Exhaustive Sets and Mutually Exclusive Events Revision Notes | GCSE AQA Higher Maths

1. The Core Idea: Probabilities That Account for Everything

Imagine you roll a fair six-sided die. The possible outcomes are 1, 2, 3, 4, 5, and 6. There is no other result — those six outcomes cover every possibility. Because they cover everything, their probabilities must add up to exactly 1. This is the heart of what we mean by an exhaustive set.

Now add one more condition: none of those outcomes can happen at the same time. You cannot roll a 3 and a 5 on a single throw. When outcomes cannot overlap and they cover every possibility, you have a set that is both mutually exclusive and exhaustive — and that combination unlocks a powerful rule for finding missing probabilities.

Think of it like a pie chart that must be completely filled. Every slice represents one outcome, and together they must account for the whole pie — a total probability of 1. If any slice is missing, the chart is incomplete and the set is not exhaustive.

2. The Key Property and Formula

When a set of events is both mutually exclusive and exhaustive, you can write a single clean rule. This rule is what examiners test directly, so it is worth pinning down precisely.

The most common application is finding a missing probability. If you know all probabilities except one, subtract the sum of the known ones from 1. The result is the missing value. This works only because the set is exhaustive and mutually exclusive — that is the condition that guarantees the total is 1.

3. Worked Examples

The rule is clean in theory — now let's see how it plays out in the style of questions you will meet in the exam. The first example uses a straightforward table of probabilities; the second introduces an algebraic unknown.

Worked example: Finding a Missing Probability from a Table

A spinner has five sections labelled 1 to 5. The table below shows the probability of landing on each number. The spinner is fair enough that the events are mutually exclusive and exhaustive. Find the missing probability $P(4)$. | Outcome | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | Probability | 0.15 | 0.20 | 0.25 |? | 0.18 |

1. $P(1) + P(2) + P(3) + P(4) + P(5) = 1$ — The five outcomes cover every possibility and cannot overlap, so their probabilities must sum to 1. [M1 — applying the exhaustive property]
2. $0.15 + 0.20 + 0.25 + P(4) + 0.18 = 1$ — Replace each known probability with its value, leaving $P(4)$ as the unknown.
3. $0.15 + 0.20 = 0.35$ $0.35 + 0.25 = 0.60$ $0.60 + 0.18 = 0.78$ — Add carefully, one step at a time, to avoid arithmetic errors.
4. $0.78 + P(4) = 1$ $P(4) = 1 - 0.78$ $P(4) = 0.22$ — Subtract the known sum from 1. [A1 — correct answer]
5. $0.15 + 0.20 + 0.25 + 0.22 + 0.18 = 1.00$ ✓ — All five probabilities sum to exactly 1, confirming the answer is correct.

Answer: $P(4) = 0.22$

That example used decimals. Exam questions also use fractions and algebraic expressions — the method is identical, but you need to handle the algebra carefully. Here is how that looks.

Worked example: Finding a Missing Probability Expressed Algebraically

Events $A$, $B$, $C$, and $D$ are mutually exclusive and exhaustive. Their probabilities are $P(A) = 2k$, $P(B) = 0.3$, $P(C) = k$, and $P(D) = 0.1$. Find the value of $k$.

1. $P(A) + P(B) + P(C) + P(D) = 1$ — The four events are mutually exclusive and exhaustive, so their probabilities sum to 1. [M1]
2. $2k + 0.3 + k + 0.1 = 1$ — Replace each probability with its given expression or value.
3. $3k + 0.4 = 1$ — $2k + k = 3k$ and $0.3 + 0.1 = 0.4$. [M1 — correct simplification]
4. $3k = 1 - 0.4$ $3k = 0.6$ $k = \frac{0.6}{3} = 0.2$ — Subtract 0.4 from both sides, then divide by 3. [A1 — correct value of $k$]
5. $P(A) = 2 \times 0.2 = 0.4$ $P(B) = 0.3$ $P(C) = 0.2$ $P(D) = 0.1$ $0.4 + 0.3 + 0.2 + 0.1 = 1.0$ ✓ — Substituting $k = 0.2$ back gives four valid probabilities that sum to 1.

Answer: $k = 0.2$

4. Retrieval Check

Before moving on to the trickier variations, check whether the core method has landed. This question targets the most common slip students make when applying the exhaustive property.

5. Complementary Events: The Two-Outcome Special Case

The exhaustive and mutually exclusive rule applies to any number of events. When there are exactly two events — an event and its complement — you get the complementary events rule, which you may already recognise from earlier probability work.

If $A$ is any event, then 'not $A$' (written $A'$ or $\bar{A}$) is the event that $A$ does not occur. Together, $A$ and 'not $A$' are mutually exclusive (they cannot both happen) and exhaustive (one of them must happen). So $P(A) + P(\text{not } A) = 1$, which rearranges to $P(\text{not } A) = 1 - P(A)$.

6. Applying the Property: Practice Problems

The worked examples above showed the method in isolation. Exam questions often embed the same skill inside a context — a biased spinner, a bag of counters, or a survey result. The method does not change, but you need to recognise when to apply it. Work through these problems before checking the solutions.

7. Exam Technique and Common Mistakes

The property itself is straightforward, but marks are lost in predictable ways. The traps below come up repeatedly in mark schemes — knowing them in advance is a genuine advantage.

One more examiner pattern worth knowing: questions sometimes give you an expression like $P(A) = 2k$ and ask you to find $k$ and then find a specific probability. Make sure you substitute your value of $k$ back into the original expression — the question is asking for the probability, not just $k$.

8. Exam-Style Question

This final question combines both specification requirements in one problem — finding a missing probability and verifying the exhaustive property. It mirrors the style and demand of an GCSE Higher exam question.