Get the basics locked down first
Probability questions follow patterns. Learn the patterns, drill them, and you'll pick up marks that other students leave on the table. That's the deal with this topic: it rewards method over intuition. Your gut feeling about chance is often wrong. The formulas are not.
Every probability answer sits on a scale from 0 (impossible) to 1 (certain). Write your answers as fractions, decimals, or percentages, but fractions are cleanest for IGCSE exam work. If your answer is negative or greater than 1, something has gone wrong. Stop. Go back.
Single event probability
The formula is straightforward:
P(event) = number of favourable outcomes / total number of possible outcomes
Worked example 1
A bag contains 4 red balls, 3 blue balls, and 5 green balls. A ball is drawn at random. Find P(blue).
- Count favourable outcomes: 3 blue balls
- Count total outcomes: 4 + 3 + 5 = 12
- P(blue) = 3/12 = 1/4
Clean. No ambiguity. The complement rule follows directly: P(not blue) = 1 - 1/4 = 3/4. This shortcut saves time whenever finding the "not" probability directly would mean counting many outcomes.
Listing outcomes systematically
Some questions ask you to list all possible outcomes. Do this in an organised way or you'll miss some.
Two coins are tossed. The possible outcomes are:
| Coin 1 | Coin 2 | Outcome |
|---|---|---|
| H | H | HH |
| H | T | HT |
| T | H | TH |
| T | T | TT |
Four outcomes total. P(at least one head) = 3/4. Sample space diagrams work the same way for dice: a 6 x 6 grid gives you all 36 outcomes for two dice. Draw the grid. Count. Done.
Relative frequency and expected frequency
Theoretical probability tells you what should happen. Relative frequency tells you what actually happened.
Relative frequency = number of times event occurred / total number of trials
Worked example 2
A spinner is spun 200 times. It lands on red 68 times. Estimate the probability of landing on red.
- Relative frequency = 68/200 = 0.34
That's it. More trials give a better estimate. If someone spun it 2000 times, their relative frequency would be closer to the true probability than yours.
Expected frequency works in reverse:
Expected frequency = probability x number of trials
If P(red) = 0.3 and the spinner is spun 500 times, expected frequency = 0.3 x 500 = 150 times. The actual result won't be exactly 150, but it should be close.
Combined events: the AND and OR rules
This is where probability gets serious. Two rules to memorise:
| Rule | When to use | Operation | Key word |
|---|---|---|---|
| AND rule | Both events must happen | Multiply | "and", "both", "followed by" |
| OR rule | Either event can happen | Add | "or", "either" |
For independent events: P(A and B) = P(A) x P(B)
For mutually exclusive events: P(A or B) = P(A) + P(B)
Worked example 3
A fair coin is tossed and a fair die is rolled. Find the probability of getting a head and a 6.
- P(head) = 1/2
- P(six) = 1/6
- Events are independent, so P(head AND six) = 1/2 x 1/6 = 1/12
Tree diagrams: your most powerful tool
Tree diagrams make combined events visible. Build them correctly every time and you won't drop marks.
How to build a tree diagram
- Each branch represents one outcome of one event
- Probabilities on branches from the same point must add to 1
- Multiply along the branches to get the probability of that path
- Add the probabilities of different paths when you need "or"
Worked example 4: with replacement
A bag contains 3 red and 7 blue balls. A ball is drawn, its colour noted, and it is replaced. A second ball is then drawn. Find P(both red).
- P(red first) = 3/10
- Ball is replaced, so P(red second) = 3/10
- P(both red) = 3/10 x 3/10 = 9/100
Worked example 5: without replacement
Same bag: 3 red and 7 blue. A ball is drawn and NOT replaced. A second ball is drawn. Find P(both red).
- P(red first) = 3/10
- Ball is not replaced. Now there are 2 red and 7 blue left, so 9 total.
- P(red second | red first) = 2/9
- P(both red) = 3/10 x 2/9 = 6/90 = 1/15
Notice the difference. Without replacement, the second probability changes because the total has dropped by one and the count of the colour drawn has dropped by one. This is the single most common source of errors in IGCSE probability questions. Read the question carefully: does it say "replaced" or not?
