Set Notation and Probability Diagrams

Mathematics (Core) revision notes · Cambridge (CIE) · 0580 · Core

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1. Sets and their elements

Describing a set

Any well-defined group of things forms a set, and each thing belonging to it is an element of that set
Elements can be anything: numbers, letters or objects
Writing the elements inside curly brackets defines a set directly
- The factors of 12 form the set {1, 2, 3, 4, 6, 12}
A rule can replace the list, written with a colon read as "such that"
- {x : x is an integer, 1 ⩽ x ⩽ 5} is another way of writing {1, 2, 3, 4, 5}
Capital letters name sets and lower-case letters name elements

The universal set

The universal set holds everything under consideration in that question, and this specification writes it as ℰ, a script capital E
Every other set in the question is contained within it
Other material uses the Greek letter xi, ξ, or the letter U for the same thing, and the diagram below is labelled that way, but ℰ is the symbol the syllabus lists and the one printed on these papers

2. The notation

Symbols required by the specification

Symbol	Meaning
n(A)	The number of elements in A
A′	The complement of A: everything in ℰ that is not in A
ℰ	The universal set
A ∪ B	The union of A and B
A ∩ B	The intersection of A and B

Union collects everything appearing in either set, or in both, and each element is written once
Intersection keeps only what appears in both sets
The complement is found by taking the universal set and removing the named set

Worked example

Reading set notation

ℰ = {x : x is an integer, 1 ⩽ x ⩽ 10}, P = {2, 3, 5, 7} and Q = {1, 3, 5, 7, 9}. Write down n(P), P ∩ Q, P ∪ Q and P′.

Solution:

n(P) counts the elements of P, so n(P) = 4
P ∩ Q keeps only what is in both: {3, 5, 7}
P ∪ Q collects everything in either, listing each once: {1, 2, 3, 5, 7, 9}
P′ removes P from the universal set: {1, 4, 6, 8, 9, 10}

3. Venn diagrams

How the diagram is built

A Venn diagram shows sets as circles inside a rectangle, where the rectangle is the universal set
Overlapping circles share the elements that lie in both sets
Anything belonging to no named set is written inside the rectangle but outside every circle
This specification limits Venn diagrams to two sets

Venn diagram with universal set as a rectangle, labelled here with the alternative symbol xi, containing two overlapping circles labelled A and B; 12, 6 and 2 lie in A alone, 14 and 28 in the overlap, 7, 21 and 35 in B alone, and 8, 5 and 1 outside both circles — Source: Set notation by Save My Exams

Regions and operations

Each operation corresponds to a region of the diagram, which is what makes the notation easy to apply
The union covers both circles entirely; the intersection covers only the overlap

Three Venn diagrams, each a rectangle labelled ℰ (script E) holding two overlapping circles A and B. In the first both circles are shaded, labelled A union B, A or B or both. In the second only the overlap is shaded, labelled A intersection B, A and B. In the third everything inside the rectangle except circle A is shaded, labelled A complement, not A — Source: Set notation by Save My Exams

A complement shades everything outside the named set but still inside the rectangle
Combined expressions are read from the inside out, so (A ∪ B)′ means shade everything the union does not cover

Counting from a diagram

n(A) counts every element inside circle A, including those in the overlap
Elements in the overlap belong to both sets and are counted once in a union

Exam tip

Fill every region first

The parts after a Venn diagram, such as n(A ∩ B), a list or a probability, are usually marked from your own diagram, so a slip in the diagram need not cost them as well. That follow-through needs something written in each region the answer uses, so fill every region before you move on, then read each answer off your diagram rather than working it out again from the question.

Worked example

Placing elements in a Venn diagram

ℰ = {x : x is an integer, 20 ⩽ x ⩽ 30}, A = {odd numbers} and B = {multiples of 3}. Show A and B on a Venn diagram, then find n(A ∪ B) and list the elements of A′.

Solution:

List each set from ℰ first: A = {21, 23, 25, 27, 29} and B = {21, 24, 27, 30}
Fill the overlap first: 21 and 27 are both odd and multiples of 3
A only: 23, 25, 29; B only: 24, 30
Outside both circles, inside the rectangle: 20, 22, 26, 28
Check the count: 3 + 2 + 2 + 4 = 11, and ℰ holds the 11 integers from 20 to 30, so each is placed exactly once
A ∪ B covers both circles, so n(A ∪ B) = 3 + 2 + 2 = 7, with 21 and 27 counted once
A′ is everything in ℰ outside circle A: {20, 22, 24, 26, 28, 30}

The completed Venn diagram: a rectangle labelled ℰ holding overlapping circles A and B, with 23, 25 and 29 in A only, 21 and 27 in the overlap, 24 and 30 in B only, and 20, 22, 26 and 28 inside the rectangle outside both circles

Worked example

Counting from a Venn diagram

In a group of 30 students, 18 study art (A) and 14 study biology (B). 7 study both. Show this on a Venn diagram, then work out how many study neither.

