Study Guides
Cambridge IGCSE Mathematics 0580: Introduction to probability, relative and expected frequencies – Study Guide
Step-by-step guide to the probability scale, single events, relative frequency and expected frequency for Cambridge IGCSE Maths 0580, Core and Extended.
- Subject
- Mathematics
- Level
- IGCSE
- Topic
- Introduction to probability, relative and expected frequencies
- Author
- Marlbridge Academic Team
- Updated
Aligned to Cambridge IGCSE Mathematics (0580), 2025-2027. Official specification .
Syllabus page (what it covers and how it is assessed): Cambridge IGCSE Mathematics.
Syllabus points this page covers, with Core and Extended
0580
- 8 Probability (whole topic)
- 8.1 Introduction to probability · Core and Extended
- 8.2 Relative and expected frequencies · Core and Extended
"Core and Extended" means part of that syllabus point is Extended only. The page's own tier notes say which part.
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This study guide covers subtopics 8.1 Introduction to probability and 8.2 Relative and expected frequencies for Cambridge IGCSE Mathematics 0580. It is aligned to the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027 (Version 3) and teaches both the Core outcomes (C8.1, C8.2) and the Extended outcomes (E8.1, E8.2). The one Extended-only outcome is probability notation, P(A) and P(A′). Core candidates sit Papers 1 (non-calculator) and 3 (calculator); Extended candidates sit Papers 2 (non-calculator) and 4 (calculator).
Combined events (8.3) and conditional probability (E8.4) are not taught here; see the topic-level Probability study guide and the Statistics and Probability (Extended) practice set. Tick off outcomes on the printable 0580 checklist.
What this guide covers
| Syllabus section | What you must be able to do | Tier |
|---|---|---|
| C8.1 / E8.1 (1) | Understand and use the probability scale from 0 to 1 | Core and Extended |
| E8.1 (2) | Understand and use probability notation: P(A) and P(A′) | Extended only |
| C8.1 (2) / E8.1 (3) | Calculate the probability of a single event, using tables, graphs or Venn diagrams where given | Core and Extended (Core Venn diagrams limited to two sets) |
| C8.1 (3) / E8.1 (4) | Use: probability of an event not occurring = 1 − probability of it occurring | Core and Extended |
| C8.2 / E8.2 (1) | Understand relative frequency as an estimate of probability | Core and Extended |
| C8.2 / E8.2 (2) | Calculate expected frequencies; understand fair, bias and random | Core and Extended |
The syllabus says probabilities should be given as a fraction, decimal or percentage. Do not write a ratio such as 3 : 5, or words such as “3 out of 5”, as your final answer.
8.1 The probability scale
Every probability is a number from 0 to 1.
- 0 means the event is impossible.
- 1 means the event is certain.
- 0.5 (or 1/2, or 50%) means the event is as likely to happen as not, often called even chance.
- Values between 0 and 0.5 are unlikely; values between 0.5 and 1 are likely.
An answer such as 1.2 or −0.3 is always wrong.
Worked example 1: placing events on the scale (non-calculator)
A bag contains 4 green, 6 yellow and 10 white beads. One bead is taken at random. Mark these on a probability scale: (a) the bead is white, (b) the bead is black, (c) the bead is not green.
Total beads = 4 + 6 + 10 = 20
(a) P(white) = 10/20 = 1/2 -> mark at 0.5 (even chance)
(b) P(black) = 0/20 = 0 -> mark at 0 (impossible)
(c) not green = 6 + 10 = 16 beads
P(not green) = 16/20 = 4/5 -> mark at 0.8 (likely)
Probability notation (Extended only)
- P(A) means “the probability of event A”.
- P(A′) means “the probability of not A”. A′ is read “A prime” or “A dash”, and it is called the complement of A.
So “the probability the bead is not green is 4/5” becomes P(G′) = 4/5. Core candidates need the idea, not the notation.
8.1 Probability of a single event
When all outcomes are equally likely, for example a fair dice or an item chosen at random:
probability of an event = number of outcomes in the event / total number of possible outcomes
Questions often give the information in a table, chart or Venn diagram: find the right count for the top and the right total for the bottom.
