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Probability - Unit 1 - Mathematics Year 9

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Introduction

If six numbered cards are even and four are multiples of three, does that mean ten different cards satisfy at least one condition? Not necessarily: some cards may satisfy both. In this course you will learn to organise the complete collection, enumerate its members and count a union or intersection without omissions or double counting.

You will connect set notation, systematic lists, Venn regions, two-way tables and outcome grids. The main goal is to explain why your enumeration is complete. Probability calculations then use those outcomes only when the model justifies treating them as equally likely.

This Year 9 placement is an editorial sequence within England's Key Stage 3 mathematics programme. The worked card collections, survey figures and practice situations are invented teaching examples, not real pupil data or recorded experiments. You need addition, subtraction, multiplication, fractions and the ability to test a number against a stated rule.

By the end, you should be able to move between representations of the same information, explain inclusive “or”, distinguish an intersection from an exclusive region, identify inconsistent totals and state the assumptions behind a probability. Your final product will include a fully checked enumeration in more than one representation.

Define a universe and list its sets systematically

Decide exactly what can be included

A set is a collection of distinct elements, with membership determined clearly. An element is one of its members. A universal set U defines the collection under discussion; it does not have to mean everything in existence. Defining U prevents an unstated change in the problem.

For our first example, imagine twelve distinct cards numbered 1 to 12:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

Let A be the even-numbered cards in U, and let B be the cards in U whose numbers are multiples of 3. Testing the members in increasing order gives:

A = {2, 4, 6, 8, 10, 12}; B = {3, 6, 9, 12}.

You can write 6 ∈ A to say that 6 is a member of A, and 5 ∉ A to say that 5 is not. Here n(A) means the number of elements in A, so n(A) = 6 and n(B) = 4. A set is not a sequence: changing the written order does not create a different set, and writing the same member twice does not create another distinct member.

A precise condition matters. “Numbers I like” needs a particular person's decisions before it can be used to classify the cards. “Even whole numbers from 1 to 12” supplies a test. Also notice the boundary: if you change U to 1 through 20, the same property produces a different list.

Use a membership table to avoid guessing

A systematic method tests every member of U once against each rule. You can mark membership with yes and no, then check the resulting lists against the table.

Card	In A: even?	In B: multiple of 3?
1	No	No
2	Yes	No
3	No	Yes
4	Yes	No
5	No	No
6	Yes	Yes
7	No	No
8	Yes	No
9	No	Yes
10	Yes	No
11	No	No
12	Yes	Yes

The cards 6 and 12 appear in both lists because they meet both rules. They are still only two physical cards. Counting memberships and counting distinct cards are different operations. This distinction is the reason you cannot automatically add the two set sizes to find how many cards meet at least one rule.

Check your method: identify one card that meets both rules, one that meets only A, one that meets only B and one that meets neither. For example, 6, 2, 3 and 1 respectively pass these four tests. A classification is complete only when every card has a place.

Recognise special relationships

A set C is a subset of A when every member of C belongs to A. In the same universe, the multiples of 6 form C = {6, 12}; this C is a subset of both A and B. A subset can equal the larger set: the word does not always mean strictly smaller.

Two sets are disjoint when they have no common members. The even and odd cards in U are disjoint. Their intersection is the empty set, written ∅, which contains no elements. The set {0} is different: it contains one element, zero, if zero is allowed in the stated universe. Zero is not a member of our current U.

These definitions help you check a diagram. An overlapping drawing allows a region for common members, but that region may contain none. A conventional Venn diagram can show possible membership combinations even when a particular collection leaves one region empty.

Read Venn regions, unions and intersections

Intersection means both conditions

The intersection A ∩ B contains the elements that belong to both A and B. For the numbered cards, it is {6, 12}, so n(A ∩ B) = 2. You must test both conditions; being a member of only one is not enough.

A common mistake is to treat the non-overlapping part of the left circle as the whole of A. In fact, the overlap belongs to A as well as B. For our cards, the region called A only is {2, 4, 8, 10}; adding the intersection {6, 12} reconstructs the complete A.

