IGCSE Mathematics: Probability

Cambridge IGCSE 0580 & Edexcel 4MA1 · Updated March 2026

Probability is the study of chance and uncertainty. In this chapter you will learn to calculate the likelihood of events, represent combined events using sample spaces, tree diagrams, and Venn diagrams, and use set notation to describe relationships between events. These skills underpin data analysis, decision-making, and statistical reasoning.

Specification Note

Content labelled Extended is required for Extended tier (Cambridge) or Higher tier (Edexcel) only.

1. Basic Probability

The Probability Scale

Probability is always a number between $0$ and $1$ inclusive. A probability of $0$ means the event is impossible; a probability of $1$ means it is certain.

$$0 \leq P(A) \leq 1$$

Theoretical Probability

When all outcomes are equally likely:

$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$$

Complementary Events

The complement of event $A$, written $A'$ (or $\bar{A}$), consists of all outcomes where $A$ does not occur.

$$P(A') = 1 - P(A)$$

Experimental Probability and Relative Frequency

When we cannot calculate probability theoretically, we estimate it through experiment. The relative frequency of an event after $n$ trials is:

$$\text{Relative frequency} = \frac{\text{number of times event occurs}}{n}$$

As $n$ increases, the relative frequency tends towards the theoretical probability. This is the basis of the law of large numbers.

Example 1 — Basic probability calculations

A bag contains 5 red, 3 blue, and 2 green counters. One counter is selected at random. Find:

(a) $P(\text{red})$ (b) $P(\text{not blue})$ (c) $P(\text{yellow})$

Step 1 Total counters: $5 + 3 + 2 = 10$

(a) $P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}$

(b) $P(\text{blue}) = \dfrac{3}{10}$, so $P(\text{not blue}) = 1 - \dfrac{3}{10} = \dfrac{7}{10}$

Example 2 — Relative frequency

A biased coin is flipped 200 times. It lands heads 130 times. Estimate the probability of getting heads on the next flip.

Step 1 Relative frequency $= \dfrac{130}{200} = \dfrac{13}{20} = 0.65$

The estimated probability of heads is $0.65$. Since this is greater than $0.5$, the coin appears biased towards heads.

Figure 8.1 — Probability distribution for the colours of counters in Example 1. Each bar shows the probability of selecting that colour.

Practice 1a

A standard six-sided die is rolled. Find: (i) $P(4)$, (ii) $P(\text{even})$, (iii) $P(\text{greater than 4})$, (iv) $P(\text{not 1})$.

Show Solution

Total outcomes: 6 equally likely

(i) $P(4) = \dfrac{1}{6}$

(ii) Even numbers: $\{2, 4, 6\}$. $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$

(iii) Numbers greater than 4: $\{5, 6\}$. $P(>4) = \dfrac{2}{6} = \dfrac{1}{3}$

(iv) $P(\text{not 1}) = 1 - \dfrac{1}{6} = \dfrac{5}{6}$

Practice 1b

A spinner is spun 400 times. It lands on red 96 times. Estimate $P(\text{red})$ and suggest the number of red sections if the spinner has 10 equal sections.

Show Solution

Relative frequency $= \dfrac{96}{400} = 0.24$

Expected number of red sections $\approx 0.24 \times 10 = 2.4$

Since sections must be whole numbers, we estimate approximately $2$ or $3$ red sections.

2. Combined Events and Sample Spaces

When two or more events occur together, we need to list all possible combined outcomes. This is called the sample space.

Mutually Exclusive Events

Events $A$ and $B$ are mutually exclusive if they cannot both occur at the same time. If $A$ and $B$ are mutually exclusive:

$$P(A \text{ or } B) = P(A) + P(B)$$

More generally, if $A_1, A_2, \ldots, A_n$ are mutually exclusive and exhaustive (one must occur):

$$P(A_1) + P(A_2) + \cdots + P(A_n) = 1$$

Sample Space Diagrams

A sample space diagram lists all possible outcomes of two combined events in a grid. Each cell represents one outcome.

