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 formulawhich 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 confidencethat 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.
Climb platforms by answering probability questions. Lighter alternative to the boss.
Mark lesson as complete
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