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Probability combined events and sets - Mathematics Year 9

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Introduction

Why are two dice sums not all equally likely? Build ordered outcomes before assigning probabilities, then distinguish independent stages from a changing sample space.

Calculate four combined probabilities and solve three complete set-counting problems. Use tables, a labelled Venn region model and independent checks; state fairness, independence and replacement assumptions.

Build equally likely ordered sample spaces

An outcome describes the entire result of an experiment. For two distinguishable fair six-sided dice, write an ordered pair(first die, second die). Assume independent throws, so every one of the 6×6=36 pairs has probability 1/36. The pair(2, 5) differs from(5, 2), even though both have sum 7. Record both rather than merging them into one outcome.

An exhaustive sample-space table. The row is the first die and the column is the second die. A cell contains their sum, not a new probability. Each cell represents one equally likely ordered outcome under the stated assumptions.

First die / second die	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

The sum 7 cells correspond to(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Six favourable cells out of 36 give 6/36=1/6. Counting by rows supplies a separate check: for each first die a from 1 to 6, exactly one legal second value 7−a completes the sum.

The possible sums are 2 to 12, but these eleven labels are not equally likely. Sum 2 has one ordered outcome(1, 1), while sum 7 has six. Therefore “one wanted sum out of eleven” is not a valid equally likely outcome count. The first error is treating sum labels as equally weighted before checking the underlying pairs producing each label.

For sum at least 10, count sum 10: (4, 6), (5, 5), (6, 4); sum 11: (5, 6), (6, 5); and sum 12: (6, 6). There are 3+2+1=6 favourable pairs, so this different event also has probability 1/6. Having the same probability does not make events identical: sum 7 and sum at least 10 have no common outcomes.

For two independent fair coins, HH, HT, TH, TT are four equally likely ordered outcomes. Each has probability 1/4, which the four equal pieces in the opening image can represent. Heads on the first toss and tails on the second is one outcome HT; “exactly one head in either order” includes HT and TH and has probability 1/2.

An exhaustive table is useful only if its rows and columns match the model. A weighted die or dependent stages changes cell probabilities; the same 36 cells may remain possible without equal weight 1/36. State the model rather than assuming that a rectangular table guarantees equal likelihood.

Calculate four probabilities and check changing branches

Four complete combined probabilities. Cover the final column and identify what is being counted or multiplied. The experiments in different rows are separate models; the changing bag composition in row 4 does not apply to dice.

Task	Experiment and event	Working	Probability
1	Two independent fair dice: sum 7	6/36	1/6
2	Same dice: sum at least 10	6/36	1/6
3	Independent fair coin and die: head and die at least 5	1/2 × 2/6	1/6
4	Bag 3 red 2 blue; two draws without replacement: both red	3/5 × 2/4	3/10

For task 3, the fair coin and fair die are independent. Head has probability 1/2 and die 5 or 6 has probability 2/6. Multiplying gives 1/6. Alternatively list twelve equally likely outcomes: H 1 to H 6 and T 1 to T 6. Favourable outcomes H 5, H 6 give 2/12=1/6, checking the product through a different representation.

Without replacement changes the second stage. The bag contains three individually distinct red counters and two individually distinct blue counters. Every remaining counter is equally likely on each draw. After red first, only two red counters remain among four total, so the second-red probability is 2/4. The complete RR path is 3/5×2/4=3/10.

After blue first, the remaining bag instead has 3 red 1 blue. The second probabilities are 3/4 red and 1/4 blue. Retain both branches rather than reusing red-first probabilities everywhere.

Ordered colour path	First stage	Second given first	Path probability
RR	3/5	2/4	3/10
RB	3/5	2/4	3/10
BR	2/5	3/4	3/10
BB	2/5	1/4	1/10

Read each row as a tree path: first choose R or B, then the second colour with the displayed conditional probability. At the first fork 3/5+2/5=1. After red, the fork 2/4+2/4=1; after blue, 3/4+1/4=1. Complete path probabilities 3/10+3/10+3/10+1/10=1 check exhaustiveness.