The four mutually exclusive regions are A only, both, B only and neither. Each card goes in exactly one of these regions. “Mutually exclusive regions” does not mean the sets A and B themselves are disjoint: here they are not.

Region	Exact members	Count
A only	{2, 4, 8, 10}	4
A ∩ B	{6, 12}	2
B only	{3, 9}	2
Neither	{1, 5, 7, 11}	4

The regional counts sum to 4 + 2 + 2 + 4 = 12, matching n(U). This is a useful check, although a correct total alone would not prove that every individual card was placed correctly. Test both the totals and the membership rules.

Union means at least one, including both

The union A ∪ B contains every element belonging to A, B or both. In mathematical set language, “or” is inclusive unless the question explicitly says otherwise.

A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}; n(A ∪ B) = 8.

You can count the union by adding the exclusive regions: 4 + 2 + 2 = 8. Alternatively, add n(A) and n(B), then subtract the overlap counted twice:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 6 + 4 − 2 = 8.

Subtracting the overlap once does not remove it from the union. It corrects the duplicate count, leaving each shared member counted once. “Exactly one of A or B” selects the two outer regions only, giving 4 + 2 = 6 cards here. “At least one” gives 8; “both” gives 2.

Complement depends on the universe

The complement of A, written A^c here, contains members of U that do not belong to A. For the cards, A^c = {1, 3, 5, 7, 9, 11}. The complement of the union contains cards meeting neither condition: (A ∪ B)^c = {1, 5, 7, 11}.

Compare two phrases carefully. “Not in A and not in B” means neither. “Not in both” means outside the intersection: it includes A only, B only and neither. For our universe, “not in both” selects 10 cards, while “neither” selects 4. The scope of “not” changes the selection.

The verified video below provides further worked Venn-diagram explanations. Pause to translate a shaded region into a membership rule, then test that rule against the numbered cards. The native examples and tasks remain available without external playback.

Convert between Venn regions and two-way tables

A table represents the same four cases

A two-way table classifies elements by two attributes. Its internal cells can correspond exactly to the four exclusive Venn regions. The row and column totals are useful summaries; they are not extra elements to add to the internal cells.

Membership	In B	Not in B	Row total
In A	2	4	6
Not in A	2	4	6
Column total	4	8	12

Read the cell at the intersection of the In A row and In B column: it gives the two cards in both sets. The top-right cell gives A only, and the bottom-left internal cell gives B only. The bottom-right internal cell, 4, gives neither. The grand total 12 is outside the four internal cells, in the totals row and column.

To find A ∪ B, add the three internal cells representing A only, both and B only: 4 + 2 + 2 = 8. Or subtract neither from the universe: 12 − 4 = 8. These two routes agree because they enumerate complementary parts of the same collection.

A two-way table and a Venn diagram should not give different results if they represent the same universe, conditions and data. If they disagree, compare definitions and exclusive regions before deciding that one representation is inherently unreliable.

Work backwards from inclusive totals

Suppose a fictional survey of 30 people records that 18 choose activity P, 17 choose activity Q and 9 choose both. These are inclusive set totals: the 9 shared people are already included in both 18 and 17.

Start with the overlap. P only is 18 − 9 = 9. Q only is 17 − 9 = 8. At least one is 9 + 9 + 8 = 26. Neither is 30 − 26 = 4. Put the four numbers in a table or Venn diagram and check that they sum to 30.

Fictional activity choice	In Q	Not in Q	Total
In P	9	9	18
Not in P	8	4	12
Total	17	13	30

Do not put 18 in the P-only region and then add 9 in the overlap. That would make the total membership of P equal to 27, contradicting the original statement. Label inclusive totals separately from exclusive regions.

Check whether the information is possible

If a claimed survey has U = 30, n(P) = 24, n(Q) = 20 and n(P ∩ Q) = 5, the claimed union is 24 + 20 − 5 = 39. That exceeds the universe, so the figures cannot all describe the same 30 people under those definitions.

The overlap must be at least 24 + 20 − 30 = 14 and no larger than the smaller set, 20. Those limits follow from the need to fit everyone into the universe and from the intersection belonging to both sets. They do not tell you the exact overlap without more information.

