Higher: applying probability models to simple problems
Translate simple contexts into probability trees, recover admissible integer contents and combine conditional branches with interpreted predictions.
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
- PR6–PR8
- shared expected frequency
- simple equations/quadratic factorisation and positive counts
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An equation for an unknown number of tokens has an algebraic negative root. What should happen next?
What you will learn
- Translate a simple context into labelled tree paths or finite conditional counts, keeping every model assumption.
- Use a given event probability to recover an unknown integer content, rejecting nonphysical algebraic roots.
- Combine disjoint conditional branches and use expected frequency as an interpreted count; no formal statistical test, independence inference from mere appearance or unsupported expected-profit theorem.
A bag contains x red and 4 blue tokens. Two uniform draws are made with replacement and independent remixing. P(two blue)=4/25. Find x and P(two red), explaining which square-root sign is physically allowed. Draw the model tree with x first, then label the recovered probabilities. Use the nonnegative integer domain for the unknown count and keep the original total positive.
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
knownBlue
4unknownColour
redunknownCountDomain
nonnegative integerdraws
2replacement
trueindependentRemixedDraws
trueP_twoBlue
4/25The 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 twoBlue; twoRed. 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
Use the original physical integer-count domain; inadmissible roots remain explicit only where requested.
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.
Use the original physical integer-count domain; inadmissible roots remain explicit only where requested.
Use the original physical integer-count domain; inadmissible roots remain explicit only where requested.
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.
Translate the context before calculating
Choose the sequence of stages that reflects the original process: a source followed by an item condition, weather followed by attendance, or successive draws from a box. Label the outcomes and the probability of each branch conditional on its own preceding history. Original marginal probabilities cannot replace stated conditional probabilities merely because the labels are familiar.
Complete the tree and identify the terminal histories belonging to the requested event. Multiply factors along each history, then add disjoint relevant paths. Reversing a condition uses the appropriate event contribution divided by the whole conditioning-event probability. Give that denominator a clear meaning and check it is positive before division.
Recover an unknown with its physical domain
An unknown number of objects is a nonnegative integer, and the total must support the stated number of draws. Write the original event probability as an equation in that count. With independent replacement, repeated category probabilities can produce a square; without replacement, both the category count and total change along the second branch. Keep these models separate when forming the equation.
Solving the equation may produce algebraic values that are not possible contents. A negative square-root value cannot be a probability, and a negative object count cannot describe the box. Show the algebraic alternatives, apply the physical domain and substitute every retained candidate into the original probability expression. Rearrangement alone does not validate a root or remove denominator restrictions.
Interpret conditional applications honestly
A mixture probability is the sum of its disjoint source-and-event paths, each weighted by that source’s probability. When conditioning on the event, compare the relevant source contribution with the total event probability. This finite path calculation does not permit unweighted averaging of conditional probabilities when sources are selected with different weights.
An expected count multiplies the event probability by the number of comparable modelled trials and remains a prediction rather than a guarantee. If conditional event probabilities differ between possible earlier conditions, independence must not be assumed. State the selection mechanism, physical count restrictions and the limits of the model; a correct numerical self-check does not mark the written argument or validate an actual system.
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
An item comes from A with probability 1/4 and B with 3/4. Its event probabilities are 1/5 given A and 1/3 given B. What is the overall event probability?
Worked example
A bag contains x red and 4 blue tokens. Two uniform draws are made with replacement and independent remixing. P(two blue)=4/25. Find x and P(two red), explaining which square-root sign is physically allowed. Draw the model tree with x first, then label the recovered probabilities. Use the nonnegative integer domain for the unknown count and keep the original total positive.
Show the worked solution
- Draw the original stages and label given conditional probabilities or expressions involving the unknown count. Form the requested event equation or weighted path sum, keep the physical count domain, and check the original model after solving.
- The positive single-blue probability is 4/(x+4). Independence with replacement gives [4/(x+4)]²=4/25.
- A probability cannot equal −2/5, so 4/(x+4)=2/5. Thus x+4=10 and x = 6 red tokens.
- The red probability is 6/10=3/5; P(two red)=(3/5)²=9/25. The recovered integer count satisfies the original model and nonnegative domain.
- Begin with first-stage red x/(x+4),blue 4/(x+4), and repeat those fractions after each branch only under the stated independent replacement model. After solving, substitute x=6 and verify all four terminal probabilities sum to 1.
Answer The positive blue probability is 2/5, so total 10 and x=6 red. The negative probability root−2/5 is inadmissible. The recovered replacement tree has red 3/5,blue 2/5 at each split: RR 9/25,RB 6/25,BR 6/25,BB 4/25. P(two red)=9/25; the original two-blue probability checks as 4/25.
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 random item comes from supplier A with probability 2/5 and B with probability 3/5. The faulty probabilities are 1/20 given A and 1/10 given B. Draw the two-level tree. Find P(faulty), P(B given faulty), and the expected faulty count among 500 comparable selections.
Give me a hint
Draw the original conditional stages, form the event equation or weighted path sum, apply the count/probability domain and check the original expression.
