Higher: independent events and tree diagrams
Construct complete Higher two- and three-stage trees for explicitly independent events, multiply along paths and add disjoint successful paths.
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
- shared PR1–PR5
- multiplication of fractions
- disjoint final paths
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which condition permits unchanged second-stage probabilities on an independent tree?
What you will learn
- Draw complete two- and three-stage trees with labelled outcomes and branch probabilities.
- Use explicitly independent events: later branch probabilities remain unchanged; multiply along paths and add mutually exclusive final paths.
- Distinguish independent from mutually exclusive: positive-probability disjoint events are not independent; replacement alone needs uniform remixing and a stated independent sampling mechanism.
A fair coin and an independent equal-sector red/blue/green spinner are used. Draw the complete two-stage tree. Find P(H and red) and P(exactly one of “H” and “red”), showing the disjoint paths used. Assume the spinner pointer angle is uniform over a full turn.
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
stages
- labels
- H
- T
- probabilities
- 1/2
- 1/2
- labels
- red
- blue
- green
- probabilities
- 1/3
- 1/3
- 1/3
independent
trueThe 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.
Construct my complete labelled probability tree
No actual representation is recorded.
The question’s event names are HAndRed; exactlyOne. 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.
Choose an explicit my unknown-count proposal statement.
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.
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.
Build every branch before calculating
A tree records successive stages from left to right. Start with every possible outcome at the first split, then add every possible next-stage outcome after each first-stage history. Label the stages and actual branch probabilities. For a two-outcome split, a stated success probability supplies its failure complement; the complete set of outgoing probabilities at each reachable split must total one.
For independent events, the next-stage probability does not change with the earlier history. Repeat that same next-stage distribution after each earlier branch. Fairness or independence must belong to the model, not to the visual symmetry of a tree. Branch lengths and angles organise the drawing; they do not encode numerical probability unless a separate convention explicitly says so.
Multiply within a path and add between paths
A terminal path represents one ordered sequence of stage outcomes. For independent stages, multiply the branch probabilities along that path. Keep each stage in its own position, even if the same success or failure label appears several times. A complete three-stage two-outcome tree has all its ordered terminal histories; do not omit an inconvenient mixed order.
Different complete terminal paths are mutually exclusive, so add their path probabilities when they satisfy the same requested event. Exactly two successes includes a separate path for failure at each possible stage. At least one success can use the complement of the all-failure path once that path probability is correct. Check all terminal probabilities together account for the whole model.
Keep the independence assumption visible
Replacing an object restores the possible contents, but replacement alone does not prove every later choice is independent. The task must also specify a uniform remixed sampling mechanism or explicitly independent draws. Under that model, later branch probabilities are unchanged. A biased mechanism or dependence through an external condition could produce different joint probabilities despite the same physical contents.
Independence and mutual exclusivity are different conditions. Two independent positive-probability events can occur together, while disjoint positive-probability events cannot be independent. In an application, identify a realistic shared influence that would challenge independence rather than treating the product calculation as an unconditional real-world guarantee. The tree is a mathematical model and still requires written reasoning.
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
Two explicitly independent events have probabilities 5/6 and 3/7. What is the probability that both occur?
Worked example
A fair coin and an independent equal-sector red/blue/green spinner are used. Draw the complete two-stage tree. Find P(H and red) and P(exactly one of “H” and “red”), showing the disjoint paths used. Assume the spinner pointer angle is uniform over a full turn.
Show the worked solution
- Draw the complete successive-stage tree with named outcomes at every split. Label the stated independent probabilities and their failure complements; multiply along complete histories and select the disjoint paths meeting the event.
- Put H/T on the first split, each 1/2, then red/blue/green after each branch, each 1/3 under the stated independence.
- H-and-red is one path with product 1/2×1/3=1/6.
- Exactly one named event uses H/blue, H/green and T/red. These are disjoint paths, so their total is 3×1/6=1/2.
Answer Complete coin-first tree: H/T each 1/2; after either, red/blue/green each 1/3. All six terminal paths have probability 1/6. P(H and red)=1/6. Exactly one of H/red selects H/blue,H/green,T/red, giving 1/2.
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 bag has 3 red and 2 blue tokens. Two uniform draws are made, replacing and remixing after the first. Draw the full tree and find P(different colours) and P(at least one red). Explain why the second split is unchanged. Use the explicitly independent uniform remixed model for the two draws.
Give me a hint
Draw all stage histories with their success/failure complements, multiply within each requested path and add only disjoint completed paths.