A negative “neither” count is not a strange kind of person or a probability to accept. It is a signal to check the data, universe or interpretation. Explain the contradiction instead of silently changing a supplied figure.

Enumerate combined outcomes with a complete grid

Keep distinct outcomes distinct

Imagine an ideal fair red spinner numbered 1 to 4 and an ideal fair blue spinner numbered 1 to 3. Assume the spins are independent. An outcome is the ordered pair (red, blue). Listing by red row and blue column creates 4 × 3 = 12 distinct equally likely ordered outcomes.

Each cell below shows its ordered pair and sum. The pair is the elementary outcome; the sum is a value calculated from it.

Red / Blue	1	2	3
1	(1,1): sum 2	(1,2): sum 3	(1,3): sum 4
2	(2,1): sum 3	(2,2): sum 4	(2,3): sum 5
3	(3,1): sum 4	(3,2): sum 5	(3,3): sum 6
4	(4,1): sum 5	(4,2): sum 6	(4,3): sum 7

The ordered pairs (1,2) and (2,1) are different outcomes because the red and blue results differ, even though both have sum 3. Conversely, the six possible sums 2 to 7 are not six equally likely outcomes: sum 2 occurs in one cell, while sum 4 occurs in three.

A complete grid prevents two common errors: omitting a possible pair and counting an ordered pair twice. Before counting a selected event, check that every allowed row and column is present and that every cell follows the model.

Find unions and intersections on the grid

Let E be the event that the sum is even. Its outcomes are {(1,1), (1,3), (2,2), (3,1), (3,3), (4,2)}, giving 6 cells. Let F be the event that the sum is at least 6. Its outcomes are {(3,3), (4,2), (4,3)}, giving 3 cells.

The intersection E ∩ F is {(3,3), (4,2)}, giving 2 cells. The union E ∪ F therefore has 6 + 3 − 2 = 7 cells. You can also list it directly: {(1,1), (1,3), (2,2), (3,1), (3,3), (4,2), (4,3)}.

The exclusive counts are E only 4, both 2, F only 1 and neither 5. These sum to 12. If you draw a Venn diagram of the outcomes, place ordered pairs in the regions, not just the sums: otherwise repeated sums can hide distinct outcomes.

Under the stated fair, independent model, P(E) = 6/12 = 1/2, P(F) = 3/12 = 1/4 and P(E ∪ F) = 7/12. The probability rule uses equally likely elementary outcomes. Listing possibilities is still useful when outcomes are unequal, but simply dividing the number selected by the number listed is then not justified.

Change the model and rebuild the grid

Now consider four distinct cards labelled R1, R2, B1 and B2. Draw two in order without replacement, with each remaining card equally likely at each draw. There are 4 × 3 = 12 equally likely ordered card pairs, not 16: a card cannot be drawn again after it has been removed.

The colour patterns RR, RB, BR and BB are not equally likely. There are 2 card pairs of colour RR, 4 of RB, 4 of BR and 2 of BB. For example, RR consists of (R1,R2) and (R2,R1). Thus P(RR) = 2/12 = 1/6, rather than 1/4.

Draw a four-by-four grid with the first card on the rows and the second on the columns. Cross out the diagonal cells where the same card would appear twice. The remaining twelve cells form the outcome set. A two-by-two colour grid hides the unequal numbers of elementary outcomes behind its four labels.

The lesson: a grid is a representation of a model, not a guarantee of equal likelihood. State what an outcome records, whether order matters, whether replacement occurs and why any equal-likelihood assumption is reasonable.

Extend the method and audit an explanation

Three sets require eight exclusive regions

Return to U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, with A even, B a multiple of 3 and C at most 6. Each member has three yes/no membership decisions. A three-set Venn diagram therefore has eight possible membership regions, including the region outside all three circles.

Test the members systematically rather than guessing the counts:

Exclusive region	Members	Count
A, B and C	{6}	1
A and B but not C	{12}	1
A and C but not B	{2, 4}	2
B and C but not A	{3}	1
A only	{8, 10}	2
B only	{9}	1
C only	{1, 5}	2
None of the three	{7, 11}	2

The eight counts sum to 12. The inclusive intersection A ∩ B still contains both 6 and 12: it includes the central triple intersection as well as the A-and-B-only region. The phrase “exactly A and B” must be interpreted as excluding C when that is what the problem means.

