A set is a collection with a precise way of writing it down. The notation looks unfamiliar for about ten minutes, and then it makes every probability question shorter.
Today's hook, Probability is built on sets. Every event is a set of outcomes, so the notation you learn here is the notation the rest of this module speaks in.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
+5 XP warm-up
Write down the set of even numbers between 1 and 11. How many elements does it have? Now write down the set of even numbers between 1 and 3 that are also odd.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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A collection, written precisely
Work through the core explanation before applying it.
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A collection, written precisely
+5 XP to read
A set is a collection of objects called its elements, listed inside braces: $A = \{2, 4, 6, 8, 10\}$. Order does not matter and repeats are not counted twice. Writing $n(A)$ means how many elements it has, so here $n(A) = 5$.
Sets use curly braces. Round brackets mean something else entirely, so the notation matters.
The empty set is a set
$\varnothing$ has no elements, so $n(\varnothing) = 0$. It is not nothing; it is a set that happens to be empty.
Complement needs a universal set
$A'$ means everything in the universal set $\xi$ that is not in $A$. Without knowing $\xi$, the complement is undefined.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
A set is a collection of elements, written inside braces.
$n(A)$ is the number of elements in a finite set $A$.
The empty set $\varnothing$ has no elements, so $n(\varnothing) = 0$.
The complement $A'$ is everything in the universal set not in $A$, and $A \subseteq B$ means every element of $A$ is in $B$.
Understand
Concepts
Why order and repetition do not affect a set.
Why the complement is meaningless without a stated universal set.
Why every set is a subset of itself and of the universal set.
Can do
Skills
Write a set in braces from a description and count its elements.
Find the complement of a set given a universal set.
Decide whether one set is a subset of another.
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Key terms
SetA collection of objects written inside braces. Like this: $A = \{2, 4, 6\}$ is the set of the first three even numbers.
ElementOne of the objects in a set. Like this: 4 is an element of $\{2,4,6\}$, written $4 \in A$.
$n(A)$The number of elements in a finite set. Like this: if $A = \{2,4,6,8\}$ then $n(A) = 4$.
Empty setThe set with no elements at all, written $\varnothing$. Like this: the set of numbers that are both even and odd is empty, so $n(\varnothing) = 0$.
Universal setThe set of everything under consideration, written $\xi$. Like this: if $\xi$ is the numbers 1 to 10, complements are taken within that range.
ComplementEverything in the universal set not in the given set, written $A'$. Like this: with $\xi = \{1,...,6\}$ and $A = \{2,4,6\}$, $A' = \{1,3,5\}$.
SubsetA set whose every element also belongs to another set, written $A \subseteq B$. Like this: $\{2,4\} \subseteq \{2,4,6\}$.
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Writing sets and counting elements
Work through the core explanation before applying it.
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Writing sets and counting elements
core concept
A set is written by listing its elements inside braces: $A = \{2, 4, 6, 8, 10\}$. The elements can be numbers, names, or outcomes of an experiment.
Order does not matter, so $\{2,4,6\}$ and $\{6,2,4\}$ are the same set. Repeats are not counted twice either: $\{2,2,4\}$ is simply $\{2,4\}$.
The notation $n(A)$ means the number of elements. For $A = \{2,4,6,8,10\}$, $n(A) = 5$.
A set can be described rather than listed. "The set of even numbers between 1 and 11" and $\{2,4,6,8,10\}$ are the same set. For large sets a description is the only practical option.
Quick check: if $A = \{3, 5, 7, 9\}$, what is $n(A)$?
A set is a collection of elements written inside braces. Order and repetition make no difference. $n(A)$ counts the elements of a finite set. A set may be listed or described in words.
Pause, copy the braces notation, the fact that order and repeats do not matter, and the meaning of $n(A)$ with a worked count, into your book.
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The empty set, the universal set and complements
core concept
We just saw how to write a set and count it. That raises a question: what if there are no elements at all, or you want everything left over? This card answers it → the empty set covers the first and the complement the second, once you say what everything means.
The empty set $\varnothing$ has no elements. The set of numbers that are both even and odd is empty. It is still a set, so $n(\varnothing) = 0$.
The universal set $\xi$ is everything under consideration in a particular problem. It has to be stated, because it changes every complement.
The complement $A'$ is everything in $\xi$ not in $A$. With $\xi = \{1,2,3,4,5,6\}$ and $A = \{2,4,6\}$, $A' = \{1,3,5\}$. Notice $n(A) + n(A') = n(\xi)$, always.
Change the universal set and the complement changes. With the same $A = \{2,4,6\}$ but $\xi$ the numbers 1 to 10, $A' = \{1,3,5,7,8,9,10\}$. The complement is never a property of $A$ alone.
Fill the blank: if $\xi = \{1,2,3,4,5\}$ and $A = \{1,3,5\}$, then $n(A') = $ .