For a second count, label the counters R 1, R 2, R 3, B 1, B 2. There are 5×4=20 equally likely ordered draws without replacement; 3×2=6 are red-red, so 6/20=3/10. Unordered counting also works: there are 10 distinct pairs, of which 3 are all-red. Every unordered pair has exactly two possible orders, so it receives equal weight.

With replacement, the counter is returned and the composition restored. Both-red probability becomes 3/5×3/5=9/25, not 3/10. Multiplication still describes a path, but its second factor must match the actual conditional model. Independence differs from mutual exclusivity: two possible positive-probability events that cannot occur together are not independent.

The original Corbettmaths preview combines a four-value spinner 2, 4, 6, 8 with a five-value spinner 1–5. Predict its twenty ordered cells, then compare this representation with our six-by-six dice table. Copy the method, not the spinner labels. Original: corbettmaths, Sample Space Diagrams - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Solve three Venn problems with exact region checks

A hundred grid containing one hundred equal square cells. — A hundred grid models hundredths. Probability1/6 or27/40 cannot be shown exactly by shading only whole cells on this grid. Use labelled region counts for exact fractions;the blank grid is not a completed Venn diagram.

Let U be a finite set of equally likely numbered cards. A and B are labels a card may carry. These are invented mathematical cards, no learner records. Intersection A∩B means both labels; union A∪B means at least one, including both. A complement is relative to U.

Counting n(A)+n(B) counts intersection cards twice. Correct this with $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ . Subtract the overlap once, not twice. Fill the intersection first, then subtract it from each set total to find the exclusive regions.

Three complete Venn problems. In case 3, union 40 is supplied and intersection is unknown: 28+23−40=11. Hide the worked region columns and reconstruct them.

Case	U total	A total	B total	Both	A only	B only	Neither	Union
1	40	18	14	5	13	9	13	27
2	30	16	12	4	12	8	6	24
3	50	28	23	11	17	12	10	40

For source case 1, the union 18+14−5=27. A-only is 18−5=13, B-only 14−5=9, and neither 40−27=13. A labelled region schematic separates membership from individual set totals:

Case 1: universe U contains 40 cards
       A                         B
   +-----------+------------+-----------+
   | A only 13 | Both A,B 5 | B only 9  |
   +-----------+------------+-----------+
Outside both sets: neither 13
Regions total: 13 + 5 + 9 + 13 = 40
Schematic regions; not area-proportional.

These regions partition U. Their sum 13+5+9+13=40 independently checks conservation, while A-only plus both 13+5=18 and B-only plus both 9+5=14 rebuild the set totals. The boxes provide a labelled membership alternative to overlapping circles; their visual widths do not encode probabilities.

For a random card from case 1, P(A∩B)=5/40=1/8, P(A∪B)=27/40 and P(exactly one)=22/40=11/20. Complement of the union is neither, 13/40. Complement of A is B-only plus neither, 22/40; it is not B-only alone.

In case 2, overlap 4 gives A-only 12, B-only 8 and neither 6. Union 24 agrees with 12+4+8, and all four regions sum 30. In unfamiliar case 3, intersection 11 gives 17 A-only, 12 B-only and 10 neither. Rebuild A 28, B 23, union 40 and total 50 to check every supplied constraint.

To repair answer 32 for source union, identify that 18+14 counts five cards twice. Remove one extra copy to obtain 27; subtracting ten would wrongly delete the overlap completely. Keep four probability calculations, three region records, assumptions, a corrected first error and independent checks.

Interactive Tasks

Quiz: Test Your Knowledge

How many equally likely ordered outcomes for two independent fair dice? (36) (!11) (!12) (!21)

Probability that their sum is 7? (1/6) (!1/11) (!1/36) (!7/36)

Which event also has probability 1/6? (Sum at least 10) (!Sum exactly 2) (!Sum at least 12) (!Sum exactly 6)

Independent fair coin and die: head and die 5 or 6? (1/6) (!1/3) (!2/3) (!1/12)

Bag 3 red 2 blue, two draws without replacement: both red? (3/10) (!9/25) (!3/5) (!2/5)

After blue first, probability of red second? (3/4) (!3/5) (!2/4) (!1/4)

Source sets A 18, B 14, both 5: union? (27) (!32) (!22) (!5)

Same sets in U 40: neither? (13) (!27) (!22) (!18)