To find A ∪ B ∪ C, combine every region except none: there are 10 members. To find exactly two of the sets, combine the three pair-only regions, giving 1 + 2 + 1 = 4. To find at least two, include the triple intersection too, giving 5. These are different selections.

Explain a counting rule instead of memorising blindly

For two finite sets, adding n(A) + n(B) counts the overlap twice; subtracting it once corrects the union. For three sets, each pairwise intersection includes the triple intersection. A checked extension is:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C).

In this example, 6 + 4 + 6 − 2 − 3 − 2 + 1 = 10. The final addition matters: a member in all three sets was first counted three times and then subtracted three times, so it must be added back once. You can always verify a small example by listing its members instead of trusting a remembered formula.

The main skill remains systematic enumeration. Use the eight-region table to explain the triple-set rule, and the four-region table to explain the two-set rule. A correct final number without an account of what was counted is weaker evidence of understanding.

Choose the representation and test its limits

A list is useful when individual members matter. A membership table tests rules explicitly. A Venn diagram makes logical regions visible. A two-way table gives four exclusive counts and their totals. A grid helps enumerate ordered outcomes of a combined procedure. Use the representation that makes completeness and duplication easiest to check.

A conventional Venn diagram is not automatically drawn to scale. A larger-looking circle does not prove a larger set or a higher probability. Read the labels and counts. A regional count of zero does not mean you should invent an element to fill the space.

Finish any solution by checking the universe, definitions, exclusive-region total, inclusive set totals and probability assumptions. If two methods disagree, trace the actual members or outcomes. If the information is insufficient or inconsistent, say so and explain why rather than inventing a missing count.

Interactive Tasks

Quiz: Test Your Knowledge

In U from one to twelve, which card belongs to both the even cards and the multiples of three? (Six) (!Four) (!Nine) (!Eleven)

What does the union of two sets include? (Members of either set including shared members counted once) (!Only the members outside both sets) (!Only shared members counted twice) (!Only members belonging to exactly one set)

For the numbered-card example, what is the size of A union B? (Eight) (!Ten) (!Two) (!Twelve)

What does a complement require you to specify? (The universal set within which not belonging is tested) (!The physical size of the drawn circle) (!The order in which a set is written) (!A rule that every outcome must be equally likely)

Which statement correctly compares neither with not in both? (Neither excludes both sets while not in both excludes only the intersection) (!They always select exactly the same elements) (!Neither means belonging to both sets) (!Not in both always means an empty set)

Thirty people include eighteen choosing P, seventeen choosing Q and nine choosing both. How many choose neither? (Four) (!Nine) (!Five) (!Thirty-five)

Why is the claimed overlap of five impossible when two sets of sizes twenty-four and twenty come from thirty people? (It would make their union larger than the universe) (!Every intersection must be empty) (!The two set sizes must always be equal) (!Thirty people cannot be represented in a table)

Why are the spinner outcomes red one blue two and red two blue one different? (The ordered pairs record which result came from each labelled spinner) (!Their sums must be different) (!All equal sums represent only one elementary outcome) (!A grid can never contain repeated sums)

What is the probability of an even sum or a sum at least six in the stated fair independent spinner model? (Seven twelfths) (!Nine twelfths) (!Two twelfths) (!One sixth)

When two cards are drawn without replacement from R1, R2, B1 and B2, why is RR not one of four equally likely colour patterns? (The colour patterns represent different numbers of equally likely distinct card pairs) (!A removed card is immediately put back) (!There are sixteen allowed ordered card pairs) (!Every colour pattern contains exactly three distinct card pairs)

Memory Game

Universal set	The stated collection within which membership is considered
Intersection	Members belonging to both selected sets
Union	Members belonging to at least one selected set
Complement	Members of the universe outside the selected set
Disjoint sets	Sets with no common members
Elementary outcome	One distinct possible result at the model's stated level of detail

