettmaths’ original probability explanation to identify favourable and possible outcomes. Then apply the separately stated die, spinner and bag assumptions here; verify the difference between colour categories and equally likely individual outcomes. Original: corbettmaths, Probability - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Choose a list for few elementary outcomes, a tree for successive stages and a labelled share model for complements. Explain the assumptions that make each representation valid before applying a formula.

Evaluate relative frequencies and fairness claims

Relative frequency is an observed count divided by the number of trials. Suppose a fictional 100-trial record has 28 red, 47 blue and 25 other outcomes. The relative frequencies are 0.28, 0.47 and 0.25, and they sum to 1. Compared with the stated 0.30/0.45/0.25 model, the deviations are −0.02, +0.02 and 0. These are invented practice counts, not a recorded experiment performed by you or by the course author.

A difference from theoretical probability is normal in random samples and does not alone prove bias. Equally, one close match does not prove that a mechanism is fair or that every assumption is correct. More controlled trials can provide stronger evidence and often stabilise relative frequencies, but they do not guarantee exact theoretical proportions in every finite run.

Evaluate a fairness claim. A learner sees 7 heads in 10 hypothetical tosses and says a coin cannot be fair because it did not give 5 heads. The model predicts equal chance on each fair independent toss, not a forced balance after ten. State the limited sample and possible random variation. Check whether the procedure itself was consistent before interpreting more data; do not claim a definitive statistical bias test that has not been carried out.

The word fair also needs a declared meaning. Our die model means equal face chances; the three-colour model intentionally has unequal colour probabilities. Testing agreement with that declared colour model is different from demanding equal counts for every colour. A game’s payout fairness would require its reward rules as well, which are not specified by a face-probability calculation alone.

Unfamiliar transfer and error clinic. A bag has 2 green and 3 yellow counters. One uniform draw gives green 2/5, not 1/2 just because there are two colour names. The chance of not green is 3/5. For two draws with replacement, green twice is 4/25; without replacement it is 2/5×1/4=1/10. Explain the changed denominator and check by enumerating ordered individual-counter pairs.

A learner gives a probability 1.2; it fails the 0–1 constraint and needs investigation before interpretation. Another adds probabilities of successive red draws when the task requires both; a correctly labelled tree shows why the joint chance uses the stage probabilities. Retain all six simple results, a model-versus-frequency comparison, a defended fairness conclusion and an actual later retry with its assumptions.

Interactive Tasks

Quiz: Test Your Knowledge

What is the chance of even on a fair 1 to 6 die? (1/2) (!1/3) (!2/3) (!3)

What is the chance of above 4 on that die? (1/3) (!1/2) (!2/5) (!4/6)

What is the chance of rolling 7 on that stated die? (0) (!1) (!1/7) (!1/6)

If red has chance 0.3, what is the chance of not red? (0.7) (!0.3) (!1.3) (!0.03)

What remains for other in the disjoint exhaustive 0.3 red/0.45 bluemodel? (0.25) (!0.75) (!0.15) (!0.55)

What is the chance of even or multiple of 3 on that die? (2/3) (!5/6) (!1/6) (!1/3)

What is red twice from 3 red/2 bluewith replacement and independence? (9/25) (!3/10) (!6/5) (!3/5)

What is red twice without replacement from that bag? (3/10) (!9/25) (!6/5) (!1/5)

What is the observed red relative frequency in the fictional 100-trialrecord? (0.28) (!0.30) (!28) (!0.72)

What follows from 7 heads in only 10 hypotheticaltosses? (The short sample alone does not prove bias) (!A fair coin must give exactly 5 heads) (!The next three tosses must all be tails) (!The theoretical head chance must equal 0.7)

Memory Game

Probability	Modelled chance on a zero-to-one scale
Outcome	Possible result of an experiment
Complement	All outcomes outside a given event
Exclusive	Events that cannot both occur on one trial
Independent	One event does not alter the other event’s probability
Frequency	Count of recorded occurrences

