Venn Diagrams and Set Notation in Probability — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers sets, the notation used to describe them, and Venn diagrams as a tool for solving probability problems. By the end of this guide you should be able to read the union, intersection and complement symbols and shade the regions they describe.

You should also be able to fill in a Venn diagram from given information, calculate probabilities from it, handle conditional probability, and work with three sets.

The organising idea is that a Venn diagram turns overlapping groups into non-overlapping regions. The reason probability questions about two categories are confusing is that the categories share members — some students study both French and German, so adding the two counts them twice. A Venn diagram solves that by splitting everything into regions that do not overlap: French only, German only, both, and neither. Once every item sits in exactly one region, probability becomes counting. That is also why the diagram must be filled in from the middle outwards, since the overlap has to be removed from each group before the outer regions are known.

Key terms and definitions

Set — a collection of items, written inside curly brackets or shown as a circle.

Element — one item in a set. The symbol ∈ means "is an element of".

Universal set (ξ) — everything under consideration, drawn as the rectangle enclosing the circles.

Union (A ∪ B) — everything in A or B or both.

Intersection (A ∩ B) — everything in both A and B.

Complement (A′) — everything not in A.

Empty set (∅) — a set with no elements.

Mutually exclusive — sets with no overlap, so their intersection is empty.

Core concepts

The symbols

A ∪ B is the union, meaning A or B or both. Think of the ∪ as a cup gathering everything up.

A ∩ B is the intersection, meaning both at once — only the overlap.

A′ is the complement of A, meaning everything outside A.

ξ is the universal set, the rectangle containing everything.

n(A) means the number of elements in A, as opposed to the set itself.

Combinations appear too. (A ∪ B)′ is everything outside both circles, and A ∩ B′ is the part of A that is not in B — the "A only" region.

Reading these accurately is half the topic, and it is worth translating each symbol into ordinary words before shading anything.

The four regions of a two-set diagram

Two overlapping circles inside a rectangle create exactly four regions:

A only — inside A but outside B.

Both — the overlap, A ∩ B.

B only — inside B but outside A.

Neither — inside the rectangle but outside both circles.

Every item belongs to exactly one of these, and their four totals add to the size of the universal set. That last point is the check to run on every completed diagram.

Filling in a diagram: start in the middle

Always enter the intersection first, then work outwards by subtraction.

If 30 students study French, 25 study German and 12 study both, then the 12 goes in the overlap. The French-only region is 30 − 12 = 18, and the German-only region is 25 − 12 = 13.

If there are 50 students altogether, the "neither" region is 50 − (18 + 12 + 13) = 7.

Writing 30 and 25 straight into the circles is the standard error, because it counts the 12 twice and makes every later probability wrong.

Calculating probabilities

Once the diagram is complete, probability is counting: the total in the relevant regions over the total in the universal set.

Using the numbers above, P(French) = 30 ÷ 50, since French includes both the French-only 18 and the overlap 12.

P(French and German) = 12 ÷ 50, taking only the overlap.

P(French or German) = (18 + 12 + 13) ÷ 50 = 43 ÷ 50, taking all three circle regions once each.

P(neither) = 7 ÷ 50.

Notice that P(French or German) is not 30 ÷ 50 added to 25 ÷ 50, which would exceed 1. The overlap has to be counted once, not twice.

The addition rule

That last point is a general result:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Adding the two sets double-counts the overlap, so it is subtracted once. Here 30 + 25 − 12 = 43, matching the region count.

For mutually exclusive sets there is no overlap, so the subtraction disappears and the totals simply add — which is the familiar rule from basic probability.

Conditional probability

A conditional probability restricts attention to part of the diagram, which changes the denominator.

"Given that a student studies French" means only the 30 French students are now under consideration, so 30 becomes the denominator.

So the probability that a student studies German given that they study French is 12 ÷ 30, not 12 ÷ 50.

The phrase to watch for is "given that", and the rule is that the condition names the new total. Questions may also phrase it as "a French student is chosen at random", which means the same thing.

This is the most demanding part of the topic, and identifying the denominator before anything else is what makes it manageable.

Shading regions

Some questions give a shaded diagram and ask for the notation, or give the notation and ask you to shade.

Work symbol by symbol. For A ∩ B′, take A, then keep only the part that is not in B — the "A only" region.

For (A ∪ B)′, first find everything in either circle, then take everything outside that, which is the region beyond both circles inside the rectangle.

Going the other way, describe what is shaded in words first, then translate. A shaded overlap is "both", so A ∩ B. Everything except the overlap is (A ∩ B)′.