Drag and Drop

Match the task.	Useful representation or check
Test each numbered card against two rules	Membership table
Show both, only and neither regions	Labelled Venn diagram
Connect exclusive counts with row and column totals	Two-way table
List labelled first and second results systematically	Ordered outcome grid
Test whether everyone has been accounted for	Sum the exclusive regions and compare with the universe

Crossword Puzzle

UNION	What operation combines all members of either set without duplicating shared members?
INTERSECTION	What operation selects common members of sets?
COMPLEMENT	What is the selection outside a set but inside its universe called?
OUTCOME	What term describes a possible result of a procedure?
DISJOINT	What word describes sets with no common members?
UNIVERSAL	What adjective describes the stated set containing everything under consideration?

LearningApps

This optional external search can change independently. The full native tasks and feedback below remain available if it cannot be loaded.

Cloze Text

The collection under consideration is the

. An individual member of a set is an

. Members common to two sets form their

. Members in at least one of two sets form their

. The mathematical word or is normally

. To correct a double count in a two-set union, subtract the

. The elements outside a set but inside its universe form its

. Sets with no common elements are

. A table can cross-check membership using row and column

. A grid of labelled first and second results records ordered

. Removing a drawn card changes a model without

. Simple favourable-outcome counting gives probabilities only when the elementary outcomes are equally

.

Open Tasks

Easy

Make a complete membership record: Use U from 1 to 15, A the even numbers and B the multiples of 5. List A and B in increasing order, then test every member of U in a yes/no table. Explain how you know no card was omitted.
Translate a membership phrase: For the same universe and sets, list both, at least one, exactly one and neither. Write a plain-English membership rule for each and identify one example showing why at least one includes both.
Explain a changed universe: Compare the complement of the even numbers when U is 1 to 12 with the complement when U is 1 to 15. Explain what changes and what stays the same, using exact members rather than just a circle sketch.
Annotate a mathematical image: Choose an actual diagram in the course. Explain what its regions mean and give a compatible example. State why its apparent area does not by itself tell you a probability or count. Include the creator and source.

Standard

Convert and verify a survey: A fictional group of 40 people has 22 in P, 19 in Q and 8 in both. Construct a two-way table and a labelled Venn diagram. Explain every exclusive count and use two methods to find the union. Check all margins and the grand total.
Build a complete spinner grid: Use independent fair spinners numbered 1 to 3 and 1 to 4. Define outcomes as labelled ordered pairs. Find the intersection and union of the events sum is odd and sum is greater than 5. Supply actual outcome lists and explain any probability calculation.
Diagnose impossible totals: A claimed universe has 25 members, with 20 in A, 18 in B and 6 in both. Explain the contradiction, determine the possible overlap range if the universe and inclusive set totals are retained, and give one consistent replacement overlap with all four regional counts. Label it as a proposed repair, not observed data.
Compare two card models: Enumerate the ordered pairs from R1, R2, B1 and B2 both with and without replacement. Compare the number of outcomes and the probability of two red cards. Explain why colour labels alone can hide unequal outcome counts.

Advanced

Audit eight membership regions: In U from 1 to 12, use A even, B a multiple of 3 and C at most 6. Draw a three-set diagram from the eight-region table, list A intersection B including the triple overlap, and calculate exactly two, at least two and the full union by separate checks.
Explain the correction term: Explain the two-set union formula by following one shared member through the addition and subtraction. Then follow a triple-shared member in the three-set formula. Give a complete numerical example and show why the final addition is required.
Create and test a counting model: Invent a small two-stage procedure with clearly labelled outcomes. Enumerate it by a grid and another method, define overlapping events and calculate their union and intersection. State order, replacement and likelihood assumptions, then change one assumption and explain which parts must be rebuilt.
Teach a misconception with evidence: Create a short explanatory poster, written dialogue or recording correcting one misconception from the course. Include a counterexample, two consistent representations, a worked answer and a check. Use fictional labels instead of collecting classmates' personal information; if you do not record or test a real procedure, state that the work is a mathematical simulation.

