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How do you use tree diagrams to find probabilities of combined events, with and without replacement?

Draw tree diagrams for two or more events, multiply along branches and add between branches, and handle independent events and conditional probability without replacement (Higher tier).

A CCEA GCSE Mathematics answer on tree diagrams, covering drawing trees for combined events, multiplying along branches and adding between them, independent events, and conditional probability without replacement.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. Building a tree diagram
  3. Multiply along, add between
  4. Independent events
  5. Conditional probability without replacement (Higher)
  6. Why this matters

What this dot point is asking

Tree diagrams organise the probabilities of two or more events happening in sequence, a key CCEA Probability tool. You must draw a tree, multiply along branches to find the probability of a path, and add between branches to combine paths. At Higher tier this extends to independent events (such as repeated coin flips) and to conditional probability without replacement, where the second probability depends on the first outcome. The without-replacement case is the most examined and the easiest to slip on.

Building a tree diagram

A tree diagram has a set of branches for the first event, and from the end of each of those, a set of branches for the second event, and so on. Each branch is labelled with the outcome and its probability. The branches from any single point must add to 1, which is a quick check that the tree is correct.

The strength of a tree is that it lays out every possible combined outcome as a path from left to right, so nothing is missed.

Multiply along, add between

The two rules for using a tree are simple but must not be mixed up.

So "multiply along, add between" is the slogan: AND means multiply, OR means add.

Independent events

Two events are independent if the outcome of one does not affect the other, such as flipping a coin twice or rolling two dice. For independent events the branch probabilities are the same at each stage, because nothing changes between events.

Conditional probability without replacement (Higher)

When items are drawn without replacement, the situation changes after the first draw, because the item is not put back. The total falls by one, and the favourable count falls too if a favourable item was taken. This makes the second event conditional on the first.

Why this matters

Tree diagrams are the standard way to handle combined and conditional probability, used in reliability, genetics and decision making, and they are a dependable Higher-tier topic. The "multiply along, add between" rule and the without-replacement adjustment are reused constantly, and the complement shortcut for "at least one" saves time. CCEA rewards a clearly labelled tree with branch probabilities, so draw it fully before calculating.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20203 marksA coin is flipped twice. Use a tree diagram to find the probability of getting two heads. (Non-calculator.)
Show worked answer →

Each flip has P(head)=12P(\text{head}) = \tfrac{1}{2} and P(tail)=12P(\text{tail}) = \tfrac{1}{2}, and the flips are independent.

To find the probability of heads then heads, multiply along the branches: 12×12=14\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}.

Marks are for the branch probabilities, for multiplying along the path, and for 14\tfrac{1}{4}. Adding the probabilities instead of multiplying is the standard error for combined events.

CCEA 20224 marksA bag has 4 red and 6 blue balls. Two are drawn without replacement. Find the probability that both are red. (Higher, calculator.)
Show worked answer →

First draw: P(red)=410P(\text{red}) = \dfrac{4}{10}.

Without replacement, one red is removed, so for the second draw there are 3 red out of 9: P(red second)=39P(\text{red second}) = \dfrac{3}{9}.

Multiply along the branch: 410×39=1290=215\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}.

Marks are for the first probability, for the adjusted second probability, for multiplying, and for 215\tfrac{2}{15}. Keeping 410\tfrac{4}{10} for the second draw (as if replaced) is the key mistake.

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