A weather forecast says there is a 30% chance of rain. A medical test is 95% accurate. A poker player calculates the odds of a flush. Probability is the language of uncertainty, and it powers everything from insurance premiums to artificial intelligence. In this lesson you'll build the foundations: sample spaces, Venn diagrams, and the addition rule.
Today's hook, You roll a die and flip a coin. Which is more likely: getting a 6 and heads, or getting an even number or heads? Instinct can fool you here. By the end of this lesson you'll have the tools to answer this instantly, and understand exactly why.
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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
You roll a standard six-sided die and flip a coin. Without using a formula, which do you think is more likely: getting a 6 and heads, or getting an even number or heads? Make a prediction before reading on.
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The two moves
Work through the core explanation before applying it.
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The two moves
+5 XP to read
There are only two core formulas in this lesson, and everything else flows from them. Lock $P(A) = \dfrac{n(A)}{n(S)}$ and $P(A \cup B) = P(A)+P(B)-P(A \cap B)$ into muscle memory.
Every probability question in this module lives on one of two roads: count favourable outcomes over total outcomes to find a basic probability, or apply the addition rule to combine events without double-counting.
$P(A) = \dfrac{n(A)}{n(S)}$
Basic probability
$P(A) = \dfrac{n(A)}{n(S)}$ only when all outcomes are equally likely. Always check this first.
Complement shortcut
$P(A') = 1 - P(A)$. When "at least one" appears, flip to the complement, it's nearly always faster.
Addition rule always
$P(A \cup B) = P(A)+P(B)-P(A \cap B)$. Never simply add, subtract the overlap every time.
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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
$P(A) = \dfrac{n(A)}{n(S)}$ for equally likely outcomes
$0 \leq P(A) \leq 1$ and $P(A') = 1 - P(A)$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Understand
Concepts
The difference between outcomes, events and the sample space
How Venn diagrams represent unions, intersections and complements
Why probabilities sum to 1 across all possible outcomes
Can do
Skills
List the sample space for a multi-stage experiment
Calculate probabilities from Venn diagrams
Apply the addition rule for combined events
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Key terms
Sample space $S$The set of all possible outcomes of a probability experiment.
Event $A$Any subset of the sample space; a collection of outcomes of interest.
Union $A \cup B$Outcomes in $A$ or $B$ or both, the total combined region.
Intersection $A \cap B$Outcomes in both $A$ and $B$, the overlap region.
Complement $A'$All outcomes not in $A$; satisfies $P(A') = 1 - P(A)$.
Mutually exclusiveEvents with no overlap: $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.
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Sample space and events
Work through the core explanation before applying it.
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Sample space and events
core concept
Every probability experiment has a sample space $S$, the set of all possible outcomes. An event is any subset of $S$.
Roll a die: $S = \{1, 2, 3, 4, 5, 6\}$
Flip a coin: $S = \{\text{H}, \text{T}\}$
Roll two dice: $S = \{(1,1),(1,2),\ldots,(6,6)\}$, 36 outcomes total
For example, "rolling an even number" is the event $A = \{2, 4, 6\}$, so $n(A) = 3$. When all outcomes are equally likely:
$$P(A) = \frac{n(A)}{n(S)}$$
where $n(A)$ is the number of outcomes in $A$ and $n(S)$ is the total number of outcomes.
Lottery odds. In Oz Lotto you choose 7 numbers from 1–47. The total combinations are $C(47,7) = 62{,}891{,}499$. Your chance of winning is roughly $P(\text{win}) = 1.59 \times 10^{-8}$. This is why mathematicians say "the lottery is a tax on people who cannot do probability."
Sample space $S$ = all possible outcomes; an event $A$ is any subset of $S$; $P(A) = \dfrac{n(A)}{n(S)}$, only valid when outcomes are equally likely
Pause, copy the three key definitions: sample space $S$ (all outcomes), event $A$ (any subset of $S$), and the probability formula $P(A) = \frac{n(A)}{n(S)}$, valid only when outcomes are equally likely into your book.
Quick check: A fair die is rolled. How many outcomes are in the event "rolling a number less than 4"?
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Venn diagrams for probability
core concept
We just saw that $P(A) = \frac{n(A)}{n(S)}$ counts favourable outcomes, but what if we need to reason about combinations of events like "A or B" or "A and B"? That raises a question: how do we visualise and calculate these compound events? This card answers it → Venn diagrams use regions for $\cap$ (AND), $\cup$ (OR), and $'$ (NOT) to organise probability calculations.
Venn diagrams use overlapping circles to show how events relate within the sample space.
$A \cup B$ (union): outcomes in $A$ or $B$ or both, the total shaded area
$A \cap B$ (intersection): outcomes in $A$ and $B$, the overlap
$A'$ (complement): outcomes not in $A$, everything outside circle $A$
The overlap $A \cap B$ is subtracted once so it is counted exactly once in the union.