Case 3: A 28, B 23, union 40: both? (11) (!51) (!17) (!12)

Independence and mutual exclusivity are? (Different conditions) (!Always identical) (!Names for equal probability) (!Guaranteed by a table)

Memory Game

Outcome	Complete experimental result
Sample space	All possible outcomes for the stated model
Intersection	Membership in both sets
Union	Membership in at least one set, including both
Complement	Inside U but outside the stated event
Replacement	Returning an item before the next draw

Drag and Drop

36	Two fair dice ordered outcomes
3/10	Both red without replacement
27	Source union count
13	Source neither count
11	Case 3 intersection count

Crossword Puzzle

OUTCOME	Complete experimental result
UNION	At least one set including its overlap
INTERSECTION	Membership in both sets
COMPLEMENT	Inside the universe but outside the event
INDEPENDENT	Describes stages whose outcomes do not affect one another probabilities
REPLACEMENT	Returning an item before another draw

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Read the dice model: Name row, column and cell meaning; state fairness and independence.
Count sum 7: List all six ordered pairs instead of counting sum labels.
Read the changing bag: Write both second-stage probabilities after red and after blue.
Name the four regions: Distinguish both, A-only, B-only and neither inside U.

Standard

Calculate four combined probabilities: Retain a full counting or path method for every row.
Solve three Venn problems: Rebuild every region, each set total and the universe.
Compare independent methods: Use twelve coin/die cells and twenty labelled counter draws.
Repair a first error: Correct union 32 and dice probability 1/11 with reasons.

Advanced

Compare replacement models: Explain 3/10 without replacement versus 9/25 with replacement.
Recover a missing intersection: Use A 28, B 23, union 40 to find 11 and all other regions.
Distinguish event relationships: Compare equal probabilities for sum 7 and sum at least 10.
Retain real evidence: Keep four probabilities, three Venn solutions, assumptions and an actual later response.

Learning Check

Four probabilities: Show exhaustive counts or conditional paths.
Changing-bag branches: Account for every ordered colour path.
Source Venn case 1: Rebuild all regions and probabilities.
Two new Venn cases: Recover missing counts and check every constraint.
First errors and assumptions: Identify why shortcuts fail.
Portfolio and visual limits: Retain complete reasoning and exact representations.

Worked checks and assessment guidance

Dice sum 7 gives 6/36=1/6; sum at least 10 gives 3+2+1=6 cells and 1/6. Head and die 5/6 gives 1/2×2/6=1/6 or 2/12. Both-red without replacement gives 3/5×2/4=3/10 or 6/20 labelled draws.
RR, RB, BR, BB have probabilities 3/10, 3/10, 3/10, 1/10. After red, the next fork is 2/4 each; after blue, it is 3/4 red and 1/4 blue. Paths sum 1. With replacement, RR becomes 9/25.
U 40, A 18, B 14, both 5 gives A-only 13, B-only 9, neither 13, union 27. Regions total 40 and reproduce both sets. Intersection 1/8, union 27/40, exactly one 11/20, neither 13/40.
Case 2 U 30, A 16, B 12, both 4 gives A-only 12, B-only 8, neither 6, union 24. Case 3 U 50, A 28, B 23, union 40 gives both 11, A-only 17, B-only 12, neither 10.
Eleven dice sums have unequal cell counts, so 1/11 is invalid. Union 18+14=32 counts five overlap cards twice; subtract 5 once to get 27. Path multiplication uses the actual conditional second factor. Independence is not mutual exclusivity.
Keep four probabilities, three labelled Venn records, exhaustive models, independent checks and corrected first errors. A hundred grid cannot express 1/6 with whole cells. Schematic widths are not probabilities; actual later retrieval remains learner work.

Learning evidence

Retain four combined probabilities, three complete Venn solutions, ordered sample spaces, a labelled region diagram, model assumptions, independent checks, first-error correction and unfamiliar transfer.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, Source area: Key stage 3 > Subject content > Probability. Fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

PieChartFractionFourthsSplit.svg, Ezra Katz, CC BY-SA 3.0. Unchanged original; caption and learning activities are original.

100 grid.PNG, The original uploader was Leightwing at English Wikiversity., Public domain. Unchanged original; caption and learning activities are original.