The empty set has no elements and $n(\varnothing) = 0$. The complement $A'$ is everything in $\xi$ not in $A$, so it depends on $\xi$ as much as on $A$, and $n(A) + n(A') = n(\xi)$.
Pause, copy the empty set, the universal set, the complement with a worked example, and the identity $n(A) + n(A') = n(\xi)$, into your book.
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Subsets
core concept
We just saw how to take everything outside a set. That raises a question: how do you describe one set sitting entirely inside another? This card answers it → that is what a subset is.
$A$ is a subset of $B$, written $A \subseteq B$, if every element of $A$ is also an element of $B$. So $\{2,4\} \subseteq \{2,4,6\}$.
Two edge cases follow from the definition rather than being extra rules. Every set is a subset of itself, since every element of $A$ is trivially in $A$. And $\varnothing$ is a subset of every set, since it has no element that could fail the test.
Being a subset is not the same as being smaller. $\{1,9\}$ is not a subset of $\{1,2,3\}$ even though it has fewer elements, because 9 is not in the larger set.
One element out of place breaks it. $A \subseteq B$ requires every element of $A$ to be in $B$. A single exception makes it false.
Which statement is FALSE?
$A \subseteq B$ means every element of $A$ is in $B$. Every set is a subset of itself, and the empty set is a subset of every set, both from the definition. One element of $A$ outside $B$ makes it false.
Pause, copy the subset definition, the two edge cases with their reasons, and the counter-example $\{1,9\}$ against $\{1,2,3\}$, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · LISTING AND COUNTING
Let $\xi$ be the whole numbers from 1 to 12. Write $A$, the set of multiples of 3 in $\xi$, and find $n(A)$.
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Multiples of 3 up to 12: 3, 6, 9, 12
Work through the universal set.
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$A = \{3, 6, 9, 12\}$
List inside braces.
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$n(A) = 4$
Count the elements.
PROBLEM 2 · COMPLEMENT
With $\xi = \{1,2,3,4,5,6,7,8\}$ and $B = \{2,4,6,8\}$, find $B'$ and verify $n(B) + n(B') = n(\xi)$.
Let $A = \{1,3\}$, $B = \{1,2,3,4\}$ and $C = \{1,3,5\}$. State which of $A \subseteq B$ and $C \subseteq B$ are true.
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$A$: is 1 in $B$? yes. Is 3 in $B$? yes
Test every element of $A$.
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So $A \subseteq B$ is true
All elements pass.
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$C$: 5 is not in $B$, so $C \subseteq B$ is false
One failure is enough.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
Write the set of odd numbers between 2 and 10, and state $n$ of it.
If $\xi = \{1,...,6\}$ and $A = \{1,2\}$, find $A'$.
Is $\{4\} \subseteq \{2,4,6\}$?
What is $n(\varnothing)$?
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Revisit your two sets
Run the quick drill and copy the summary into your book.
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Revisit your two sets
At the start you wrote the even numbers between 1 and 11, then the even numbers between 1 and 3 that are also odd. Name the second set using the correct notation, and say what its $n$ value is.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 33 marks
Q1. Let $\xi = \{1, 2, 3, ..., 10\}$ and $A$ be the set of multiples of 4 in $\xi$. Write $A$ in set notation, state $n(A)$, and write $A'$. (3 marks)
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ApplyBand 43 marks
Q2. Let $P = \{2,3,5,7\}$ and $Q = \{2,3,4,5,6,7\}$. State whether $P \subseteq Q$, justifying your answer, and find $n(Q) - n(P)$. (3 marks)
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UnderstandBand 42 marks
Q3. Explain why the complement of a set cannot be determined without knowing the universal set, using an example. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $\{3,5,7,9\}$, $n = 4$. Practice 2: $A' = \{3,4,5,6\}$. Practice 3: yes, 4 is in the larger set. Practice 4: 0.
Q1 (3 marks): Multiples of 4 up to 10 are 4 and 8, so $A = \{4, 8\}$ [1]. $n(A) = 2$ [1]. $A' = \{1,2,3,5,6,7,9,10\}$ [1].
Q2 (3 marks): Every element of $P$, namely 2, 3, 5 and 7, is also in $Q$ [1]. So $P \subseteq Q$ is true [1]. $n(Q) = 6$ and $n(P) = 4$, so the difference is 2 [1].
Q3 (2 marks): The complement is everything not in the set, and everything is whatever the universal set says it is [1]. With $A = \{2,4,6\}$ and $\xi = \{1,...,6\}$ the complement is $\{1,3,5\}$, but with $\xi = \{1,...,10\}$ the same $A$ has complement $\{1,3,5,7,8,9,10\}$ [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Set Sorter
earn bronze · silver · gold
Read set notation, count elements and take complements at speed. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.