Combined Events, Possibility Spaces and Mutually Exclusive Events — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers probability when more than one thing happens: two dice, a coin and a spinner, two counters drawn from a bag. By the end of this guide you should be able to build a possibility space, count outcomes from it, and calculate the probability of a combined event.

You should also be able to recognise mutually exclusive events and add their probabilities, use the complement rule for "not" and "at least one" questions, and know when probabilities may be multiplied.

The organising idea is that "and" multiplies while "or" adds — but each only under a condition. Multiplying works when the events are independent, so the first outcome does not change the second. Adding works when the events are mutually exclusive, so they cannot both happen and nothing gets counted twice. Almost every error in this topic is one of these rules applied where its condition fails: multiplying probabilities that affect each other, or adding probabilities for events that overlap. Checking the condition before reaching for the rule is the habit that matters.

Key terms and definitions

Combined event — an event involving more than one trial or more than one thing happening.

Possibility space (sample space diagram) — a table or grid listing every possible outcome.

Mutually exclusive — events that cannot both happen at the same time.

Exhaustive — a set of outcomes covering every possibility, so their probabilities total 1.

Independent — events where one outcome has no effect on the other.

Complement — the event "not A", whose probability is 1 − P(A).

With replacement — the item is returned, so probabilities stay the same.

Without replacement — the item is kept, so the second probability changes.

Core concepts

Possibility spaces

A possibility space lists every outcome, and for two events a grid is the fastest way to build one.

Put one event along the top and the other down the side, then fill each cell with the combined result.

Two dice give a 6 by 6 grid with 36 cells. A coin and a four-sided spinner give 2 by 4, so 8 outcomes. The number of cells is the two totals multiplied, not added.

Once the grid is filled, probability is counting: favourable cells over total cells.

The crucial point is that the cells are equally likely, but what fills them may not be. Filling a two-dice grid with totals shows 7 appearing six times and 2 appearing once, so the totals are not equally likely at all. Always count cells.

Listing systematically

Where a grid does not suit — three coins, say — list in a fixed order instead.

For three coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Eight outcomes, found by running through the first coin, then the second, then the third.

Listing at random almost guarantees a missed outcome, and a missing outcome makes the denominator wrong, which spoils every answer that follows.

Mutually exclusive events

Two events are mutually exclusive when they cannot both happen in one trial: rolling a 2 and rolling a 5, or drawing a red card and drawing a black card.

For these, P(A or B) = P(A) + P(B).

So the probability of rolling a 2 or a 5 is 1/6 + 1/6 = 2/6.

The condition matters. Drawing a red card and drawing a king are not mutually exclusive, because the two red kings satisfy both. Adding 26/52 and 4/52 would count those two cards twice, giving 30/52 instead of the correct 28/52.

Before adding, ask whether the two events could ever happen together. If they could, adding is wrong.

Exhaustive events and the total of 1

A set of mutually exclusive outcomes that covers every possibility is exhaustive, and its probabilities total 1.

This gives two standard techniques.

A missing probability is found by subtracting the rest from 1. If a spinner's colours have probabilities 0.3, 0.25 and 0.15 with one unknown, the missing one is 1 − 0.7 = 0.3.

The complement rule: P(not A) = 1 − P(A). If the probability of rain is 0.35, the probability of no rain is 0.65.

"At least one" questions

These are where the complement rule earns its keep.

"At least one" covers several outcomes, but its opposite — "none" — is usually a single case, so calculating that and subtracting from 1 is far quicker.

For two coins, "at least one head" covers HH, HT and TH. Its opposite is TT alone, with probability 0.25, so the answer is 1 − 0.25 = 0.75.

The saving grows with the number of trials. For five coins, "at least one head" covers 31 of the 32 outcomes, while "no heads" is one.

Independent events and multiplying

Two events are independent when one has no effect on the other — separate dice, or drawing with replacement.

For independent events, P(A and B) = P(A) × P(B).

So the probability of two heads on two coins is 0.5 × 0.5 = 0.25.

Multiplying is only valid under independence. If a counter is drawn and not replaced, the second probability changes, and the conditional probability must be used instead — which is the subject of tree diagrams.

With and without replacement

With replacement, the item goes back, the total stays the same, and the two draws are independent. Drawing two reds from a bag of 3 red and 7 others is (3/10) × (3/10).

Without replacement, one item has gone. Both the numerator and the denominator fall, so the same calculation becomes (3/10) × (2/9).

Reading which case applies is the first thing to do in any two-draw question, and the phrase "and does not replace it" is the signal.

Relative frequency and estimating probability

Where outcomes are not equally likely — a bent coin, a weighted spinner — the formula for equally likely outcomes cannot be used, and the probability has to be estimated from results.