Do/don't checklist for tree diagrams
| Do | Don't |
|---|---|
| Check branches from each node add to 1 | Assume replacement unless stated |
| Update fractions for "without replacement" | Forget to reduce both numerator and denominator |
| Label every branch clearly | Leave branches unlabelled |
| Multiply along, add across | Add along branches |
| Circle the paths you need before calculating | Try to do everything in your head |
Conditional probability (Extended)
Conditional probability asks: what's the probability of event A, given that event B has already happened? The notation is P(A|B), read as "probability of A given B".
You've already used it. In worked example 5, P(red second | red first) = 2/9 is conditional probability. The condition (red first) changed the available outcomes.
Worked example 6
In a class of 30 students, 18 study French, 12 study Spanish, and 6 study both. A student is chosen at random. Given that the student studies French, find the probability they also study Spanish.
- We know the student studies French, so our total is now 18 (not 30)
- Of these 18 French students, 6 also study Spanish
- P(Spanish | French) = 6/18 = 1/3
The "given" condition shrinks the sample space. You're no longer looking at all 30 students. You're only looking at the 18 who do French.
Venn diagrams for probability (Extended)
Venn diagrams organise overlapping groups. For probability, fill in the numbers and divide by the total.
Worked example 7
Of 50 students: 28 play football, 22 play basketball, 10 play both. Find P(football or basketball).
- Football only: 28 - 10 = 18
- Basketball only: 22 - 10 = 12
- Both: 10
- Neither: 50 - 18 - 12 - 10 = 10
- P(football or basketball) = (18 + 12 + 10)/50 = 40/50 = 4/5
The addition rule for events that are not mutually exclusive:
P(A or B) = P(A) + P(B) - P(A and B)
Check: 28/50 + 22/50 - 10/50 = 40/50. Same answer. The subtraction stops you counting the overlap twice.
Common mistakes and how to fix them
| Mistake | Why it costs marks | Fix |
|---|---|---|
| Giving a probability greater than 1 | Impossible by definition. Signals a fundamental error. | Always check: is your answer between 0 and 1? If not, recalculate. |
| Using replacement fractions when the question says "without replacement" | Produces the wrong probability for the second event | Underline the word "replaced" or "not replaced" in the question before you start. |
| Adding when you should multiply (or vice versa) | AND = multiply, OR = add. Mixing them up changes the entire answer. | Circle the key word in the question: "and/both" means multiply, "or/either" means add. |
| Forgetting to subtract the overlap in Venn diagram OR questions | Counts the intersection twice, inflating the probability | Use the formula P(A or B) = P(A) + P(B) - P(A and B) every time. |
| Not simplifying fractions | May not lose marks but signals rushed work and can cause errors in follow-up parts | Simplify as a final step. Check numerator and denominator share no common factor. |
| Drawing tree diagrams with branches that don't add to 1 | The entire tree is wrong from that point onwards | After writing each set of branches, add them. If the total isn't 1, fix it before continuing. |
Self-check questions
- A bag contains 5 red, 3 green, and 2 yellow counters. One counter is picked at random. Find P(green) and P(not green).
- A spinner has sections numbered 1 to 8, all equally likely. Find P(even number) and P(number greater than 5).
- A coin is tossed three times. List all possible outcomes. Find P(exactly two heads).
- A spinner is spun 400 times and lands on blue 92 times. Estimate the probability of blue. How many times would you expect blue in 1000 spins?
- Two fair dice are rolled. Using a sample space diagram, find P(total = 7) and P(total greater than 10).
- A bag has 6 red and 4 blue marbles. Two are drawn without replacement. Draw a tree diagram and find P(both blue).
- Using the same bag as question 6 but WITH replacement, find P(one of each colour).
- In a group of 40 people, 25 like coffee, 18 like tea, and 8 like both. Draw a Venn diagram. Find P(coffee or tea) and P(neither).
- Given that a person from question 8 likes coffee, find the probability they also like tea.
- A box contains 3 faulty light bulbs and 12 working ones. Two bulbs are tested without replacement. Find P(at least one faulty). Hint: use the complement.
A practical, exam-focused guide to probability for Cambridge IGCSE Mathematics (0580), covering single events, relative frequency, combined events with tree diagrams, and conditional probability with Venn diagrams, plus worked examples and common mistakes.
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