Cambridge IGCSE Mathematics 0580

Probability and tree diagrams

Probability is a number from 0 to 1 measuring how likely an event is; the probability an event does not happen is 1 minus the probability it does. For two events, and means multiply (when they are independent, or along a branch of a tree diagram) and or means add (when they cannot both happen). Cambridge IGCSE Maths 0580 sets probability on 36 of the 42 papers Quanta has mapped, for 182 marks, and the majority of those marks are tree diagrams — with the without-replacement case, where the second denominator drops by one, doing most of the grading.

Updated 15 September 2026

Single events

P(A) = \frac{\text{number of outcomes in } A}{\text{total number of equally likely outcomes}} \qquad P(\text{not } A) = 1 - P(A)

Give answers as fractions, decimals or percentages — the question usually says which, and a fraction need not be simplified unless asked. Relative frequency estimates a probability from an experiment (times it happened ÷ trials), and the expected number of times an event occurs in $n$ trials is $n \times P(A)$ — a 0.15 chance of rain on each of 40 days is 6 expected rainy days.

Combined events: AND and OR

P(A \text{ and } B) = P(A) \times P(B) \quad \text{(independent events)}

P(A \text{ or } B) = P(A) + P(B) \quad \text{(mutually exclusive events)}

Independent means one outcome does not change the other’s probability — two dice, a coin and a spinner, picking with replacement. Mutually exclusive means they cannot both happen — a card is a heart or a spade, not both. When events are not mutually exclusive, adding double-counts the overlap: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ (Extended).

Tree diagrams

A tree diagram lists every sequence of outcomes for two or more events, with the probability on each branch. Two rules do all the work: multiply along a branch to get the probability of that sequence; add the branches that satisfy the question. And the pair of branches from any point must sum to 1.

With replacement, the second set of branches repeats the first. Without replacement, the second event depends on the first: the total drops by one, and the count of whatever was taken drops by one. That is the whole of what “conditional” means at IGCSE, and it is where the marks are.

Worked example

A bag holds 5 red and 3 blue counters. Two counters are taken at random without replacement. Find the probability that (a) both are red, (b) they are different colours, (c) at least one is red.

First pick: $P(R) = \tfrac{5}{8}$ , $P(B) = \tfrac{3}{8}$ . Second pick, 7 counters left: after a red, $P(R) = \tfrac{4}{7}$ , $P(B) = \tfrac{3}{7}$ ; after a blue, $P(R) = \tfrac{5}{7}$ , $P(B) = \tfrac{2}{7}$ .

(a) Both red — one branch, multiply:

P(RR) = \tfrac{5}{8} \times \tfrac{4}{7} = \tfrac{20}{56} = \tfrac{5}{14}

(b) Different colours — two branches (red then blue, or blue then red), add:

P(RB) + P(BR) = \tfrac{5}{8}\times\tfrac{3}{7} + \tfrac{3}{8}\times\tfrac{5}{7} = \tfrac{15}{56} + \tfrac{15}{56} = \tfrac{30}{56} = \tfrac{15}{28}

(c) At least one red — everything except both blue:

1 - P(BB) = 1 - \tfrac{3}{8}\times\tfrac{2}{7} = 1 - \tfrac{6}{56} = \tfrac{50}{56} = \tfrac{25}{28}

Check: (a) + (b) + $P(BB)$ should be 1: $\tfrac{20}{56} + \tfrac{30}{56} + \tfrac{6}{56} = 1$ . ✓ Marks are usually one for the correct branch probabilities, one for the multiplication, one for adding the right branches — so an unsimplified $\tfrac{30}{56}$ scores fully; a decimal rounded to 0.54 may not.

Conditional probability and Venn diagrams

Extended papers ask “given that” questions: the probability of $A$ given that $B$ has happened restricts the outcomes to those in $B$ :

P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)} = \frac{n(A \cap B)}{n(B)}

In a Venn diagram or a two-way table this is a change of denominator: from the whole group to the group that satisfies the condition. In a tree, it is reading the second-stage branch directly — the $\tfrac{4}{7}$ above is already “red given the first was red”.

Venn questions use set notation — $A \cap B$ (both), $A \cup B$ (either), $A'$ (not $A$ ) — and are usually counting problems in disguise: fill the overlap first, then the rest of each set, then the region outside both, and check the total. Probabilities are then region ÷ total.

Common mistakes

1.Same denominator on the second pick without replacement
One counter has gone: the total is 7, not 8, and the colour taken has one fewer. This is the mark that separates the tree-diagram questions.
2.Adding where the question says 'and'
A sequence of outcomes on one branch is multiplied. Probabilities are added only across different branches that each satisfy the question.
3.One order counted for 'one of each'
Red-then-blue and blue-then-red are different branches. Both count, so the answer is the sum of both.
4.'At least one' built up case by case
It works, but it takes three branches and invites a slip. One minus the probability of none is one line.
5.A probability greater than 1, or a rounded decimal
Above 1 means something was added that should have been multiplied. And keep fractions: $\tfrac{15}{28}$ is exact; 0.54 may lose the accuracy mark.

Common questions

When do you multiply and when do you add probabilities?

Multiply for a sequence of outcomes that all happen — “and”, along a branch of a tree. Add for alternatives that cannot both happen — “or”, across branches.

What changes in a tree diagram without replacement?

The second-stage probabilities depend on the first outcome: the denominator falls by one, and the numerator for the colour (or item) already taken falls by one. With replacement, the second stage repeats the first.

How do you find the probability of 'at least one'?

One minus the probability of none. “At least one red” is $1 - P(\text{no red})$ , which is a single branch.

Is conditional probability on the Core paper?

Tree diagrams with and without replacement appear on both tiers. Formal conditional probability with set notation and Venn diagrams of three sets are Extended.

Practise probability and tree diagrams against real mark schemes

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