Example 3 — Two dice sample space

Two fair six-sided dice are rolled. A sample space diagram shows all 36 equally likely outcomes. Find the probability that the sum equals 7.

Step 1 List pairs that sum to 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — that is 6 pairs.

Step 2 $P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$

Example 4 — Mutually exclusive events

A card is drawn from a standard pack of 52. Find $P(\text{ace or king})$.

Step 1 An ace and a king cannot be the same card, so these events are mutually exclusive.

Step 2 $P(\text{ace}) = \dfrac{4}{52}$; $P(\text{king}) = \dfrac{4}{52}$

Step 3 $P(\text{ace or king}) = \dfrac{4}{52} + \dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13}$

Practice 2a

A fair coin is tossed and a fair four-sided die (faces 1–4) is rolled. List the full sample space and find: (i) $P(\text{head and even number})$, (ii) $P(\text{tail and number} > 2)$.

Show Solution

Sample space (8 equally likely outcomes): H1, H2, H3, H4, T1, T2, T3, T4

(i) Head and even: H2, H4 — 2 outcomes. $P = \dfrac{2}{8} = \dfrac{1}{4}$

(ii) Tail and greater than 2: T3, T4 — 2 outcomes. $P = \dfrac{2}{8} = \dfrac{1}{4}$

Practice 2b

$P(A) = 0.3$, $P(B) = 0.45$. Events $A$ and $B$ are mutually exclusive. Find $P(A \text{ or } B)$ and $P(\text{neither } A \text{ nor } B)$.

Show Solution

$P(A \text{ or } B) = 0.3 + 0.45 = 0.75$

$P(\text{neither}) = 1 - 0.75 = 0.25$

3. Tree Diagrams

A tree diagram shows all possible outcomes for two or more events happening in sequence. Each branch shows one possible outcome and its probability. The rules are:

Rules for Tree Diagrams

Multiply along branches to find the probability of a combined outcome.
Add across branches to find the probability of events that can happen in different ways.
Probabilities on branches from the same point must always sum to $1$.
The sum of all end-branch probabilities equals $1$.

With and Without Replacement

When selecting items from a set:

With replacement: the item is returned before the next selection, so probabilities remain the same at each stage.
Without replacement: the item is not returned, so both the numerator and denominator change at each stage.

Example 5 — Tree diagram without replacement

A bag contains 4 red and 3 blue balls. Two balls are selected at random without replacement. Find the probability that both balls are the same colour.

Step 1 Draw the tree diagram. Total balls at start = 7.

First selection:

$P(\text{red}) = \dfrac{4}{7}$
$P(\text{blue}) = \dfrac{3}{7}$

Second selection (given first was red, now 6 balls remain: 3 red, 3 blue):

$P(\text{red} \mid \text{first red}) = \dfrac{3}{6} = \dfrac{1}{2}$
$P(\text{blue} \mid \text{first red}) = \dfrac{3}{6} = \dfrac{1}{2}$

Second selection (given first was blue, now 6 balls remain: 4 red, 2 blue):

$P(\text{red} \mid \text{first blue}) = \dfrac{4}{6} = \dfrac{2}{3}$
$P(\text{blue} \mid \text{first blue}) = \dfrac{2}{6} = \dfrac{1}{3}$

Step 2 Find $P(\text{same colour})$:

$P(\text{RR}) = \dfrac{4}{7} \times \dfrac{3}{6} = \dfrac{12}{42} = \dfrac{2}{7}$

$P(\text{BB}) = \dfrac{3}{7} \times \dfrac{2}{6} = \dfrac{6}{42} = \dfrac{1}{7}$

$P(\text{same colour}) = \dfrac{2}{7} + \dfrac{1}{7} = \dfrac{3}{7}$

Example 6 — Tree diagram with replacement

A biased coin has $P(\text{heads}) = 0.6$. It is flipped twice. Find: (a) $P(\text{two heads})$, (b) $P(\text{exactly one tail})$.

