Complements and mutually exclusive addition
Identify complements within an exhaustive sample space and add probabilities only for events that cannot occur together.
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
- PR1–PR3
- fractions and exhaustive categories
- set complements
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What justifies P(A or B)=P(A)+P(B) under the mutually-exclusive addition rule?
What you will learn
- Identify the complement relative to the stated exhaustive sample space.
- Use 1−P(A), and add event probabilities only when the alternatives are mutually exclusive.
- Distinguish overlap from disjointness and from independence; defer multiplication/conditional rules to Higher and never silently assume missing probabilities.
A single spinner result is red, blue or green with probabilities 7/20, 1/4 and 2/5. Find P(not red) and P(red or blue). Explain the different rules used.
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
- green
probabilities
- 7/20
- 1/4
- 2/5
categoriesExclusiveAndExhaustive
trueThe 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 notRed; redOrBlue. 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
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.
Define the whole space and its complement
The complement of event A consists of all outcomes in the stated sample space that are not in A. The event and its complement are disjoint and together cover the whole space. Their probabilities therefore add to one, so the complement probability is one minus P(A). The reference sample space must stay the same throughout the subtraction.
A missing category probability can be recovered by subtraction only when the stated categories are mutually exclusive and exhaustive. Real-world words such as weather categories do not guarantee those properties unless the model defines them. Check the result lies between zero and one; do not use a negative remainder or quietly assume that unmentioned outcomes cannot occur.
Establish non-overlap before adding
Two events are mutually exclusive when no elementary outcome belongs to both. In that case, their either-or event contains their separate outcome counts without overlap, so its probability is the sum of the two probabilities. Show a set list, Venn membership or a clear description that establishes the non-overlap condition before applying the rule.
If an outcome belongs to both descriptions, adding their separate probabilities counts that outcome twice. The mutually-exclusive shortcut is then unavailable. A finite sample space can still be used to select the union directly, counting each elementary outcome once. Distinct names or different verbal descriptions do not establish that events are disjoint.
Reason carefully with negative descriptions
Events described as not red and not green can overlap: an outcome in a third category may satisfy both statements. Work from actual membership rather than assuming that two complements are opposites. Draw or list a shared outcome as a counterexample when non-overlap fails. A complement belongs to one specified event, not automatically to any other event mentioned nearby.
Mutual exclusivity concerns whether events share outcomes. Independence concerns whether information about one event changes the probability of another and is a separate Higher method. Do not swap those words or assume that unrelated-sounding labels justify a probability rule. Here complement subtraction and disjoint addition remain tied to the declared exhaustive space and actual event membership.
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
P(A)=2/7 in a stated exhaustive sample space. What is P(not A)?
Worked example
A single spinner result is red, blue or green with probabilities 7/20, 1/4 and 2/5. Find P(not red) and P(red or blue). Explain the different rules used.
Show the worked solution
- Identify the exhaustive sample space and actual event membership. Mark the complement or the disjoint union before choosing subtraction or mutually-exclusive addition; exhibit any shared comparison outcome.
- The three mutually exclusive outcomes cover the sample space and their probabilities total 1.
- P(not red)=1−7/20=13/20 uses the complement rule.
- Red and blue cannot occur on the same spin, so P(red or blue)=7/20+1/4=12/20=3/5.
Answer P(not red)=13/20 by complement subtraction. P(red or blue)=3/5 by adding the mutually exclusive red and blue outcomes. The three stated colour probabilities total 1.
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 card is chosen uniformly from labels 1 to 12. Let A be score at most 4 and B be score at least 10. Find P(not A) and P(A or B), explaining why addition is allowed. Would the same simple addition be valid for “even” or “greater than 8”?
Give me a hint
Identify the exhaustive space, mark an event complement and check for an actual shared outcome before adding probabilities.
Compare my reasoning
- Identify the exhaustive sample space and actual event membership. Mark the complement or the disjoint union before choosing subtraction or mutually-exclusive addition; exhibit any shared comparison outcome.
- A contains 1, 2, 3, 4 and B contains 10, 11, 12, so their probabilities are 4/12 and 3/12.
- P(not A)=1−4/12=2/3. The two specified events are disjoint, so P(A or B)=4/12+3/12=7/12.
- The alternative events overlap at 10 and 12. Adding their probabilities without accounting for the shared outcomes would count them twice, so the mutually-exclusive rule does not apply.
A={1,2,3,4} and B={10,11,12}; P(A)=1/3, P(B)=1/4, P(not A)=2/3 and P(A or B)=7/12. A and B are disjoint. Even and greater-than-8 overlap at 10 and 12, so simple mutually-exclusive addition is invalid for that comparison.
Look for these in your work
- List the disjoint event sets.
- Use the correct denominator 12.
- Exhibit actual overlap for the invalid comparison.
A simplified daily weather model has exactly one of dry, rain or snow. P(dry)=11/20 and P(rain)=7/20. Find P(snow), P(not rain) and P(rain or snow), giving the conditions that make the calculations valid.
Give me a hint
Identify the exhaustive space, mark an event complement and check for an actual shared outcome before adding probabilities.
Compare my reasoning
- Identify the exhaustive sample space and actual event membership. Mark the complement or the disjoint union before choosing subtraction or mutually-exclusive addition; exhibit any shared comparison outcome.
- The three categories are explicitly exhaustive, so P(snow)=1−11/20−7/20=1/10.
- The complement of rain has probability 1−7/20=13/20.
- Rain and snow are mutually exclusive in this simplified model, so their either/or probability is 7/20+1/10=9/20. Real weather need not have these disjoint categories unless the model defines them that way.
P(snow)=1/10, P(not rain)=13/20 and P(rain or snow)=9/20. Subtraction uses the declared exhaustive dry/rain/snow model; addition uses its mutually exclusive categories.
Look for these in your work
- Use the declared model, not an unstated real-world assumption.
- Check the recovered probability lies in 0–1.
- Name complement and exclusivity conditions.
A bag contains 18 tokens, each exactly one of red, blue or green. There are 6 blue and 4 green tokens. Find the red count, P(not blue) and P(red or green). Explain why “not red” and “not green” are not mutually exclusive. For each probability, select one physical token uniformly.
Give me a hint
Identify the exhaustive space, mark an event complement and check for an actual shared outcome before adding probabilities.
Compare my reasoning
- Identify the exhaustive sample space and actual event membership. Mark the complement or the disjoint union before choosing subtraction or mutually-exclusive addition; exhibit any shared comparison outcome.
- The remaining red count is 18−6−4=8.
- P(not blue)=1−6/18=2/3. Red and green are disjoint colours, so P(red or green)=(8+4)/18=2/3.
- Every blue token is both not red and not green. The shared six blue tokens show that those two complemented events overlap and cannot use the mutually-exclusive addition shortcut.
There are 8 red tokens. P(not blue)=2/3 and P(red or green)=2/3. Every one of the 6 blue tokens is both not red and not green, so those two complemented events overlap.
Look for these in your work
- Recover the missing count from an exhaustive total.
- Use either complement or disjoint counting correctly.
- Give a concrete overlap witness.
Common mistakes
- Using a remainder when categories are not known to cover the whole model.
- Adding overlapping probabilities as though the events were mutually exclusive.
- Treating two not statements as opposite events without checking shared outcomes.
What can you explain now?
Give a membership reason why not A and not B need not be mutually exclusive.
Compare with the explanation
An outcome outside both A and B belongs to both complements.
That shared outcome is enough to refute the claim that the complements have an empty intersection.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State the complement rule and the exact non-overlap condition for disjoint addition, with a membership counterexample when it fails.
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.