Feedback for your open work

Task	Checking guidance
Membership record	A = {2,4,6,8,10,12,14}; B = {5,10,15}. The table should contain all fifteen universe members once. Membership in both is allowed.
Membership phrases	Both = {10}; at least one = {2,4,5,6,8,10,12,14,15}; exactly one removes 10 from that list; neither = {1,3,7,9,11,13}. Check counts 1,9,8 and 6 respectively.
Changed universe	The odd cards 13 and 15 are added to the complement when U grows to 15; the even membership rule is unchanged. A complement is always relative to its stated universe.
Diagram annotation	Interpret the actual labels and shading. A compatible example must obey the represented logical relationship. Do not invent a measured probability from a conventional circle's area.
Survey conversion	P only 14, both 8, Q only 11, neither 7. Union 33 equals 22+19−8 and 40−7. The table row totals are 22 and 18; column totals 19 and 21.
New spinner grid	There are 12 equally likely ordered cells. Odd sum has 6; sum greater than 5 has 3; both has 1, namely the pair giving sum 7. Union 8 and probability 8/12=2/3. List the cells to establish completeness rather than trusting the counts alone.
Impossible data	Claimed union 20+18−6=32 exceeds 25. Feasible overlap ranges from 13 to 18. Choosing 14 as a proposed repair gives A only 6, both 14, B only 4 and neither 1.
Card models	With replacement there are 16 ordered pairs and 4 are RR, so P(RR)=1/4. Without replacement there are 12 ordered pairs and 2 are RR, so P(RR)=1/6. The four colour patterns conceal how many labelled-card pairs realise each.
Eight regions	Counts in the displayed order are 1,1,2,1,2,1,2,2. A intersection B includes 6 and 12. Exactly two gives 4, at least two gives 5, and the full union gives 10.
Counting rule	A shared member is counted twice by the two set totals and subtracted once. A triple member is added three times, subtracted three times, then added once in the three-set rule. Use actual members to check the argument.
New model	Assess whether the sample space is complete, individual outcomes are distinct and the event selections match their definitions. Count-based probability needs justified equal likelihood. A changed procedure can invalidate the old grid.
Explanation	A successful correction shows where an incorrect inference fails, not just that a teacher says it is wrong. Match the same universe and definitions across both representations; distinguish actual testing from a simulation.

Learning Check

Distinguish logical selections: For the numbered-card example, explain the difference between union, intersection, exactly one, neither and not in both. Give the exact selected members for at least three of these and show a card that distinguishes two easily confused phrases.
Reconcile a diagram and a table: A learner writes 6 in A only,2 in the overlap,4 in B only and 0 outside for the original U of twelve cards. The total is twelve. Explain why the classification is still wrong and repair it using both the membership rules and the correct two-way table.
Reason about an unknown overlap: A universe has 35 members, with 21 in A and 18 in B but no overlap supplied. Determine the possible overlap range, give two different valid regional arrangements and explain why a unique union cannot be concluded.
Distinguish sums from outcomes: In the stated red 1–4 and blue 1–3 spinner model, explain why six possible sums are not equally likely. Compare the probabilities of sum 2 and sum 4, then explain how overlapping events are counted on the twelve-cell grid.
Interpret an inclusive intersection: Explain why A intersection B in the three-set example contains two members even though the A-and-B-only region has one. Use the triple region and verify the count through the three-set union rule.
Evaluate a proposed probability method: A learner draws four boxes labelled RR,RB,BR,BB for the card procedure without replacement and assigns probability 1/4 to each. Explain the hidden aggregation error, replace the model with a complete labelled-card grid and state the corrected probabilities for all four colours.

Assessment feedback

Logical selections: the union is {2,3,4,6,8,9,10,12}; the intersection is {6,12}; exactly one is {2,3,4,8,9,10}; neither is {1,5,7,11}. Not in both is U without 6 and 12. For example, card 2 is not in both but is not in neither. Explain the membership tests, not just the symbols.

Incorrect diagram: the total alone hides wrong set memberships. A only must contain 4 cards, both 2, B only 2 and neither 4. The proposed regions would give A membership 8 rather than 6, and B membership 6 rather than 4. Use individual examples such as 1 outside both and 6 inside both to expose the error.

