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Year 11 Methods (Unit 1 & 2) Probability

Sets And Venn Diagrams

20 practice questions 1 video lesson Theory + worked examples

Master sets and Venn diagrams for Queensland Year 11 Mathematical Methods (QCAA). A sample space is a set of outcomes, and events are sets within it, giving clear language for describing events.

You will learn set notation for the complement of an event, the intersection and union of two events, recognise mutually exclusive events, and use Venn diagrams to illustrate these descriptions and region counts — the foundation of QCAA probability.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1 Topic 5), a set is a collection of outcomes and a Venn diagram pictures how events overlap. This page covers set notation \(A\cap B\), \(A\cup B\), \(A'\), reading region counts, the complement and De Morgan's laws, and the counting addition rule \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\).

A set is a collection of distinct objects called elements. The universal set \(\xi\) (or \(U\)) holds every outcome under consideration. We write \(x\in A\) for ‘\(x\) is in \(A\)’ and \(x\notin A\) for ‘\(x\) is not in \(A\)’.

The intersection \(A\cap B\) is the elements in both \(A\) and \(B\); the union \(A\cup B\) is the elements in either set (or both). The complement \(A'\) is everything in \(\xi\) that is not in \(A\). Two events are mutually exclusive when they cannot both happen, so \(A\cap B=\varnothing\).

De Morgan's laws connect complements to unions and intersections: \((A\cup B)'=A'\cap B'\) and \((A\cap B)'=A'\cup B'\).

Addition rule for counts: \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Subtract the overlap once so the elements in both sets are not counted twice.
Two-set Venn diagram with region countsRectangle for the universal set with circles A and B; A only 8, both 6, B only 5, outside 4. A B 8 6 5 4
A two-set Venn diagram splits \(\xi\) into four regions: \(A\) only, \(A\cap B\), \(B\) only, and outside both. Here \(n(A\cup B)=8+6+5=19\).
Three-set Venn diagram with region countsThree overlapping circles A, B and C with a count written in each of the seven regions and seven outside. A B C 5 4 6 3 2 1 2 7
A three-set Venn diagram has seven inner regions. Read each count directly; the centre \(A\cap B\cap C\) is shared by all three circles.

The counting addition rule for the size of a union:

\[n(A\cup B)=n(A)+n(B)-n(A\cap B)\]
n(AB)=n(A)+n(B)-n(AB)

For mutually exclusive events \(A\cap B=\varnothing\), so the overlap is zero:

\[n(A\cup B)=n(A)+n(B)\]
n(AB)=n(A)+n(B)

De Morgan's laws for complements:

\[(A\cup B)'=A'\cap B'\qquad (A\cap B)'=A'\cup B'\]
(AB)=AB
The four regions always add to the whole: \(n(\xi)=n(A\cap B')+n(A\cap B)+n(A'\cap B)+n(A'\cap B')\).

How to solve a two-set Venn problem

  1. Draw two overlapping circles inside a rectangle for \(\xi\).
  2. Fill the overlap first: write \(n(A\cap B)\) in the middle, then subtract it to get \(A\) only \(=n(A)-n(A\cap B)\) and \(B\) only \(=n(B)-n(A\cap B)\).
  3. Find ‘neither’ by taking the three inner counts from \(n(\xi)\).
  4. Answer the question by reading or adding the required regions; for a probability, divide the favourable count by \(n(\xi)\).
Example 1 — Set notation and listing
Let \(\xi=\{1,2,\dots,10\}\), \(A=\{\text{even numbers}\}\) and \(B=\{\text{multiples of }3\}\). List \(A\cap B\), \(A\cup B\) and \(A'\).
Solution

List each set from the universal set:

\(A\)\(=\)\(\{2,4,6,8,10\}\)
\(B\)\(=\)\(\{3,6,9\}\)

Intersection — elements in both:

\(A\cap B\)\(=\)\(\{6\}\)

Union — elements in either (list once):

\(A\cup B\)\(=\)\(\{2,3,4,6,8,9,10\}\)

Complement — everything in \(\xi\) not in \(A\):

\(A'\)\(=\)\(\{1,3,5,7,9\}\)

\(A\cap B=\{6\}\), \(A\cup B=\{2,3,4,6,8,9,10\}\), \(A'=\{1,3,5,7,9\}\).