Worked example 2: probability from a table (non-calculator)
A class of students records how they travel to school.
| Method | Walk | Bus | Car | Cycle |
|---|---|---|---|---|
| Number of students | 14 | 18 | 11 | 7 |
One student is chosen at random. Find the probability that the student (a) travels by bus, (b) walks or cycles.
Total = 14 + 18 + 11 + 7 = 50
(a) P(bus) = 18/50 = 9/25 (= 0.36 = 36%)
(b) walk or cycle = 14 + 7 = 21
P(walk or cycle) = 21/50 (= 0.42 = 42%)
Any of the three forms (fraction, decimal, percentage) is allowed. If you simplify a fraction, check the simplified version is still equal to the original: 18/50 and 9/25 are both 0.36. The same method works for a bar chart: read each frequency, add for the total, then divide.
Worked example 3: probability from a two-set Venn diagram (non-calculator)
A sports club has 32 members. Set S is members who swim and set T is members who play tennis. The Venn diagram shows:
- S only: 9
- both S and T (the overlap): 5
- T only: 11
- outside both circles: 7
One member is chosen at random. Find the probability that the member (a) plays tennis, (b) does exactly one of the two sports, (c) does neither.
Check total: 9 + 5 + 11 + 7 = 32
(a) T = T only + overlap = 11 + 5 = 16
P(tennis) = 16/32 = 1/2
(b) exactly one = S only + T only = 9 + 11 = 20
P(exactly one) = 20/32 = 5/8
(c) P(neither) = 7/32
In (a), everyone inside the T circle plays tennis, so include the overlap.
Worked example 4: three-set Venn diagram (Extended only)
Core Venn diagrams here are limited to two sets; the Extended outcome has no such limit.
A group of 60 students is surveyed on three clubs, A, B and C. The regions of the Venn diagram are:
| Region | Number |
|---|---|
| A only | 8 |
| B only | 10 |
| C only | 6 |
| A and B, not C | 4 |
| A and C, not B | 3 |
| B and C, not A | 5 |
| A, B and C | 2 |
| none of the three | 22 |
A student is chosen at random. Find (a) the probability the student is in exactly two clubs, (b) P(C′).
Check total: 8 + 10 + 6 + 4 + 3 + 5 + 2 + 22 = 60
(a) exactly two = 4 + 3 + 5 = 12 (do NOT include the 2 in all three)
P(exactly two) = 12/60 = 1/5
(b) inside C = 6 + 3 + 5 + 2 = 16
P(C) = 16/60, so P(C′) = 1 − 16/60 = 44/60 = 11/15
8.1 The probability of an event not occurring
An event either happens or it does not, so the two probabilities add to 1:
probability of not A = 1 − probability of A
P(A′) = 1 − P(A) (Extended notation)
So the probabilities of all the outcomes of one event add up to 1, which lets you find a missing value in a table.
Worked example 5: missing probabilities (non-calculator)
A spinner can land on red, blue, green or yellow.
| Colour | Red | Blue | Green | Yellow |
|---|---|---|---|---|
| Probability | 0.35 | 0.2 | x | 2x |
Find x, and the probability the spinner lands on yellow.
0.35 + 0.2 + x + 2x = 1
0.55 + 3x = 1
3x = 0.45
x = 0.15
P(yellow) = 2x = 0.3
Complements in context:
- The probability it rains tomorrow is 35%. The probability it does not rain is 100% − 35% = 65%.
- (Extended) A train is late with probability P(L) = 0.08. Then P(L′) = 1 − 0.08 = 0.92.
8.2 Relative frequency as an estimate of probability
When outcomes are not equally likely (a badly made spinner, a drawing pin landing point up), you estimate the probability from an experiment.
relative frequency = number of times the outcome happened / total number of trials
Relative frequency is an estimate of the probability. With few trials it can be far from the true value; it tends to settle as the number of trials grows, so more trials give a more reliable estimate.
