Higher: changing probabilities without replacement
Construct conditional trees from remaining contents, retain every order and distinguish changed conditional probabilities from unchanged marginals.
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 trees
- shared complements and disjoint sums
- subtraction from integer counts
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A uniform first draw removes a gold token from 2 gold and 5 grey tokens. What remains before the next draw?
What you will learn
- Draw conditional branches from the actual remaining contents after each named history.
- Retain order, multiply conditional path factors and add disjoint successful paths; include two- and simple three-draw applications.
- Contrast with replacement, guard exhausted/zero-probability histories and avoid falsely multiplying unchanged marginal probabilities.
A bag has 5 red and 3 blue tokens. Draw two uniformly without replacement. Draw the complete conditional tree and find P(two red), P(different colours), P(second red) and P(second red given first red). Compare the second split after a red with the split after a blue.
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
counts
- 5
- 3
draws
2replacement
falseThe 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 twoRed; different; secondRed; secondRedGivenFirstRed; secondRedGivenFirstBlue. 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.
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.
Update contents after the actual history
Without replacement, the selected object is removed before the next draw. At each reachable history, subtract the objects actually removed from their own categories and reduce the total number of remaining objects. The next branch probability is the remaining category count divided by that remaining total. Different first outcomes can therefore produce different second-stage splits.
Draw the complete requested tree with outcome labels and probabilities after every history. A three-colour first split needs all three later distributions, not only a binary summary that hides the requested colours. An aggregated event such as nonwhite may be useful later, but justify it from the underlying contents and preserve the original detailed tree where the question requires it.
Retain order and combine lawful paths
Multiply successive conditional branch factors along one history. A red-then-blue path and a blue-then-red path are different ordered outcomes; include both when the requested event is different colours. Add their probabilities because complete terminal histories are disjoint. A complement such as at least one selected category can instead subtract the appropriate all-other path from one.
For three draws, update both the total and the relevant category count after each of the two earlier outcomes. Exactly one faulty object has a separate path for each possible faulty position. A complete tree and a check that terminal probabilities sum to one help reveal a forgotten order or an unchanged denominator that should have fallen.
Separate conditional, marginal and unreachable cases
The probability of the second outcome without a first-outcome condition is a marginal probability obtained by adding all relevant terminal paths. It may match the first-draw marginal even when the second-stage conditional probabilities differ. That numerical equality does not establish independence. Compare the probabilities after particular first outcomes to see the remaining-content dependence.
A category can become exhausted, making a next-category branch have probability zero. A later conditional on a history that itself cannot occur has no nonzero conditioning denominator; do not invent a negative count or interpret zero divided by zero as a probability. Drawing more objects than the box contains is not a valid completed sampling model. Replacement is a separate comparison, with its own stated mechanism.
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
Using that remaining box of 1 gold and 5 grey tokens, what is P(next grey)?
Worked example
A bag has 5 red and 3 blue tokens. Draw two uniformly without replacement. Draw the complete conditional tree and find P(two red), P(different colours), P(second red) and P(second red given first red). Compare the second split after a red with the split after a blue.
Show the worked solution
- Draw every stage outcome after each preceding history and write the actual remaining contents alongside its split. Reduce the correct category count and total before multiplying each path; retain every qualifying order.
- The first split is red 5/8, blue 3/8. After red there are 4 red/3 blue among 7; after blue there are 5 red/2 blue among 7.
- RR=5/8×4/7=5/14. RB+BR=15/56+15/56=15/28.
- Second red is RR+BR=5/8, but its conditional probability is 4/7 after first red and 5/7 after first blue. Equal first/second marginal probabilities do not make the draws independent.
Answer First red/blue probabilities 5/8,3/8. After red: red 4/7,blue 3/7; after blue: red 5/7,blue 2/7. RR 5/14,RB 15/56,BR 15/56,BB 3/28. P(two red)=5/14, P(different)=15/28, P(second red)=5/8, P(second red given first red)=4/7; given first blue it is 5/7. The equal marginal 5/8 does not imply independence.
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 contains 4 green and 2 yellow tokens. Draw two uniformly without replacement. Draw the full tree and calculate P(at least one yellow), showing either its disjoint paths or its complementary path.
Give me a hint
Write the remaining contents after each history, then multiply that history’s factors and include every qualifying order.
Compare my reasoning
- Draw every stage outcome after each preceding history and write the actual remaining contents alongside its split. Reduce the correct category count and total before multiplying each path; retain every qualifying order.
- First probabilities are 4/6 and 2/6. After green the remaining split is 3/5 green, 2/5 yellow; after yellow it is 4/5 green, 1/5 yellow.
- The no-yellow path GG has probability 4/6×3/5=2/5.