The addition rule: To find $P(A \cup B)$, we add $P(A)$ and $P(B)$, but this double-counts the overlap, so we subtract it once:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Mutually exclusive special case: If $A$ and $B$ have no overlap, then $P(A \cap B) = 0$ and the rule simplifies to $P(A \cup B) = P(A) + P(B)$.
$\cup$ means OR (union); $\cap$ means AND (intersection); $'$ means NOT (complement); Mnemonic: $\cap$ looks like an "A"nd gate; $\cup$ looks like a "U"nion container
Pause, copy the three Venn notation rules: $\cup$ = OR (union), $\cap$ = AND (intersection), $'$ = NOT (complement), with the mnemonics ($\cap$ looks like an "And" gate; $\cup$ looks like a "Union" container) into your book.
Did you get this? True or false: for mutually exclusive events $A$ and $B$, $P(A \cup B) = P(A) + P(B)$.
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 · BASIC PROBABILITY
In a class of 30 students: 18 play basketball ($B$), 15 play tennis ($T$), and 8 play both. Find $P(B)$, $P(T)$, $P(B \cap T)$, and $P(B \cup T)$.
Apply the addition rule. Without subtracting the 12 who own both, we'd count them twice.
Fill the gap: If $P(A) = 0.4$, $P(B) = 0.3$, and $P(A \cap B) = 0.15$, then $P(A \cup B) = $ .
Misconceptions to fix · the 3 traps that cost marks
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Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
Trap 01
Forgetting the intersection
Writing $P(A \cup B) = P(A) + P(B)$ without subtracting $P(A \cap B)$ is the most common probability error. The result can exceed 1, a dead giveaway that you've made this mistake.
Trap 02
Confusing $\cup$ and $\cap$
Students swap "OR" and "AND" under exam pressure. $\cup$ is OR (larger set); $\cap$ is AND (smaller set, can't be bigger than either event alone).
Trap 03
Using $P = n(A)/n(S)$ without checking equal likelihood
The basic formula only works when outcomes are equally likely. A biased coin, a weighted die, or survey data with unequal frequencies all break this assumption.
Did you get this? True or false: if $P(A) = 0.6$ and $P(B) = 0.5$, then $P(A \cup B)$ can equal $1.1$.
Activities · practice with the ideas
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Drill it, then lock it in
Run the quick drill and copy the summary into your book.
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A card is drawn from a standard 52-card deck. Find $P(\text{heart})$.
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Two coins are flipped. List the sample space and find $P(\text{at least one head})$.
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In a group of 40 students, 22 study Biology, 18 study Chemistry, 10 study both. Find $P(\text{Biology} \cup \text{Chemistry})$.
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A die is rolled twice. How many outcomes are in $S$? Find $P(\text{sum} = 7)$.
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Explain why $P(A \cup B)$ can never be greater than $P(A) + P(B)$.
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Revisit your thinking
Earlier you were asked: which is more likely, a 6 AND heads, or an even number OR heads?
Getting a 6 and heads: $P(6 \cap \text{H}) = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12} \approx 0.083$.
Getting an even number or heads: $P(\text{even} \cup \text{H}) = \dfrac{3}{6} + \dfrac{1}{2} - \dfrac{3}{12} = \dfrac{9}{12} = 0.75$. The OR condition is far more likely because it includes many more outcomes, the AND condition is highly restrictive.
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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 43 marks
Q1. A bag contains 5 red, 3 blue, and 2 green marbles. One marble is drawn at random. Find: (a) $P(\text{red})$; (b) $P(\text{not green})$; (c) $P(\text{red or blue})$. (3 marks)
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ApplyBand 43 marks
Q2. In a survey of 80 people: 45 own a smartphone, 50 own a laptop, and 30 own both. (a) Draw a Venn diagram showing this information. (b) Find the probability that a randomly selected person owns a smartphone or a laptop but not both. (3 marks)
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AnalyseBand 53 marks
Q3. A fair coin is flipped four times. (a) How many outcomes are in the sample space? (b) Find $P(\text{at least one tail})$. (c) Explain why the complement method is more efficient than direct counting for part (b), and generalise this insight to other probability problems. (3 marks)
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Comprehensive answers (click to reveal)
Activity 1: 1. $P(\text{heart}) = \frac{13}{52} = \frac{1}{4}$ · 2. $S = \{\text{HH,HT,TH,TT}\}$; $P \geq 1\text{H} = \frac{3}{4}$ · 3. $\frac{22+18-10}{40} = \frac{30}{40} = \frac{3}{4}$ · 4. $n(S)=36$; $P(\text{sum}=7)=\frac{6}{36}=\frac{1}{6}$ · 5. Because $P(A \cap B) \geq 0$, so subtracting it from $P(A)+P(B)$ can only make the union smaller.