Relative frequency = number of times the event happened ÷ number of trials.

Spinning a spinner 200 times and landing on red 84 times gives a relative frequency of 84 ÷ 200 = 0.42.

The more trials, the closer the estimate usually gets to the true probability. That is why a judgement about bias needs the number of trials mentioned: twelve spins showing eight reds proves nothing, whereas twelve hundred spins showing eight hundred reds is strong evidence.

Comparing a relative frequency with the theoretical probability is the standard way these questions test fairness. A fair six-sided die should give each number a relative frequency near 1/6, and a persistent, large departure over many trials suggests bias.

Two-way tables

Some combined-event questions present the data in a two-way table rather than as probabilities, with one characteristic across the top and another down the side.

Probabilities are read straight from the counts: the number in a cell over the overall total.

The table also answers "given that" questions, since restricting attention to one row or column changes the denominator. Asking for the probability that a student walks to school given that they are in Year 11 means dividing the relevant cell by the Year 11 total rather than by the whole sample.

Filling in any missing entries from the row and column totals is usually the first step, and it is often worth a mark by itself.

Expected frequency

To estimate how often an event occurs over many trials, multiply its probability by the number of trials.

Rolling two dice 180 times, with P(total of 7) = 6/36 = 1/6, gives an expected 30 sevens.

It is an average expectation, not a guarantee, and questions often ask you to say so.

Worked examples

Example 1: A two-dice possibility space

Two fair dice are rolled and the scores added. Find the probability that the total is at least 10.

The grid has 36 equally likely cells.

Totals of at least 10 mean 10, 11 or 12. A total of 10 comes from three cells, 11 from two, and 12 from one.

That is 3 + 2 + 1 = 6 favourable cells.

So the probability is 6/36 = 1/6.

Counting the three totals as three possibilities, rather than counting cells, would give the wrong answer entirely.

Example 2: Mutually exclusive or not

A card is drawn from a standard pack. Is P(a heart or a queen) equal to P(heart) + P(queen)?

No. A card can be both — the queen of hearts satisfies each condition.

The events are therefore not mutually exclusive, so adding would count that card twice.

P(heart) = 13/52 and P(queen) = 4/52, but their sum of 17/52 over-counts by one card. The correct probability is 16/52, which simplifies to 4/13.

Example 3: "At least one"

A bag contains 4 red and 6 blue counters. A counter is drawn and replaced, then a second is drawn. Find the probability of getting at least one red.

Work with the opposite. "No reds" means blue twice.

With replacement the draws are independent, so P(blue twice) = (6/10) × (6/10) = 36/100.

Therefore P(at least one red) = 1 − 36/100 = 64/100, which simplifies to 16/25.

Adding 4/10 and 4/10 would give 8/10, which double-counts the case where both are red.

Common mistakes and how to avoid them

Adding probabilities of events that overlap. Check they are mutually exclusive first.

Multiplying probabilities that affect each other. Multiplying needs independence, so check for replacement.

Counting totals instead of cells. On two dice the 36 cells are equally likely; the totals are not.

Listing outcomes unsystematically. Work in a fixed order, or a case will be missed.

Tackling "at least one" directly. Use 1 minus the probability of none.

Using the same denominator twice without replacement. The total falls by one after the first draw.

Giving a probability above 1. Check every answer lies between 0 and 1.

Exam technique for "Combined Events"

Draw the possibility space grid whenever two events combine. It is quicker than listing and much harder to get wrong.

Write down the total number of outcomes first, since it fixes the denominator and usually earns a mark.

Before adding, ask whether the events could happen together; before multiplying, ask whether one affects the other. Stating the answer to that question shows the examiner your reasoning.

Use 1 − P(none) for any question containing "at least".

Check the wording for "replaced" or "not replaced" before doing any arithmetic.

Simplify fractions and confirm the answer sits between 0 and 1.

Quick revision summary

"And" multiplies, "or" adds — but each only under its condition: multiplying needs independence, adding needs the events to be mutually exclusive.

A possibility space grid lists every outcome; the number of cells is the two totals multiplied. Two dice give 36 cells, which are equally likely even though the totals are not.

Mutually exclusive events cannot both happen, so P(A or B) = P(A) + P(B). A heart and a queen are not mutually exclusive, so that rule fails there.

Exhaustive probabilities total 1, so a missing probability is found by subtraction and P(not A) = 1 − P(A).

For "at least one", work out the probability of none and subtract from 1.

With replacement the draws are independent and the denominator stays the same; without replacement both numerator and denominator fall.

Expected frequency = probability × number of trials, and it is an average rather than a guarantee.