Where two expressions look different but shade the same region, they are equivalent — A′ ∩ B′ shades the same area as (A ∪ B)′, since being outside both circles is the same as not being in either.

Venn diagrams from probabilities

A diagram can be built from probabilities rather than counts, and the method is identical — the four regions then total 1 instead of the population size.

If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, then the overlap holds 0.3, the A-only region holds 0.6 − 0.3 = 0.3, and the B-only region holds 0.5 − 0.3 = 0.2.

Those three total 0.8, so the "neither" region holds 0.2.

Checking that all four regions sum to 1 is the same check as before, and it catches an arithmetic slip immediately.

Three-set Venn diagrams

Three circles create eight regions: one for each single set only, three for each pair overlapping without the third, one for all three, and one outside everything.

Fill in from the centre outwards: the all-three region first, then each pair region by subtracting the centre, then each single region by subtracting everything already placed inside it, and finally the outside.

If 8 students take all three subjects and 20 take both maths and physics, then the maths-and-physics-but-not-chemistry region holds 20 − 8 = 12.

Working outwards in that order is essential, since each step depends on the ones before it.

Listing sets

Some questions give sets as lists rather than diagrams.

If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}, then A ∩ B = {4, 5} and A ∪ B = {1, 2, 3, 4, 5, 6, 7}.

Note that the union lists shared elements only once, and that the order of elements does not matter.

If the universal set is the numbers 1 to 10, then A′ = {6, 7, 8, 9, 10}.

Worked examples

Example 1: Completing a diagram

In a group of 60 people, 35 own a dog, 28 own a cat and 15 own both. Complete a Venn diagram and find the probability that a person chosen at random owns neither.

Start with the intersection: 15 in the overlap.

Dog only: 35 − 15 = 20. Cat only: 28 − 15 = 13.

Total so far: 20 + 15 + 13 = 48, so neither is 60 − 48 = 12.

P(neither) = 12 ÷ 60 = 1/5.

Check: 20 + 15 + 13 + 12 = 60. ✓

Example 2: Conditional probability

Using the same data, find the probability that a person owns a cat given that they own a dog.

"Given that they own a dog" restricts attention to dog owners, and there are 35 of them, so 35 is the denominator.

Of those, the ones who also own a cat are the 15 in the overlap.

So the probability is 15 ÷ 35 = 3/7.

Answering 15 ÷ 60 would ignore the condition, which is the error the phrase "given that" is there to test.

Example 3: Reading set notation

ξ is the numbers 1 to 12. A is the multiples of 3 and B is the even numbers. List A ∩ B and (A ∪ B)′.

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

A ∩ B is the numbers in both, so the multiples of 3 that are even: {6, 12}.

A ∪ B is everything in either: {2, 3, 4, 6, 8, 9, 10, 12}.

Its complement is everything in ξ that is left over: {1, 5, 7, 11}.

Those four are the numbers that are neither even nor multiples of 3, which is what the outside region of the diagram represents.

Common mistakes and how to avoid them

Writing the totals straight into the circles. Put the intersection in first and subtract outwards.

Adding two probabilities that overlap. Subtract the intersection once, or count the regions directly.

Using the wrong denominator for a conditional probability. "Given that" names the new total.

Confusing ∪ and ∩. Union is "or" and gathers everything; intersection is "and" and takes only the overlap.

Forgetting the "neither" region. It is part of the universal set and is often where the answer lies.

Listing a shared element twice in a union. Each element appears once.

Filling a three-set diagram from the outside. Start at the centre and work out.

Exam technique for "Venn Diagrams and Set Notation"

Draw the diagram even when the question does not ask for one. It converts an overlapping problem into simple counting and is usually faster than reasoning in symbols.

Enter the intersection first, then subtract outwards, and check the four regions total the universal set before calculating anything.

Translate each symbol into words before shading — union as "or", intersection as "and", complement as "not".

For a conditional probability, write the denominator down first, taken from the condition.

State probabilities as fractions of the total in the universal set unless the question restricts it.

For three sets, work centre-outwards and label every region as you place it.

Quick revision summary

A Venn diagram turns overlapping groups into non-overlapping regions, so that probability becomes counting.

Symbols: ∪ is union ("or"), ∩ is intersection ("and"), A′ is the complement ("not A"), ξ is the universal set, and n(A) is the number of elements in A.

Two sets create four regions — A only, both, B only, neither — which must total the universal set.

Fill in from the middle outwards: intersection first, then subtract to get each "only" region, then the outside.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B), because adding the sets double-counts the overlap.

For conditional probability, the words "given that" name the new denominator: P(German given French) = 12 ÷ 30, not 12 ÷ 50.

Three sets give eight regions, filled from the centre outwards.