Step 1 Since flips are independent (with replacement): $P(H) = 0.6$, $P(T) = 0.4$ each time.

(a) $P(HH) = 0.6 \times 0.6 = 0.36$

(b) Exactly one tail: $P(HT)$ or $P(TH)$

$P(HT) = 0.6 \times 0.4 = 0.24$

$P(TH) = 0.4 \times 0.6 = 0.24$

$P(\text{exactly one tail}) = 0.24 + 0.24 = 0.48$

Practice 3a

A box contains 5 white and 3 black counters. Two counters are chosen at random without replacement. Find $P(\text{at least one white counter})$.

Show Solution

It is easier to use $P(\text{at least one white}) = 1 - P(\text{no white}) = 1 - P(\text{BB})$.

$P(BB) = \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}$

$P(\text{at least one white}) = 1 - \dfrac{3}{28} = \dfrac{25}{28}$

Practice 3b

$P(\text{rain on any day}) = 0.3$. Find the probability that it rains on exactly one of two consecutive days.

Show Solution

Days are independent. $P(\text{rain}) = 0.3$, $P(\text{no rain}) = 0.7$.

$P(\text{rain, no rain}) = 0.3 \times 0.7 = 0.21$

$P(\text{no rain, rain}) = 0.7 \times 0.3 = 0.21$

$P(\text{exactly one day rain}) = 0.21 + 0.21 = 0.42$

4. Venn Diagrams and Set Notation

Venn diagrams use overlapping circles inside a rectangle to show relationships between sets. The rectangle represents the universal set $\xi$ (all possible outcomes).

Set Notation

$\xi$ — universal set (all outcomes)
$A \cup B$ — union of $A$ and $B$ (in $A$ or $B$ or both)
$A \cap B$ — intersection of $A$ and $B$ (in both $A$ and $B$)
$A'$ — complement of $A$ (not in $A$)
$n(A)$ — number of elements in set $A$

Addition Rule

For any two events $A$ and $B$:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

The intersection is subtracted to avoid counting it twice. If $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$ and the formula reduces to $P(A \cup B) = P(A) + P(B)$.

Example 7 — Venn diagram from given information

In a group of 40 students, 25 study French ($F$), 18 study Spanish ($S$), and 8 study both. Draw a Venn diagram and find:

(a) $P(F \cup S)$ (b) $P(F \cap S')$ (c) $P(F' \cap S')$

Step 1 Fill in the Venn diagram regions:

Both: $F \cap S = 8$
French only: $25 - 8 = 17$
Spanish only: $18 - 8 = 10$
Neither: $40 - 17 - 8 - 10 = 5$

Step 2 Calculate probabilities (out of 40):

(a) $P(F \cup S) = \dfrac{17 + 8 + 10}{40} = \dfrac{35}{40} = \dfrac{7}{8}$

(b) $P(F \cap S') = \dfrac{17}{40}$ (French only)

Example 8 — Using the addition rule

$P(A) = 0.5$, $P(B) = 0.4$, $P(A \cap B) = 0.2$. Find $P(A \cup B)$ and $P(A' \cap B)$.

$P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7$

$P(A' \cap B)$ = probability in $B$ but not $A$ = $P(B) - P(A \cap B) = 0.4 - 0.2 = 0.2$

Practice 4a

In a class of 30 pupils, 16 play football ($F$), 14 play cricket ($C$), and 5 play both. Find $P(F' \cap C')$, the probability that a randomly chosen pupil plays neither sport.

Show Solution

Football only: $16 - 5 = 11$

Cricket only: $14 - 5 = 9$

Both: $5$

Neither: $30 - 11 - 5 - 9 = 5$

$P(F' \cap C') = \dfrac{5}{30} = \dfrac{1}{6}$

Practice 4b

$P(X) = 0.45$, $P(Y) = 0.35$, $P(X \cup Y) = 0.65$. Find $P(X \cap Y)$ and state whether $X$ and $Y$ are mutually exclusive.

