Tree Diagrams
Tree Diagrams in NSW Year 10 Core maths show every outcome of a multistage experiment as branches, with a probability written on each branch (NESA outcome MA5‑PRO‑C‑01). Multiplying along a path gives the probability of that outcome, and adding paths combines outcomes.
You will complete tree diagrams, find probabilities such as exactly two heads in three tosses or two marbles of different colours, and work both with and without replacement. You will also solve problems about traffic lights, penalty shots, choosing class captains, matching socks, raffle prizes, tennis serves and whether Sam gets to school on time when his alarm might fail. It is part of Year 10 Core Probability.
Every question with a fully worked solution.
- Tree Diagrams - Video - Tree Diagrams Watch
Theory
This Year 10 maths page shows how to draw and use a tree diagram: multiply along a path, add the paths you need, change the second-stage branches when there is no replacement, extend to three stages, and handle second stages whose probabilities depend on what happened first.
A tree diagram shows a multistage experiment. Each stage is a set of branches, labelled with the outcome and its probability.
A path runs from the start to one end branch, such as red then blue. Each end branch is one outcome of the whole experiment.
The branches leaving any one point add to \(1\). The probabilities of all the end branches also add to \(1\).
With replacement the second stage looks the same as the first (independent). Without replacement the second-stage probabilities change, because an item has gone (dependent).
A branch may only appear after some outcomes. In tennis, a second serve only happens after the first is out, so that stage is drawn on the "out" branch only.
The two rules for trees:
How to use a tree diagram
- Draw the first stage: one branch per outcome, with its probability.
- Draw the second stage from each end. Change the fractions if there is no replacement, or if the question gives different probabilities after each first result.
- Check that each set of branches from a point adds to \(1\).
- Multiply along each path you need.
- Add those paths. For "at least one", find \(1 - P(\text{none})\) instead.
(i) \(P(\text{both yellow})\)
(ii) \(P(\text{different colours})\).
With replacement, both draws have \(P(\text{G}) = \dfrac{3}{8}\) and \(P(\text{Y}) = \dfrac{5}{8}\).
| \((\text{i})\quad P(\text{YY})\) | \(=\) | \(\dfrac{5}{8} \times \dfrac{5}{8} = \boldsymbol{\dfrac{25}{64}}\) |
(ii) Different colours means the path GY or the path YG. Add the two paths.
| \(P(\text{GY}) + P(\text{YG})\) | \(=\) | \(\dfrac{3}{8} \times \dfrac{5}{8} + \dfrac{5}{8} \times \dfrac{3}{8}\) |
| \(\) | \(=\) | \(\dfrac{30}{64} = \boldsymbol{\dfrac{15}{32}}\) |
(i) \(P(\text{same colour})\)
(ii) \(P(\text{at least one red})\).
After the first apple, \(6\) are left, with one fewer of the colour taken.
| \(P(\text{GG})\) | \(=\) | \(\dfrac{4}{7} \times \dfrac{3}{6} = \dfrac{12}{42}\) |
| \(P(\text{RR})\) | \(=\) | \(\dfrac{3}{7} \times \dfrac{2}{6} = \dfrac{6}{42}\) |
| \((\text{i})\quad P(\text{same})\) | \(=\) | \(\dfrac{12}{42} + \dfrac{6}{42} = \dfrac{18}{42} = \boldsymbol{\dfrac{3}{7}}\) |
(ii) The complement of "at least one red" is GG.
| \(P(\text{at least one R})\) | \(=\) | \(1 - \dfrac{12}{42} = \dfrac{30}{42} = \boldsymbol{\dfrac{5}{7}}\) |
Each question: \(P(\text{C}) = \dfrac{1}{4}\) and \(P(\text{W}) = \dfrac{3}{4}\). A full tree has \(2 \times 2 \times 2 = 8\) end branches. Exactly \(2\) correct is on three paths, CCW, CWC and WCC:
| \(\text{each path}\) | \(=\) | \(\dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{3}{4} = \dfrac{3}{64}\) |
| \(P(\text{exactly } 2)\) | \(=\) | \(3 \times \dfrac{3}{64}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{9}{64}}\) |
The three paths have the same probability because each has two C branches and one W branch.
The bus can be late in two ways. Multiply along each path, then add.
| \(P(\text{late})\) | \(=\) | \(0.25 \times 0.4 + 0.75 \times 0.1\) |
| \(\) | \(=\) | \(0.1 + 0.075 = \boldsymbol{0.175}\) |
Common pitfalls
Frequently asked questions
How do you use a tree diagram in probability?
Draw a set of branches for each stage, labelled with probabilities. Multiply along each path to get its probability, then add the paths that make up the event.
When do you multiply and when do you add on a tree diagram?
Multiply along a path (this and then that). Add different paths (this path or that path).
How does a tree diagram change without replacement?
On the second stage, the total goes down by one, and the count of the item already taken goes down by one. So the second-stage fractions are different on each branch.
How do you check a tree diagram?
The branches from each point must add to \(1\), and the probabilities of all the end branches must add to \(1\).
How many end branches does a tree have?
Multiply the number of branches at each stage. A spinner with \(3\) colours spun twice gives \(3 \times 3 = 9\) end branches.