Listing set elements in a Venn diagramEven numbers to ten in A, multiples of three in B, six in the overlap, and one five seven outside. A B 2, 4, 8, 10 6 3, 9 1, 5, 7
Example 2 — Addition rule for counts
In a class of \(30\) students, \(18\) play soccer, \(14\) play tennis and \(6\) play both. How many play at least one of the two sports, and how many play neither?
Solution

At least one — apply the addition rule:

\(n(S\cup T)\)\(=\)\(n(S)+n(T)-n(S\cap T)\)
\(=\)\(18+14-6\)
\(=\)\(26\)

Neither — subtract from the whole class:

\(n(\text{neither})\)\(=\)\(30-26\)
\(=\)\(4\)

\(26\) students play at least one sport; \(4\) play neither.

Soccer and tennis Venn diagramThirty students: soccer only 12, both 6, tennis only 8, neither 4; the union is shaded. S T 12 6 8 4
Example 3 — Complement and De Morgan
For \(n(\xi)=40\), \(n(A)=22\), \(n(B)=17\) and \(n(A\cap B)=9\), find \(n(A\cap B')\) and \(n(A'\cap B')\).
Solution

\(A\) only \(=A\cap B'\) — remove the overlap:

\(n(A\cap B')\)\(=\)\(n(A)-n(A\cap B)\)
\(=\)\(22-9\)
\(=\)\(13\)

Union first, for the outside region:

\(n(A\cup B)\)\(=\)\(22+17-9\)
\(=\)\(30\)

Outside both \(=(A\cup B)'=A'\cap B'\) by De Morgan:

\(n(A'\cap B')\)\(=\)\(n(\xi)-n(A\cup B)\)
\(=\)\(40-30\)
\(=\)\(10\)

\(n(A\cap B')=13\) and \(n(A'\cap B')=10\).

Complement of a union shadedA only 13, both 9, B only 8, outside 10; the region outside both circles is shaded. A B 13 9 8 10
Example 4 — Three-set Venn and probability
Fifty people were surveyed on three sports. The Venn counts are shown. Find how many like exactly one sport, how many like exactly two, and the probability a randomly chosen person likes at least one sport.
Solution

Exactly one — the three ‘only’ regions:

\(n(\text{exactly one})\)\(=\)\(8+6+5\)
\(=\)\(19\)

Exactly two — the three pairwise overlaps (not the centre):

\(n(\text{exactly two})\)\(=\)\(4+3+2\)
\(=\)\(9\)

At least one — total inside the circles:

\(n(\text{at least one})\)\(=\)\(8+6+5+4+3+2+1\)
\(=\)\(29\)

Probability \(=\dfrac{\text{favourable}}{\text{total}}\):

\(P(\text{at least one})\)\(=\)\(\dfrac{29}{50}\)

Exactly one: \(19\); exactly two: \(9\); \(P(\text{at least one})=\dfrac{29}{50}\).

Three-sport survey Venn diagramFifty people surveyed on sports A, B and C with each region count shown and twenty one outside. A B C 8 6 5 4 3 2 1 21

Common pitfalls

Double-counting the overlap. \(n(A)\) already includes the elements of \(A\cap B\). Add \(n(A)+n(B)\) and subtract \(n(A\cap B)\) once — do not add the middle again.
Forgetting ‘neither’. The three inner regions rarely fill the rectangle. Always check \(n(A'\cap B')=n(\xi)-n(A\cup B)\).
Muddling ‘exactly two’ with the centre. On a three-set diagram, ‘exactly two’ means the pairwise slivers only; the centre is ‘all three’.
Mixing up \(\cap\) and \(\cup\). \(\cap\) (intersection) means and; \(\cup\) (union) means or. Read the wording carefully.

Frequently asked questions

What is the difference between union and intersection?

\(A\cup B\) (union) is everything in either set; \(A\cap B\) (intersection) is only what is in both sets at once.

What does the addition rule for sets say?

\(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). You subtract the overlap once because it is counted in both \(n(A)\) and \(n(B)\).

What are mutually exclusive events?

Events that cannot both occur, so \(A\cap B=\varnothing\). Then \(n(A\cup B)=n(A)+n(B)\) with nothing to subtract.

What are De Morgan's laws?

\((A\cup B)'=A'\cap B'\) and \((A\cap B)'=A'\cup B'\). The complement of a union is the intersection of the complements.

How do you find the number in neither set?

Take the union from the universal set: \(n(A'\cap B')=n(\xi)-n(A\cup B)\).

How do you get a probability from a Venn diagram?

Divide the count in the favourable region by the total number of outcomes \(n(\xi)\).