Compare my reasoning
- Draw the original stages and label given conditional probabilities or expressions involving the unknown count. Form the requested event equation or weighted path sum, keep the physical count domain, and check the original model after solving.
- The two faulty paths have probabilities 2/5×1/20=1/50 and 3/5×1/10=3/50. The supplier events are mutually exclusive.
- Add these paths to get P(faulty)=4/50=2/25. Within the faulty event, the B path contributes 3 of those 4 parts, giving P(B | faulty)=3/4.
- Expected faulty count in 500 is 500×2/25=40 under the same selection/quality model. This is a finite path-and-restricted-total calculation, not an invocation of an unlisted formal theorem.
- Complete the nonfaulty branches too: P(not faulty given A)=19/20 and given B=9/10. The four terminal products are A/faulty 1/50,A/not-faulty 19/50,B/faulty 3/50,B/not-faulty 27/50, summing to 1.
Faulty path contributions are 1/50 from A and 3/50 from B, so P(faulty)=2/25. P(B given faulty)=(3/50)/(2/25)=3/4. Expected faulty count in 500 comparable selections is 40, not a guaranteed realised number.
Look for these in your work
- Label source and conditional outcome levels correctly.
- Sum disjoint faulty paths.
- Condition on the total faulty event and interpret expected frequency.
A bag contains n red and 5 blue tokens. Two uniform draws are made without replacement and P(two blue)=2/9. Form an equation, find all algebraic roots and select the physical value of n. Draw the corresponding conditional tree and check the event probability.
Give me a hint
Draw the original conditional stages, form the event equation or weighted path sum, apply the count/probability domain and check the original expression.
Compare my reasoning
- Draw the original stages and label given conditional probabilities or expressions involving the unknown count. Form the requested event equation or weighted path sum, keep the physical count domain, and check the original model after solving.
- The event has probability 5/(n+5)×4/(n+4)=2/9. Rearranging gives (n+5)(n+4)=90.
- Expanding and factoring yields n²+9 n−70=(n−5)(n+14)=0, so the algebraic roots are 5 and −14.
- Only n = 5 is a nonnegative count. With ten tokens, the original event is 5/10×4/9=2/9, confirming the value.
- On the nonnegative integer count domain, n+5 and n+4 are positive. After the valid root n=5, first red/blue are each 5/10; after red the split is 4/9,5/9 and after blue it is 5/9,4/9. This produces the four checked path products.
The equation is 5/(n+5)×4/(n+4)=2/9, equivalent on the count domain to(n+5)(n+4)=90. It gives n²+9 n−70=(n−5)(n+14)=0, so roots 5 and−14. Reject−14 as a count; n=5 gives total 10 and original probability 5/10×4/9=2/9. The recovered tree has RR 2/9,RB 5/18,BR 5/18,BB 2/9.
Look for these in your work
- Reduce the second total and blue count.
- Preserve both algebraic roots before applying context.
- Check the valid integer count in the original probability.
A simple daily model gives dry weather probability 3/4 and wet probability 1/4. A person attends with probability 2/3 given dry and 2/5 given wet. Draw the weather/attendance tree. Find P(attends), P(wet given attends), and the expected attendance count across 600 comparable modelled days. State why independence of weather and attendance must not be assumed.
Give me a hint
Draw the original conditional stages, form the event equation or weighted path sum, apply the count/probability domain and check the original expression.
Compare my reasoning
- Draw the original stages and label given conditional probabilities or expressions involving the unknown count. Form the requested event equation or weighted path sum, keep the physical count domain, and check the original model after solving.
- The attendance paths are 3/4×2/3=1/2 for dry and 1/4×2/5=1/10 for wet. Add them to obtain 3/5.
- Within all attendance paths, the wet part is (1/10)/(3/5)=1/6. Expected attendance count is 600×3/5=360.
- Attendance has different stated probabilities after dry and wet weather, so it is not independent of weather in this model. The expected count is not a guaranteed real sequence of days.
- Complete both outcomes after each weather branch: attend/not-attend are 2/3,1/3 after dry and 2/5,3/5 after wet. The four terminal products are 1/2,1/4,1/10,3/20 and sum to 1; reverse conditioning uses only the two attendance paths.
Dry-and-attends contributes 1/2; wet-and-attends 1/10. P(attends)=3/5, P(wet given attends)=1/6 and expected attendance across 600 comparable modelled days is 360. Attendance is not independent of weather because the stated conditional attendance probabilities differ.
Look for these in your work
- Use the supplied conditional probabilities.
- Condition on attendance when reversing the question.
- Explain model dependence and prediction limits.
Common mistakes
- Averaging source-conditional probabilities without their source weights.
- Accepting an algebraic root that is a negative count or probability.
- Replacing a conditional model by an unsupported independent product or a guaranteed future count.
What can you explain now?
Why must an admissible recovered count be substituted into the original probability model?
Compare with the explanation
It checks the original denominators, sampling rule and event probability after the physical-domain filter.
Algebraic manipulation can introduce inadmissible alternatives or obscure a denominator restriction; the original event provides the final check.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain the event model, all needed path factors, any rejected algebraic value and the meaning and limits of a predicted count.
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.