Compare my reasoning
- Draw the complete successive-stage tree with named outcomes at every split. Label the stated independent probabilities and their failure complements; multiply along complete histories and select the disjoint paths meeting the event.
- The first draw has red 3/5 and blue 2/5. Replacement plus the explicitly independent remixed mechanism restores the same branch probabilities.
- The different-colour paths are RB=3/5×2/5 and BR=2/5×3/5, giving 12/25.
- At least one red includes every path except BB, so 1−4/25=21/25; the full four path products sum to 1.
Both stage splits are red 3/5,blue 2/5 under the independent replacement/remix model. Terminal probabilities RR 9/25,RB 6/25,BR 6/25,BB 4/25. P(different)=12/25 and P(at least one red)=21/25. Both mixed orders are included.
Look for these in your work
- Show replacement and remixing.
- Include both colour orders.
- Use an appropriate complement only after the path probability is correct.
Independent tests A, B, C succeed with probabilities 2/3, 3/4, 4/5. Draw a three-stage success/failure tree. Find P(all succeed), P(exactly two succeed) and P(none succeed).
Give me a hint
Draw all stage histories with their success/failure complements, multiply within each requested path and add only disjoint completed paths.
Compare my reasoning
- Draw the complete successive-stage tree with named outcomes at every split. Label the stated independent probabilities and their failure complements; multiply along complete histories and select the disjoint paths meeting the event.
- Use complementary failure probabilities 1/3, 1/4, 1/5 on each split; independence keeps the relevant probability unchanged across earlier histories.
- All succeed has probability 2/3×3/4×4/5=2/5. Exactly two succeed has three disjoint paths: failure at A, B or C.
- Their probabilities are 1/5, 2/15 and 1/10, totalling 13/30. None succeeds has probability 1/3×1/4×1/5=1/60.
- For the complete terminal reference, SSS=2/5,SSF=1/10,SFS=2/15,SFF=1/30,FSS=1/5,FSF=1/20,FFS=1/15,FFF=1/60. The eight probabilities sum to 1; exactly two successes selects only FSS,SFS,SSF.
At stages A,B,C, success probabilities are 2/3,3/4,4/5 and failures 1/3,1/4,1/5. P(all succeed)=2/5, P(exactly two)=13/30 and P(none)=1/60. Exactly-two paths are FSS 1/5,SFS 2/15,SSF 1/10; show all eight final paths on the complete tree.
Look for these in your work
- Construct all eight success/failure paths.
- Use the correct failure probability at each stage.
- Add all three exactly-two paths.
In an explicitly independent model, three alarms work with probabilities 4/5, 9/10 and 3/4. Draw the success/failure tree. Find P(at least one works) and P(exactly one fails). Explain why a shared power failure would challenge the model.
Give me a hint
Draw all stage histories with their success/failure complements, multiply within each requested path and add only disjoint completed paths.
Compare my reasoning
- Draw the complete successive-stage tree with named outcomes at every split. Label the stated independent probabilities and their failure complements; multiply along complete histories and select the disjoint paths meeting the event.
- The failure probabilities are 1/5, 1/10, 1/4. The all-fail path has product 1/200, so P(at least one works)=199/200.
- Exactly one failure has three paths with products 27/200, 12/200 and 36/200, totalling 75/200=3/8.
- The multiplication uses stated independence. A shared power source could connect failures, so marginal working probabilities alone would no longer justify these joint calculations.
- All eight ordered terminal products are SSS 27/50,SSF 9/50,SFS 3/50,SFF 1/50,FSS 27/200,FSF 9/200,FFS 3/200,FFF 1/200. They sum to 1. Exactly one failure selects FSS,SFS,SSF, not the one-success paths.
Complete alarm tree uses success 4/5,9/10,3/4 and failure 1/5,1/10,1/4. P(none works)=1/200, so P(at least one works)=199/200. Exactly one failure has contributions 27/200,12/200,36/200, giving 3/8. A shared power cause can invalidate the independence assumption.
Look for these in your work
- Use all three independent stages.
- Separate at-least-one from exactly-one failure.
- State the independence limitation.
Common mistakes
- Adding branch probabilities along one path rather than multiplying them.
- Keeping only one mixed order in an exactly-two event.
- Assuming replacement or a balanced drawing proves independence.
What can you explain now?
Why is a replacement statement alone insufficient to prove independence of repeated draws?
Compare with the explanation
It restores the contents but does not specify an independent unbiased selection mechanism.
An explicit uniform remixed or independent mechanism is needed to justify unchanged conditional branch probabilities.
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Make it stick
Explain the complete tree, unchanged branch probabilities, path products, disjoint path sums and an application limit on independence.
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.