Systematic sample spaces and two successive events
Build complete ordered sample spaces for one and two events, retain repeated physical identities and select event outcomes without double counting.
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
- PR1
- systematic lists and Cartesian grids
- fractions
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Two stages are recorded as (letter, score). What must every ordered pair preserve?
What you will learn
- List all elementary outcomes for a single event and two successive events in a consistent order.
- Represent an event as a subset of the sample space and count favourable outcomes only when elementary outcomes are equally likely.
- Retain order and repeated physical labels; a grid cell must represent a unique physical elementary outcome, not collapse different equally likely outcomes into one.
First list the single-toss sample space and P(H). Then construct the ordered sample space for two independent fair coin tosses. Find the probabilities of exactly one head and at least one head by marking the relevant outcomes.
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
first
- H
- T
second
- H
- T
mechanism
two independent fair coin tossessingleEvent
HThe 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.
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.
Build my ordered outcomes and select the events
No actual representation is recorded.
The question’s event names are exactlyOneHead; atLeastOneHead. 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.
No current explicit planar target is selected. Grid identities, event memberships, row indices, scalar weights and ordered histories use their own symbolic fields; none becomes a fake x/y coordinate.
My quantities, conclusions and explanation
Keep the original event support, probability/conditioning assumptions and requested format. No denominator or answer is inserted.
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.
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.
Fix an order before listing
A sample space lists every elementary outcome permitted by the model. For one event, write each possible elementary result once. For two successive events, choose a fixed coordinate order, such as first result followed by second result, and keep it throughout. Reversing the coordinates can describe a different outcome even when the same two labels appear.
Use a systematic list or a grid to prevent omissions. Hold the first result fixed while listing every second result, then move to the next first result. A grid row and column must refer to the same named stages as the written pair. Check completeness against the number of available combinations, while remembering that counting combinations alone does not establish equal probabilities.
Mark an event as actual outcomes
An event is a subset of the sample space. Read the condition and mark exactly the cells or pairs that satisfy it. Exactly one excludes outcomes satisfying both named properties; at least one includes them. The word or includes the shared outcomes unless the context explicitly states otherwise. A cell that satisfies two descriptions is still one elementary outcome.
When every listed elementary pair is equally likely, the event probability is the number of selected pairs divided by the whole number of pairs. Establish fairness and independence from the stated mechanism before using that equal-pair count. If outcome weights differ, retain the weights rather than treating a convenient list of distinct labels as a uniform space.
Keep physical identity when labels repeat
Different physical cards may carry the same visible value. Uniform card selection makes their physical identities equally likely. Pair each card identity with the second event first, then select all identities that satisfy the requested label condition. Collapsing repeated cards into one label can merge several equally likely outcomes into a heavier single label.
A complete written space is part of the mathematical method, not decoration after the final fraction. Show which outcomes belong to the event and why no outcome was counted twice. Check that disjoint exhaustive events account for the whole space. This lesson uses direct finite outcome counting; general multiplication and conditional tree methods are treated separately at Higher Tier.
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 fair six-sided die and an independent equal two-label spinner are used. How many equally likely ordered elementary pairs are there?
Worked example
First list the single-toss sample space and P(H). Then construct the ordered sample space for two independent fair coin tosses. Find the probabilities of exactly one head and at least one head by marking the relevant outcomes.
Show the worked solution
- Choose the declared coordinate order and construct the complete single-event list and ordered two-event grid. Mark the requested event cells on that original space, retaining physical identities and any shared cell only once.
- One toss has outcomes H, T with P(H)=1/2. For two tosses, keep first and second positions distinct.
- The four equally likely ordered pairs are HH, HT, TH, TT. Exactly one head occurs in HT and TH.
- Thus P(exactly one H)=2/4=1/2. At least one head adds HH, giving 3/4; HT and TH are different sequences.
- A complete grid has rows H,T and columns H,T; its cells are(H,H),(H,T),(T,H),(T,T). Select the two mixed cells for exactly one and every cell except(T,T)for at least one.
Answer Single-toss space {H,T}, with P(H)=1/2. Ordered two-toss space {(H,H),(H,T),(T,H),(T,T)}. Exactly one head selects(H,T),(T,H), giving 1/2; at least one head also includes(H,H), giving 3/4.
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.
List the four single-die outcomes. Construct all ordered pairs (die score, coin result) for a fair four-sided die and an independent fair coin. Find P(H and an even score), and P(die score at least 3).
Give me a hint
Fix the pair order, complete every row, and mark the requested cells before forming a favourable-to-total fraction.
Compare my reasoning
- Choose the declared coordinate order and construct the complete single-event list and ordered two-event grid. Mark the requested event cells on that original space, retaining physical identities and any shared cell only once.