- At least one yellow has probability 1−2/5=3/5, also 4/15+4/15+1/15. The denominator of the second draw is 5, not 6.
First green/yellow 4/6,2/6; after green 3/5,2/5; after yellow 4/5,1/5. GG 2/5,GY 4/15,YG 4/15,YY 1/15. P(at least one yellow)=1−2/5=3/5; the mixed orders are distinct.
Look for these in your work
- Draw conditional fractions for both histories.
- Identify the complementary all-green event.
- Check the four terminal probabilities total 1.
A bag contains 4 red, 3 blue and 2 white tokens. Draw two uniformly without replacement. Construct the three-outcome conditional tree. Find P(first red then white), P(exactly one white), P(two white), and P(second white given first white).
Give me a hint
Write the remaining contents after each history, then multiply that history’s factors and include every qualifying order.
Compare my reasoning
- Draw every stage outcome after each preceding history and write the actual remaining contents alongside its split. Reduce the correct category count and total before multiplying each path; retain every qualifying order.
- The first total is 9; each second total is 8. A red then white path has probability 4/9×2/8=1/9.
- Exactly one white combines W/non W and non W/W: 2/9×7/8+7/9×2/8=7/18. This aggregation is legitimate because every nonwhite first removal leaves 2 white among 8.
- Two whites has probability 2/9×1/8=1/36. The conditional second-white probability after white is 1/8, versus 2/8=1/4 after a nonwhite first draw.
- The complete terminal products are RR 1/6,RB 1/6,RW 1/9,BR 1/6,BB 1/12,BW 1/12,WR 1/9,WB 1/12,WW 1/36. They sum to 1. Exactly one white selects RW,BW,WR,WB; it excludes WW.
First red/blue/white 4/9,3/9,2/9. After red use 3/8,3/8,2/8; after blue use 4/8,2/8,2/8; after white use 4/8,3/8,1/8. P(red then white)=1/9, P(exactly one white)=7/18, P(two white)=1/36 and P(second white given first white)=1/8. Given a nonwhite first removal, the second-white probability is 1/4.
Look for these in your work
- Keep all three first-stage outcomes.
- Use the actual remaining counts.
- Explain any aggregated nonwhite branch.
A box has 6 good and 4 faulty items. Inspect three uniformly without replacement. Draw the three-stage tree. Find P(all good), P(at least one faulty) and P(exactly one faulty). Compare P(all good) with a different model that replaces and remixes after every draw. In the separate replacement comparison, assume independent uniform remixed draws.
Give me a hint
Write the remaining contents after each history, then multiply that history’s factors and include every qualifying order.
Compare my reasoning
- Draw every stage outcome after each preceding history and write the actual remaining contents alongside its split. Reduce the correct category count and total before multiplying each path; retain every qualifying order.
- All-good probability is 6/10×5/9×4/8=1/6, so at least one faulty has probability 5/6.
- Exactly one faulty uses FGG, GFG and GGF. Each ordered path has probability 1/6 by its actual conditional factors, giving total 1/2.
- With replacement and independent remixing, all-good probability would be(6/10)^3=27/125. The different contents rule changes the model and cannot be interchanged silently.
- Label the first split G 6/10,F 4/10; after G use G 5/9,F 4/9, and after F use G 6/9,F 3/9. At the third split, after GG use G 4/8,F 4/8; after GF or FG use G 5/8,F 3/8; after FF use G 6/8,F 2/8. Complete no-replacement paths are GGG 1/6,GGF 1/6,GFG 1/6,GFF 1/10,FGG 1/6,FGF 1/10,FFG 1/10,FFF 1/30. Label each intermediate split from its actual remaining contents; these eight disjoint probabilities sum to 1.
Without replacement, P(all good)=1/6, P(at least one faulty)=5/6 and P(exactly one faulty)=1/2. The three one-fault paths FGG,GFG,GGF each have probability 1/6. With independent replacement/remixing, the separate all-good probability is 27/125. The two sampling models must not be interchanged.
Look for these in your work
- Draw all eight final paths with valid remaining counts.
- Include all three one-fault positions.
- Keep replacement and no-replacement models separate.
Common mistakes
- Reducing the total but not the count of the colour actually removed.
- Using unchanged marginal probabilities on a without-replacement path.
- Conditioning on an unreachable history as though its probability were zero rather than undefined.
What can you explain now?
What is wrong with assigning a numerical conditional probability after a history that cannot occur?
Compare with the explanation
The conditioning event has probability zero, so the finite restricted-population ratio is undefined.
A zero-probability branch can be labelled zero, but a later conditional on that unreachable history cannot be manufactured from 0/0.
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Make it stick
Explain each reduced numerator/denominator, the full set of ordered successful paths and the distinction between a marginal and a conditional 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.