Drag and Drop

1/2	Fair die even chance
1/3	Fair die above 4 chance
0	Chance of 7 on the stated 1–6 die
1	Chance of any result in the full set 1 through 6
0.25	Other chance in the partitioned colour model

Crossword Puzzle

OUTCOME	Possible elementary result
COMPLEMENT	Everything outside the stated event
EXCLUSIVE	Cannot occur together on one trial
INDEPENDENT	One event leaves the other chance unchanged
FREQUENCY	Recorded occurrence count
MODEL	Declared assumptions and probability assignments

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

List die outcomes: State all six equally likely faces and identify the even ones.
Complete six simple cases: Use the six original table events with an explanation each.
Translate probability formats: Show why 1/2, 0.5 and 50% describe the same chance.
Declare a spinner model: Label four equal sectors and distinguish sector chance from colour chance.

Standard

Use complements and partitions: Explain why the exhaustive colour model leaves 0.25 forother.
Compare observed and theoretical shares: Compute 28/100, 47/100 and 25/100 and compare with the stated model.
Correct an overlapping addition: List the union of even and multiples of 3 without counting 6 twice.
Evaluate the short-run claim: Explain why 7 heads in 10 does not alone establish bias or force future tails.

Advanced

Compare replacement assumptions: Use a two-stage representation to obtain 9/25 versus 3/10 forredtwice.
Transfer to another bag: Obtain 2/5, 3/5, 4/25 and 1/10 for the green/yellow example.
Distinguish meanings of fairness: Separate equal face chances, agreement with a nonuniform model and an unspecified payout rule.
Retrieve and defend later: Keep an actual later solution with event definition, model, representation and a corrected first error.

Learning Check

Six simple probabilities: Calculate all six original event probabilities.
Models and observed frequencies: Compare the fictional 100-trialrecord with the colour model.
Alternatives and overlap: Find even or multiple of 3 on one die roll.
Successive stages: Compare two red draws with and without replacement.
Unfamiliar transfer and errors: Solve the 2 green/3 yellowcase and correct category-count reasoning.
Fairness and later evidence: Give a limited conclusion and retain actual later work.

Worked checks and assessment guidance

In order 1/2, 1/3, 0, 1, 0.7, 0.25. State equally likely die faces, complement and disjoint exhaustive colour categories; probabilities stay within 0–1.
28/100=.28, 47/100=.47, 25/100=.25, sum 1. Differences−.02, +.02, 0 from.30/.45/.25 are sample deviations, not proof of fairness or bias; no real experiment is claimed.
Union 2, 3, 4, 6 has 4 of 6 outcomes, giving 2/3. Naive 1/2+1/3 counts 6 twice. Addition without subtracting overlap needs mutually exclusive events.
Replacement/independencegives 3/5×3/5=9/25; withoutreplacementafterredthereare 2 redin 4, giving 3/5×2/4=3/10. State stage assumptions rather than adding two draw chances.
Green 2/5, notgreen 3/5; bothgreenwithreplacement 4/25, withoutreplacement 1/10. Two colour labels do not mean equal colour chances, and 1.2 cannot be a probability.
Seven hypothetical heads in ten does not force exact balance or future tails. State assumptions, sample limitations and what fairness means. Retain six calculations, an appropriate representation and genuine later retrieval.

Learning evidence

Keep six simple probabilities, a model/frequency comparison, a stage-aware bag calculation and a defended limited fairness conclusion. Actual experiments and later retrieval require actual retained observations.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, probability,statistics,geometry and mathematical reasoning. Year 7 sequencing and the survey enrichment are editorial within KS3. All course examples,models,fictional data and worked answers are original. Actual Commons sources retain their creator/licence credits below; original-publisher videos supplement the written explanations. Reviewed 12 November 2026.

PieChartFractionFourthsSplit.svg, Ezra Katz, CC BY-SA 3.0. Unchanged original; caption and learning activities are original.

100 grid.PNG, The original uploader was Leightwing at English Wikiversity., Public domain. Unchanged original; caption and learning activities are original.