- There are four die outcomes and two coin outcomes; construct each die row with H then T.
- The eight ordered pairs are equally likely under the independent fair model. H with an even score selects (2, H), (4, H).
- That event has probability 2/8=1/4. Scores 3 or 4 select four cells, so the other probability is 4/8=1/2.
- The complete rows are(1,H),(1,T);(2,H),(2,T);(3,H),(3,T);(4,H),(4,T). Mark(2,H),(4,H)for H-and-even and every pair in rows 3 and 4 for score at least 3.
Single-die outcomes {1,2,3,4}. Ordered space {(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T)}. H-and-even selects(2,H),(4,H), so P=1/4. Scores at least 3 select the four pairs in rows 3 and 4, so P=1/2.
Look for these in your work
- Construct all eight cells once.
- Keep the coordinate order.
- Count each event independently.
An equal-sector red/blue/green spinner is used with an independent fair die labelled 1 to 3. List the single-spinner sample space, then all nine ordered pairs. Find P(green or a die score of 1) by explicitly selecting cells without counting any cell twice. Assume the spinner pointer angle is uniform over a full turn.
Give me a hint
Fix the pair order, complete every row, and mark the requested cells before forming a favourable-to-total fraction.
Compare my reasoning
- Choose the declared coordinate order and construct the complete single-event list and ordered two-event grid. Mark the requested event cells on that original space, retaining physical identities and any shared cell only once.
- The spinner sample space is red, blue, green. Each can be paired with 1, 2 or 3, giving nine equally likely pairs.
- Green selects (green, 1), (green, 2), (green, 3); score 1 adds(red, 1), (blue, 1), since (green, 1) is already selected.
- There are five selected cells out of nine, so the probability is 5/9. This is direct event counting, not an unsupported addition of overlapping event probabilities.
- The complete rows are(red,1),(red,2),(red,3);(blue,1),(blue,2),(blue,3);(green,1),(green,2),(green,3). The requested union selects the entire green row and the score 1 column, with their shared cell once.
Single-spinner space {red,blue,green}. Ordered space {(red,1),(red,2),(red,3),(blue,1),(blue,2),(blue,3),(green,1),(green,2),(green,3)}. Green-or-1 selects all three green pairs plus(red,1),(blue,1), giving 5/9; (green,1) is counted once.
Look for these in your work
- Complete the 3×3 sample space.
- Show the shared cell only once.
- Justify equal likelihood before dividing.
Four physical cards a, b, c, d carry labels 0, 0, 1, 2. A card is chosen uniformly and an independent fair coin tossed. List the single-card physical outcomes and all eight ordered pairs (card ID, coin result). Find P(label 0 and H). Explain why the six distinct (label, coin) pairs are not equally likely.
Give me a hint
Fix the pair order, complete every row, and mark the requested cells before forming a favourable-to-total fraction.
Compare my reasoning
- Choose the declared coordinate order and construct the complete single-event list and ordered two-event grid. Mark the requested event cells on that original space, retaining physical identities and any shared cell only once.
- The single physical-card outcomes are a, b, c, d; pair each with H and T to produce eight equally likely physical pairs.
- The requested event contains (a, H) and (b, H), so its probability is 2/8=1/4.
- Collapsing a/b into label 0 leaves only six distinct label pairs but gives (0, H)two elementary outcomes and (1, H)only one. Those six labels therefore cannot be counted with equal weights.
- The eight physical pairs are(a,H),(a,T),(b,H),(b,T),(c,H),(c,T),(d,H),(d,T). The event uses(a,H)and(b,H); collapsing a/b gives the 0 label pair double the weight of either nonzero-label pair.
Physical single-card outcomes {a,b,c,d}; ordered space {(a,H),(a,T),(b,H),(b,T),(c,H),(c,T),(d,H),(d,T)}. Label 0-and-H selects(a,H),(b,H), so P=1/4. The six distinct label/coin pairs are not equally likely: each 0-labelled pair merges two physical outcomes, whereas each 1- or 2-labelled pair merges one.
Look for these in your work
- Retain physical card identities.
- Select both zero-labelled H outcomes.
- Explain the weight lost by collapsing duplicates.
Common mistakes
- Omitting a mixed order because its reversed labels already appear.
- Counting a shared cell twice in an or event.
- Collapsing repeated physical cards and then assuming equal label-pair weights.
What can you explain now?
Why can a list of distinct value pairs be unsuitable for equally likely counting when physical cards repeat a value?
Compare with the explanation
Some value pairs merge more physical outcomes than others and therefore have larger probability weights.
Retain the equally likely physical identities, or attach correct weights after merging. Distinct labels alone do not establish equal likelihood.
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Make it stick
Construct a systematic sample space, explain its elementary weights and show the selected event cells before finding a 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.