Worked example 6: a spinner experiment (calculator)
A four-sided spinner numbered 1 to 4 is spun 200 times. A running record of the number of 4s is kept.
| Number of spins | 10 | 50 | 100 | 200 |
|---|---|---|---|---|
| Number of 4s so far | 5 | 19 | 37 | 71 |
| Relative frequency of 4 | 0.5 | 0.38 | 0.37 | 0.355 |
After 10 spins the estimate is 0.5, but by 200 spins it has settled near 0.36. The best estimate is the one from the most trials: 0.355.
The full results after 200 spins are:
| Score | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Frequency | 31 | 52 | 46 | 71 |
| Relative frequency | 0.155 | 0.26 | 0.23 | 0.355 |
The relative frequencies add to 1, as they must.
8.2 Fair, bias and random
- Fair: every outcome is equally likely. A fair four-sided spinner lands on each number with probability 1/4 = 0.25.
- Biased: the outcomes are not equally likely. Some come up more often than they should.
- Random: chosen by chance, with no pattern and no outcome favoured. “A student is chosen at random” means every student has the same chance of being picked.
To judge bias, compare relative frequencies from a large number of trials with the fair-case probabilities. In worked example 6, a fair spinner would give each score a relative frequency near 0.25. After 200 spins, 4 has 0.355 and 1 has 0.155. Both are a long way from 0.25, so the results suggest the spinner is biased (towards 4). Say “suggests”, not “proves”. After only 10 spins you could not draw a sensible conclusion.
8.2 Expected frequencies
If you know or have estimated a probability, you can predict how often the outcome should happen:
expected frequency = probability × number of trials
This is what you expect on average; the real result will usually be close, not exact.
Worked example 7: a fair dice (non-calculator)
A fair ten-sided dice is numbered 1 to 10. It is rolled 150 times. Calculate the expected number of times the score is a multiple of 3.
Multiples of 3: 3, 6, 9 -> 3 outcomes out of 10
P(multiple of 3) = 3/10
Expected number = 3/10 × 150 = 45
Worked example 8: using a relative frequency (calculator)
The spinner from worked example 6 is spun another 600 times. Estimate how many times it lands on 4.
Best estimate of P(4) = 0.355
Expected number = 0.355 × 600 = 213
Worked example 9: a population (calculator)
A town has 18 400 adults. A survey suggests the probability that an adult chosen at random cycles to work is 0.065. Estimate the number of adults in the town who cycle to work.
Expected number = 0.065 × 18 400 = 1196
Worked example 10: working backwards (non-calculator)
At a school fair, the probability of winning a prize on a game is 0.15. The organiser expects to give out 36 prizes. How many people does the organiser expect to play?
0.15 × n = 36
n = 36 ÷ 0.15 = 3600 ÷ 15 = 240
Calculator and non-calculator skills
Every outcome here can be tested on a non-calculator paper, so be fluent with simplifying fractions, converting between fractions, decimals and percentages (3/20 = 0.15 = 15%), subtracting from 1, and dividing by a decimal (multiply both numbers by 100 first, as in worked example 10). On Papers 3 and 4, write the calculation down before the answer so method marks are still available if you key it in wrongly.
Common errors
- Using the wrong total, such as the number of categories instead of the number of people.
- Missing the overlap in a Venn diagram. “Plays tennis” means the whole tennis circle, overlap included.
- Counting the centre region in “exactly two” on a three-set diagram (Extended).
- Giving a probability as a ratio or in words. Use a fraction, decimal or percentage.
- Choosing the wrong relative frequency. The best estimate uses the largest number of trials, or all the trials combined.
- Claiming certainty about bias. Results can suggest bias; a small sample tells you very little.
- Rounding the probability before multiplying when finding an expected frequency. Round only the final answer, if at all.
Where to go next
Revise the whole topic with the Probability revision notes, then test yourself with the Probability practice questions. Check your gaps with the free Core diagnostic or Extended diagnostic, and find other resources for syllabus 0580 on the course hub.
Official syllabus
Cambridge University Press & Assessment (Cambridge International), Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027, Version 3: Subject content, Core sections C8.1 Introduction to probability and C8.2 Relative and expected frequencies; Extended sections E8.1 and E8.2.
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