Higher: simple conditioning in finite diagrams
Find simple finite conditional probabilities by restricting Venn, table or sample-space populations, and preserve undefined zero-condition cases.
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 Venn/sample spaces
- PR7 conditional meaning
- nonzero denominator
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
To find P(A given B) in a uniform finite population, which nonzero denominator is required?
What you will learn
- Interpret a stated condition as restricting the possible population/outcomes before finding the favourable fraction.
- Use Venn/two-way/sample-space counts to form simple conditional probabilities and distinguish direction of conditioning.
- If the conditioning event has no possible outcomes, the requested conditional probability is undefined; it is not automatically zero. No formal Bayes theorem or continuous conditional model is claimed.
The supplied Venn diagram has A-only 9, both 6, B-only 5 and neither 4. Find P(A given B), P(B given A) and P(A and B). Explain why these three denominators differ. One member of the stated population is chosen uniformly; mark each condition before restricting the denominator.
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
regions
- Aonly
- 9
- AB
- 6
- Bonly
- 5
- outside
- 4
selection
uniform population memberSupplied atomic regions (schematic areas)
- Aonly: 9
- AB: 6
- Bonly: 5
- outside: 4
No requested event is highlighted. These are symbolic population regions; clicks never create Cartesian work points or calculate a denominator.
The 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.
Select my favourable and conditioning populations
No actual representation is recorded.
The question’s event names are A_given_B; B_given_A; A_and_B. 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
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.
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.
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.
Restrict the population first
The phrase given B changes the reference population to outcomes belonging to B. Within that restricted population, count the outcomes also satisfying A. For a uniform finite population, P(A given B) is the shared A-and-B count divided by the B count, provided the B count is positive. Mark the conditioning set before choosing a denominator.
Without conditioning, the same shared count would be divided by the whole population to find the joint probability. Reversing the condition divides by the A count instead. The numerator may look the same while the denominator changes, so joint probability and the two reversed conditional probabilities generally differ. Read the grammatical direction of given carefully.
Use actual diagrams and systematic outcomes
In a Venn diagram, select the entire conditioning set, including its relevant intersections. In a two-way table, identify the stated conditioning row or column and use its total. Original cell or region counts are sufficient; do not assume that a drawn area or a missing margin is itself a probability. Show the restricted total you calculated from those original counts.
For ordered equally likely outcomes, first construct the complete original space, then keep only the outcomes meeting the condition. Count the requested event inside that restricted list. A condition about a sum, for example, need not leave every possible first score equally weighted; the allowed ordered pairs, rather than a shorter list of distinct labels, determine the conditional weights.
Check existence and explain the model
If the conditioning event has no possible outcomes in the model, the requested finite conditional probability is undefined. It is not automatically zero: zero would mean that a nonempty reference population contains no favourable outcomes. State the impossible condition explicitly and keep that case separate from a well-defined probability of an impossible event within a possible conditioning population.
This method uses simple finite ratios and ordinary event/path counting. It does not require a named general theorem about reversing conditions, nor does it establish a real-world causal relationship. Retain the original uniformity or outcome weights, show which population was restricted and explain why its denominator is lawful before reducing the final fraction.
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 finite population has 13 members in B,10 in A and 4 in both. What is P(A given B)?
Worked example
The supplied Venn diagram has A-only 9, both 6, B-only 5 and neither 4. Find P(A given B), P(B given A) and P(A and B). Explain why these three denominators differ. One member of the stated population is chosen uniformly; mark each condition before restricting the denominator.
Show the worked solution
- Keep the original Venn regions, table cells or full ordered outcome space. Mark the condition first, count its restricted population, and only then count favourable outcomes inside it; check the denominator is positive.
- The whole population is 24, with 15 in A and 11 in B. The intersection has 6 members.
- Given B restricts the population to those 11 members, so P(A | B)=6/11. Given A restricts it to 15, so P(B | A)=6/15=2/5.
- Without conditioning, P(A and B)=6/24=1/4. The event numerator may match while the reference population changes.
Answer Whole population 24; A count 15; B count 11. P(A given B)=6/11, P(B given A)=2/5 and P(A and B)=1/4. The respective denominators are 11,15,24.
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 table records walk/club 8, walk/no-club 7, bus/club 4 and bus/no-club 11 students. Find P(club given walk), P(walk given club) and P(bus given no-club). Show the restricted totals. Select one recorded student uniformly before applying each stated condition, and mark the conditioning row or column.
Give me a hint
Construct or identify the original outcomes, mark the conditioning population and count favourable members within that nonempty population.