Solution:

Fill the overlap first, because the two given totals each already include it
The intersection holds 7
Art only = 18 − 7 = 11
Biology only = 14 − 7 = 7
Those three regions hold 11 + 7 + 7 = 25 students
The universal set holds 30, so the region outside both circles holds 30 − 25 = 5
Check by adding all four regions: 11 + 7 + 7 + 5 = 30
Writing 18 in the art circle and 14 in the biology circle would count the 7 twice and total 39

The completed Venn diagram: a rectangle labelled ℰ holding overlapping circles A and B, with 11 in A only, 7 in the overlap, 7 in B only and 5 inside the rectangle outside both circles

4. Combining two events

The two rules

Where two things happen one after the other, two rules cover almost everything
- "And" means multiply, so the probability of A happening and then B is P(A) × P(B), where P(A) is short for the probability of A
- "Or" means add, so the probability of one case or another, when the two cannot both happen, is the sum of their probabilities
Rephrasing a question in terms of "and" and "or" first is what turns it into a calculation
- "Both counters are red" is red and red
- "One of each colour" is red and blue or blue and red
The probability of all the possible cases together is 1, which is a quick check on any full list
Where a question asks for at least one, it is usually faster to work out the probability of none and subtract from 1

Independent events

Two events are independent when the first does not change the second
With replacement means the item goes back, so the bag is the same for the second pick and the two probabilities are equal
These papers combine events with replacement only, so the result of the first stage never changes the probabilities for the second
Keeping fractions unsimplified while working makes them far easier to add at the end

Worked example

Two picks with replacement

A bag holds 5 red counters and 7 blue counters. A counter is taken at random, its colour is recorded, and it is put back. A second counter is then taken. Find the probability that both are red.

Solution:

There are 5 + 7 = 12 counters, so the first pick is red with probability 5/12
The counter goes back, so the bag is unchanged and the second pick is red with probability 5/12 as well
"And" means multiply, so P(red then red) = 5/12 × 5/12
P(red then red) = 25/144

5. Tree diagrams

Building one

A tree diagram sets out two or more stages, with a branch for each outcome at each stage
The syllabus states the convention these papers follow: outcomes at the ends of the branches, probabilities beside them
Each pair or group of branches from one point must add to 1, which fills in any missing probability
When a question prints a tree diagram, it asks you to complete it first, and that alone carries marks

The layout of a two-stage tree diagram: the first experiment branches to A and not A with P(A) and P(not A) beside the branches, each of those is followed by a pair of branches to B and not B with P(B) and P(not B) beside them, and the probability of each of the four routes is written at its end as P(A and B), P(A and not B), P(not A and B) and P(not A and not B)

Reading one

Follow one complete route across the diagram and multiply the probabilities you pass, which gives that path its own probability
Add between separate paths when more than one outcome satisfies the question

A tree diagram for two sets of traffic lights: the first branches green with probability 5 over 7 and red with 2 over 7, the second green with 8 over 9 and red with 1 over 9, and each of the four routes is multiplied along to give 40, 5, 16 and 2 out of 63 — Source: Tree diagrams by Save My Exams

The probabilities of all the complete paths add to 1
In these papers the first stage never changes the second, so each second pair of branches carries the same probabilities whichever first branch it follows: the first pair again when one spinner or bag is used twice, or a new pair for a different event, like the second set of traffic lights above

Exam tip

Multiply the values on your tree

The probability of a route is marked from your own tree: writing the multiplication with the values on your branches, such as 3/8 × 3/8 below, earns the method mark even if one of them is wrong. So always write that line before the answer. It counts only while each value is a probability between 0 and 1, so a branch value such as 8/3 is a slip to fix first.

Worked example

Two spins on a tree diagram

A spinner has 3 red sections and 5 blue sections, all the same size. It is spun twice. Find the probability that both spins are red, and the probability that exactly one spin is red.

Solution:

There are 8 sections, so each spin is red with probability 3/8 and blue with probability 5/8
The spinner is unchanged between spins, so the second pair of branches repeats the first

A completed tree diagram with the headings First spin and Second spin. The first pair of branches is Red 3/8 and Blue 5/8, and after each of them the second pair is again Red 3/8 and Blue 5/8. No route products or answers are shown.

Both red is a single route, so multiply along it: P(red, red) = 3/8 × 3/8 = 9/64
Exactly one red happens two ways: red then blue, or blue then red
P(red, blue) = 3/8 × 5/8 = 15/64, and P(blue, red) = 5/8 × 3/8 = 15/64
Add the separate routes: 15/64 + 15/64 = 30/64 = 15/32
Check: the four routes come to 9/64 + 15/64 + 15/64 + 25/64 = 1

6. Sample space diagrams

Laying out every outcome

Where two experiments happen together, a table shows every combination in a grid
The syllabus calls it a sample space diagram, and a question may print it as a table of the possible outcomes, sometimes partly filled in for you to complete
Put one experiment along the top and the other down the side, then fill each cell with the combined result
The number of cells is the total number of outcomes, and counting the cells that satisfy an event gives the numerator
Complete the whole grid before counting, since a systematic table makes miscounting almost impossible