Introduction to Probability
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.
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.
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.
Key facts
- $P(A) = \dfrac{n(A)}{n(S)}$ for equally likely outcomes
- $0 \leq P(A) \leq 1$ and $P(A') = 1 - P(A)$
- $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
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
Skills
- List the sample space for a multi-stage experiment
- Calculate probabilities from Venn diagrams
- Apply the addition rule for combined events
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:
where $n(A)$ is the number of outcomes in $A$ and $n(S)$ is the total number of outcomes.
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"?
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:
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
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 $P(B \cup T)$.
What is the probability of rolling at least one 6 in two rolls of a die?
In a survey of 50 people: 30 own a dog, 25 own a cat, and 12 own both. Find $P(\text{dog} \cup \text{cat})$.
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
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
A card is drawn from a standard 52-card deck. Find $P(\text{heart})$.
Two coins are flipped. List the sample space and find $P(\text{at least one head})$.
In a group of 40 students, 22 study Biology, 18 study Chemistry, 10 study both. Find $P(\text{Biology} \cup \text{Chemistry})$.
A die is rolled twice. How many outcomes are in $S$? Find $P(\text{sum} = 7)$.
Explain why $P(A \cup B)$ can never be greater than $P(A) + P(B)$.
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
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)
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)
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.
Q1 (3 marks): Total = 10. (a) $P(\text{red}) = \frac{5}{10} = \frac{1}{2}$ [1]. (b) $P(\text{not green}) = 1 - \frac{2}{10} = \frac{4}{5}$ [1]. (c) $P(\text{red or blue}) = \frac{5+3}{10} = \frac{4}{5}$ (mutually exclusive) [1].
Q2 (3 marks): (a) Smartphone only = 15, Laptop only = 20, Both = 30, Neither = 15 [1]. (b) $P = \frac{15+20}{80} = \frac{35}{80} = \frac{7}{16}$ [2].
Q3 (3 marks): (a) $n(S) = 2^4 = 16$ [0.5]. (b) $P(\text{at least one T}) = 1 - \frac{1}{16} = \frac{15}{16}$ [1]. (c) Direct counting requires summing four cases (exactly 1T, 2T, 3T, 4T). Complement needs one calculation. Generalisation: use complement whenever "at least one" appears [1.5].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering probability questions. Lighter alternative to the boss.
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