Compare my reasoning
- Keep the original Venn regions, table cells or full ordered outcome space. Mark the condition first, count its restricted population, and only then count favourable outcomes inside it; check the denominator is positive.
- There are 15 walkers, 12 club members and 18 non-club members; the complete table totals 30.
- P(club|walk)=8/15; P(walk|club)=8/12=2/3. The reversal changes the denominator.
- Among 18 non-club members, 11 use the bus, giving 11/18. Do not divide all three numerators by 30 because each statement supplies a condition.
Walkers 15, club members 12 and non-club members 18; whole table 30. P(club given walk)=8/15, P(walk given club)=2/3 and P(bus given no-club)=11/18. Mark each conditioning row or column before dividing.
Look for these in your work
- Identify the correct row/column condition.
- Show each restricted denominator.
- Retain the original counts and no inferred probability labels.
Choose one card uniformly from 1 to 12. Find P(multiple of 3 given even) and P(even given multiple of 3) by listing the restricted outcomes. Is P(even given greater than 12) defined in this model?
Give me a hint
Construct or identify the original outcomes, mark the conditioning population and count favourable members within that nonempty population.
Compare my reasoning
- Keep the original Venn regions, table cells or full ordered outcome space. Mark the condition first, count its restricted population, and only then count favourable outcomes inside it; check the denominator is positive.
- Given even leaves 2, 4, 6, 8, 10, 12; only 6 and 12 are multiples of 3, giving 2/6=1/3.
- Given a multiple of 3 leaves 3, 6, 9, 12; two are even, giving 2/4=1/2.
- No label exceeds 12, so the last conditioning set is empty. A probability conditional on that impossible event is undefined here, not 0.
Given even: {2,4,6,8,10,12}; favourable multiples of 3 are{6,12}, so 1/3. Given multiple of 3: {3,6,9,12}; favourable evens{6,12}, so 1/2. Greater-than-12 leaves no outcomes, so the last conditional is undefined, not 0.
Look for these in your work
- List both restricted spaces separately.
- Use the actual nonzero denominator.
- Recognise the empty conditioning set.
Two independent fair four-sided dice are rolled. Construct the ordered sample space, then restrict it to a sum of 5. Find P(first score greater than 2 given sum 5) and P(second score 4 given sum 5). Also find P(sum 5 given first score greater than 2). Would conditioning on sum 9 be defined?
Give me a hint
Construct or identify the original outcomes, mark the conditioning population and count favourable members within that nonempty population.
Compare my reasoning
- Keep the original Venn regions, table cells or full ordered outcome space. Mark the condition first, count its restricted population, and only then count favourable outcomes inside it; check the denominator is positive.
- The original sample space has 16 equally likely ordered pairs. Sum 5 leaves (1, 4), (2, 3), (3, 2), (4, 1).
- Within those four, two have first score greater than 2 and one has second score 4; the probabilities are 1/2 and 1/4.
- Conditioning instead on first score greater than 2 leaves eight pairs, of which two have sum 5, giving 1/4. Sum 9 is impossible for these dice, so that alternative conditional is undefined.
- List all sixteen pairs systematically as four rows: (1,1),(1,2),(1,3),(1,4); (2,1),(2,2),(2,3),(2,4); (3,1),(3,2),(3,3),(3,4); (4,1),(4,2),(4,3),(4,4). The restricted sum 5 space has four pairs, while the separate first>2 condition has eight; do not reverse a conditional without changing that denominator.
The sum 5 condition leaves(1,4),(2,3),(3,2),(4,1). P(first>2 given sum 5)=1/2 and P(second 4 given sum 5)=1/4. Given first>2 leaves 8 of the 16 ordered outcomes and 2 have sum 5, so P(sum 5 given first>2)=1/4. Sum 9 is impossible, so conditioning on it is undefined.
Look for these in your work
- Construct then restrict the actual ordered space.
- Change denominator when the condition changes.
- Reject the impossible condition rather than inventing 0.
Common mistakes
- Using the whole population denominator after given has restricted the sample space.
- Reversing a conditional without changing its denominator.
- Returning zero for an empty conditioning population.
What can you explain now?
How does an undefined conditional on an empty set differ from a well-defined conditional probability of zero?
Compare with the explanation
An empty condition gives no denominator; zero probability requires a nonempty condition with no favourable outcomes inside it.
Distinguish 0/0, which is undefined here, from 0 divided by a positive restricted-population count.
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Make it stick
Explain the joint numerator, each possible conditioning denominator and the reason an impossible condition cannot be assigned a numerical ratio.
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.