Two-Stage Events - Tree Diagrams
Learn to use tree diagrams for two-stage events in NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-07). A tree diagram sets out every outcome of a two-part experiment as a path of branches, so outcomes that are equally likely can be listed and counted.
You will draw the tree, count the total outcomes and the favourable outcomes to find a probability, label items that look alike so every path is equally likely, and handle draws where the item is or is not replaced — such as taking two counters from a bag or tossing two coins.
Theory
A tree diagram sets out every outcome of a two-stage event as a path of branches. This Year 12 Standard 1 (NSW) guide shows how to list the outcomes, count the favourable ones to find a probability, and handle draws with and without replacement and “at least one”.
A tree diagram shows every outcome of a two-stage experiment as a path of branches. Draw one branch for each result of the first stage, then a full set of second-stage branches from the end of each first branch. Write the outcome at the end of each path, e.g. \(HT\) means head then tail.
The sample space is the list of all the paths. When every path is equally likely, the probability of an event is the number of favourable paths divided by the total number of paths.
With replacement (or for coins, spinners and dice) the second stage has the same choices as the first. Without replacement the item drawn first is missing, so each first branch has one fewer second branch. If some items look the same, label them (\(R_1\), \(R_2\)) so that every path is equally likely. “At least one” is often quickest as \(1-P(\text{none})\).
For equally likely outcomes:
Count the outcomes of a two-stage experiment:
How to use a tree diagram
- Draw the first-stage branches, then a full set of second-stage branches from the end of each one (one fewer if there is no replacement). Label identical items so every path is equally likely.
- List the outcome at the end of every path and count the total.
- Count the favourable paths and divide by the total (or use \(1-P(\text{none})\) for “at least one”).
\(2\) coin branches, each with \(3\) spinner branches: \(2\times3=6\) equally likely outcomes. Only \(TG\) is favourable.
| \(n(S)\) | \(=\) | \(6\) |
| \(n(TG)\) | \(=\) | \(1\) |
| \(P(TG)\) | \(=\) | \(\dfrac{1}{6}\) |
\(P(\text{tail and green})=\dfrac{1}{6}\).
The first card cannot be drawn again, so each first branch has only \(2\) second branches: \(3\times2=6\) outcomes. Card \(A\) appears in \(AB\), \(AC\), \(BA\) and \(CA\).
| \(n(S)\) | \(=\) | \(6\) |
| \(n(A\text{ drawn})\) | \(=\) | \(4\) |
| \(P(A\text{ drawn})\) | \(=\) | \(\dfrac{4}{6}=\dfrac{2}{3}\) |
\(P(A\text{ drawn})=\dfrac{2}{3}\).
Label the balls \(O_1\), \(O_2\) and \(K\) so every path is equally likely: \(3\times3=9\) outcomes. One of each: \(O_1K\), \(O_2K\), \(KO_1\), \(KO_2\).
| \(n(S)\) | \(=\) | \(9\) |
| \(n(\text{one of each})\) | \(=\) | \(4\) |
| \(P(\text{one of each})\) | \(=\) | \(\dfrac{4}{9}\) |
\(P(\text{one of each})=\dfrac{4}{9}\).
\(3\times3=9\) equally likely outcomes. “At least one win” is the complement of “no win”: \(L_1L_1\), \(L_1L_2\), \(L_2L_1\), \(L_2L_2\).
| \(P(\text{no win})\) | \(=\) | \(\dfrac{4}{9}\) |
| \(P(\text{at least one win})\) | \(=\) | \(1-\dfrac{4}{9}\) |
| \(\) | \(=\) | \(\dfrac{5}{9}\) |
\(P(\text{at least one win})=\dfrac{5}{9}\).
Common pitfalls
Frequently asked questions
How do you draw a tree diagram for a two-stage experiment?
Draw one branch for each result of the first stage. From the end of each of those branches draw a full set of branches for the second stage. Each path from the start to an end is one outcome; write it at the end of the path.
How do you find a probability from a tree diagram?
Count the outcomes at the ends of the branches, then count the ones that are favourable. When the outcomes are equally likely, the probability is the favourable count divided by the total count, for example 2 out of 4 for exactly one head with two coins.
How many outcomes does a two-stage experiment have?
Multiply the number of results for each stage. A coin then a three-colour spinner gives 2 times 3 = 6 outcomes; two dice give 6 times 6 = 36.
What changes on a tree diagram without replacement?
The item drawn first is not put back, so it cannot be drawn again. Each first branch has one fewer second branch: 3 cards drawn twice without replacement give 3 times 2 = 6 outcomes instead of 9.
Why label items that look the same?
With 2 red and 1 blue marble the colour pairs RR, RB, BR and BB are not equally likely. Labelling the marbles R1, R2 and B gives 9 paths that are all equally likely, so they can be counted.
What is the probability of two heads when tossing two coins?
The tree has 4 equally likely outcomes, HH, HT, TH and TT. Only HH has two heads, so the probability is 1 out of 4, which is 1 quarter.