Unknown overlap: it ranges from 4 to 18. If the overlap is 4, the union is 35 and neither 0; exclusive counts are 17 and 14. If the overlap is 10, the union is 29 and neither 6; exclusive counts are 11 and 8. Both fit the stated totals, so the union is not determined uniquely.

Sums and outcomes: sum 2 occurs only at (1,1), giving probability 1/12; sum 4 occurs at (1,3),(2,2),(3,1), giving 3/12 = 1/4. Repeated sums do not merge distinct elementary outcomes. Enumerate the event intersection first or mark each selected cell once to avoid duplicating its contribution to the union.

Three-set intersection: 6 is in all three, while 12 is in A and B but not C. Both belong to A intersection B. The three-set rule gives 6+4+6−2−3−2+1 = 10, agreeing with the eight-region enumeration. Pairwise intersections in that rule are inclusive.

Card model: count distinct allowed ordered pairs. RR has 2, RB 4, BR 4 andBB 2 out of 12, giving 1/6,1/3,1/3 and 1/6. The missing diagonal reflects no replacement. Equal selection of remaining cards justifies equally likely labelled pairs; it does not make their four aggregated colour categories equal.

Learning evidence

Your work should show a clearly stated universe, unambiguous membership rules and complete enumerations without duplicate elements. Include a membership table, a Venn diagram, a two-way count table and a labelled outcome grid that each answer a relevant question.

Explain why inclusive totals differ from exclusive regions, why union includes the overlap once, and why complement depends on the universe. Show a checked count of an intersection, a union and a neither region. Include one contradiction or insufficient-information example and explain it without inventing a missing observation.

For combined events, distinguish elementary outcomes from values or categories calculated from them. Record the procedure's assumptions about order, replacement and likelihood. Include an independently checked probability calculation and one example where counting broad labels equally would fail.

Keep an original incorrect attempt, if available, and its corrected version. Identify the precise membership, duplication, omission or modelling error you repaired. A correctly labelled simulation is valid learning work; do not claim an experiment, survey or recording took place when it did not.

OER and Wikipedia

This matching background article explores Venn diagrams and their relationship with set operations. The worked examples, invented data, tasks and feedback in this course are original teaching material. External resources supplement the native practice rather than replacing it.

Sources and media record

Source or medium	Provenance and use
Department for Education: mathematics programmes	Key Stage 3 probability objective on systematic enumeration with tables, grids and Venn diagrams. Year 9 placement is editorial. This course develops that complete objective rather than claiming to cover all probability in one unit.
OpenStax: set operations with two sets	Mathematical reference for union, intersection, inclusive or and cardinality. Course examples and explanations are newly written; no textbook diagram or exercise copied.
NRICH: representations of probability	Jenny Gage, University of Cambridge NRICH. Reference for connecting counts, proportions and representations. No NRICH exercise or worksheet republished.
AMSI: Sets and Venn diagrams	Teacher reference for definitions and checking enumeration. Its restrictive non-commercial/no-derivatives licence is not treated as permission to adapt or redistribute its diagrams or exercises; none are copied here.
File:Venn.svg	Commons source; Jnnnnn; public-domain diagram. Shows intersection, union and complement from left to right. Display resizing only.
File:Venn A intersect B.svg	Commons source; Cepheus; public-domain diagram. The overlap is labelled A intersection B; other colours still belong to the full respective sets. Display resizing only.
File:Venn diagram for A union B.svg	Commons source; He7d3r; CC0. Display resizing only.
File:Absolute complement.svg	Commons source; User:Brighterorange; CC BY-SA 3.0. The grey region is outside A but inside U. Display resizing only; no source-image changes.
xwK--rNDI9E	Venn Diagrams - Corbettmaths; corbettmaths. Actual video identity, embedding permission, playability and thumbnail verified; no additional open-video licence asserted.
Venn diagram	Verified matching English Wikipedia article; consult its attribution and licence history.

Sources and actual media checked on 10 November 2026. All card models, survey data and outcomes in the course are invented mathematical examples. No real learner information is used, and no physical experiment or complete playback of every video is claimed.