Show Solution

$P(X \cap Y) = P(X) + P(Y) - P(X \cup Y) = 0.45 + 0.35 - 0.65 = 0.15$

Since $P(X \cap Y) = 0.15 \neq 0$, the events are not mutually exclusive.

5. Conditional Probability Extended

Conditional probability is the probability of event $A$ occurring, given that event $B$ has already occurred. It is written $P(A \mid B)$ (read: "probability of $A$ given $B$").

Conditional Probability Formula

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0$$

Rearranging: $P(A \cap B) = P(A \mid B) \times P(B)$ — this is the multiplication rule.

Independent Events

Events $A$ and $B$ are independent if the occurrence of one does not affect the other:

$$P(A \mid B) = P(A) \quad \text{and} \quad P(A \cap B) = P(A) \times P(B)$$

Example 9 — Conditional probability from a Venn diagram

Using the data from Example 7 (40 students, French 25, Spanish 18, both 8):

Find the probability that a student selected at random studies French, given that they study Spanish.

Step 1 We want $P(F \mid S)$.

Step 2 $P(F \cap S) = \dfrac{8}{40}$; $P(S) = \dfrac{18}{40}$

Step 3 $P(F \mid S) = \dfrac{P(F \cap S)}{P(S)} = \dfrac{8/40}{18/40} = \dfrac{8}{18} = \dfrac{4}{9}$

Example 10 — Conditional probability without replacement

Cards labelled 1 to 10 are shuffled. Two cards are drawn without replacement. Find the probability that the second card is even, given the first card drawn was odd.

Step 1 After drawing an odd card, 9 cards remain: 5 even and 4 odd.

Step 2 $P(\text{second even} \mid \text{first odd}) = \dfrac{5}{9}$

Example 11 — Testing for independence

$P(A) = 0.4$, $P(B) = 0.3$, $P(A \cap B) = 0.12$. Are $A$ and $B$ independent?

Step 1 If independent: $P(A) \times P(B) = 0.4 \times 0.3 = 0.12$

Step 2 $P(A \cap B) = 0.12 = P(A) \times P(B)$ ✓

Yes, $A$ and $B$ are independent events.

Practice 5a

A bag has 6 red and 4 blue balls. Two are selected without replacement. Find $P(\text{second ball is red} \mid \text{first ball is red})$.

Show Solution

After drawing one red ball: 9 balls remain (5 red, 4 blue).

$P(\text{second red} \mid \text{first red}) = \dfrac{5}{9}$

Practice 5b

$P(A \cap B) = 0.15$ and $P(B) = 0.5$. Find $P(A \mid B)$. If $P(A) = 0.3$, are $A$ and $B$ independent?

Show Solution

$P(A \mid B) = \dfrac{0.15}{0.5} = 0.3$

Since $P(A \mid B) = 0.3 = P(A)$, events $A$ and $B$ are independent.

6. Mixed Practice Problems

Question 1

A fair spinner has sections labelled 1, 1, 2, 3, 3, 3. It is spun once. Find: (a) $P(3)$, (b) $P(\text{odd number})$, (c) $P(\text{not 1})$.

Show Solution

Total sections: 6

(a) Sections with 3: 3. $P(3) = \dfrac{3}{6} = \dfrac{1}{2}$

(b) Odd numbers: 1, 1, 3, 3, 3 — 5 sections. $P(\text{odd}) = \dfrac{5}{6}$

Question 2

Two fair dice are rolled. Find the probability that the product of the two numbers is greater than 20.

Show Solution

Total outcomes: 36. Products greater than 20:

$(4,6): 24$, $(5,5): 25$, $(5,6): 30$, $(6,4): 24$, $(6,5): 30$, $(6,6): 36$ — 6 outcomes.

$P(\text{product} > 20) = \dfrac{6}{36} = \dfrac{1}{6}$

Question 3

A bag contains 7 counters: 3 red, 2 green, 2 yellow. Two are drawn at random without replacement. Find $P(\text{both green})$.

Show Solution

$P(\text{first green}) = \dfrac{2}{7}$

$P(\text{second green} \mid \text{first green}) = \dfrac{1}{6}$

$P(\text{both green}) = \dfrac{2}{7} \times \dfrac{1}{6} = \dfrac{2}{42} = \dfrac{1}{21}$

Question 4

$P(A) = \dfrac{3}{5}$, $P(B) = \dfrac{1}{3}$, and $A$ and $B$ are independent. Find $P(A \cap B)$ and $P(A \cup B)$.

Show Solution

$P(A \cap B) = P(A) \times P(B) = \dfrac{3}{5} \times \dfrac{1}{3} = \dfrac{1}{5}$

$P(A \cup B) = \dfrac{3}{5} + \dfrac{1}{3} - \dfrac{1}{5} = \dfrac{9}{15} + \dfrac{5}{15} - \dfrac{3}{15} = \dfrac{11}{15}$

Question 5

In a survey of 50 people, 30 own a cat ($C$), 25 own a dog ($D$), and 10 own both. Find the probability that a randomly chosen person owns a cat but not a dog.

Show Solution

Cat only: $30 - 10 = 20$

$P(C \cap D') = \dfrac{20}{50} = \dfrac{2}{5}$

Question 6

A bag has 5 red and 3 blue balls. One ball is drawn, its colour noted, and it is not replaced. A second ball is then drawn. Complete a tree diagram and find $P(\text{one red and one blue})$.

Show Solution

$P(\text{red then blue}) = \dfrac{5}{8} \times \dfrac{3}{7} = \dfrac{15}{56}$

$P(\text{blue then red}) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$

$P(\text{one of each}) = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28}$

Question 7

$P(X \cup Y) = 0.8$, $P(X) = 0.6$, $P(Y) = 0.5$. Find $P(X \cap Y)$ and $P(X' \cap Y)$.

Show Solution

$P(X \cap Y) = P(X) + P(Y) - P(X \cup Y) = 0.6 + 0.5 - 0.8 = 0.3$

$P(X' \cap Y)$ = in $Y$ but not $X$ $= P(Y) - P(X \cap Y) = 0.5 - 0.3 = 0.2$

Question 8 Extended

A medical test for a disease gives a positive result for 95% of people who have the disease and a positive result for 4% of people who do not have the disease. In a population where 2% have the disease, find the probability that a person who tests positive actually has the disease.

Show Solution

Let $D$ = has disease, $T+$ = tests positive.

$P(D) = 0.02$, $P(D') = 0.98$, $P(T+ \mid D) = 0.95$, $P(T+ \mid D') = 0.04$

$P(T+) = P(T+ \mid D) \cdot P(D) + P(T+ \mid D') \cdot P(D')$

$= 0.95 \times 0.02 + 0.04 \times 0.98 = 0.019 + 0.0392 = 0.0582$

$P(D \mid T+) = \dfrac{P(T+ \mid D) \cdot P(D)}{P(T+)} = \dfrac{0.019}{0.0582} \approx 0.326$

So only about $32.6\%$ of those who test positive actually have the disease — illustrating the importance of disease prevalence in interpreting test results.

Question 9

A card is drawn at random from a pack of 52 playing cards. Events: $H$ = card is a heart, $K$ = card is a king. Find $P(H \cup K)$.

Show Solution

$P(H) = \dfrac{13}{52} = \dfrac{1}{4}$; $P(K) = \dfrac{4}{52} = \dfrac{1}{13}$; $P(H \cap K) = \dfrac{1}{52}$ (king of hearts)

$P(H \cup K) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$

Question 10 Extended

From Question 9, find the probability that a randomly chosen card is a king, given that it is a heart.

Show Solution

$P(K \mid H) = \dfrac{P(K \cap H)}{P(H)} = \dfrac{1/52}{13/52} = \dfrac{1}{13}$

Note that $P(K \mid H) = P(K) = \dfrac{1}{13}$, which confirms that $H$ and $K$ are independent events.