Why Draw a Tree?
A multi-stage experiment produces outcomes through a sequence of steps. Listing every possible result can be straightforward for a small experiment, but it is easy to skip a possibility or confuse the order of results. A tree diagram makes the stages visible: each branch records a possible result at one stage, and each complete path from the start to an endpoint records one outcome.
For example, choosing a meal might involve selecting a bread and then a filling. A tree keeps track of both choices and shows how they combine. The same idea works when later choices depend on earlier results, as in drawing objects without replacement. In either case, the diagram is a structured way to list outcomes and then identify which ones belong to an event.
The sample space, written \(\Omega\), is the set of all possible outcomes of the experiment. An event is a subset of the sample space: it includes the outcomes that meet a specified condition. For a tree diagram, this means an event can be found by selecting the terminal paths that satisfy the condition.
How to Build and Read a Tree
Start at the first stage and draw one branch for each possible result. From the end of each branch, draw the branches for the possible results at the second stage. Continue until all stages have been represented. Label branches with the result of the stage they represent, and keep the order of stages consistent across the whole diagram.
The branches extending from a particular point should show the possible results at the next stage given the results already recorded along that path. If earlier results affect what can happen next, the branches can differ from one part of the tree to another. At the end, follow every complete path from start to finish and write its results in stage order. These ordered sequences, not the separate branch labels, are the outcomes.
State what happens first, second, and at each later stage.
At each point, include every possible result for the next stage, given the path so far.
Read each complete path from the first stage to the last as an ordered outcome.
Count all terminal paths for the total number of outcomes, or select the paths that meet an event’s condition.
Worked Example: Building a Tree for a Meal Choice
In a hypothetical lunch-ordering exercise, a student first chooses wheat or rye bread and then chooses cheese or turkey filling. Draw the tree, list the outcomes, and identify the event “wheat bread with turkey.”
The stages are bread first and filling second. Each bread choice has two possible fillings:
- Wheat
- Cheese: (wheat, cheese)
- Turkey: (wheat, turkey)
- Rye
- Cheese: (rye, cheese)
- Turkey: (rye, turkey)
Each outcome lists bread before filling because that is the order of the stages. The sample space is \(\Omega=\{(\text{wheat},\text{cheese}),(\text{wheat},\text{turkey}),(\text{rye},\text{cheese}),(\text{rye},\text{turkey})\}\). There are four terminal paths and therefore four outcomes. The specified event contains just the path (wheat, turkey), so it has one outcome.
This tree has two branches at the first stage, and each of those branches has two branches at the second stage. The multiplication rule for sequential choices gives the same total: \(2\cdot2=4\). The tree shows what those four combinations actually are, not only how many there are.
When Later Choices Depend on Earlier Results
A tree is especially useful when the possible choices change from one stage to the next. In a draw without replacement, for instance, an object selected at the first stage is no longer available at the second. The second-stage branches therefore depend on which first-stage branch was followed. A tree represents this changing set of possibilities directly.
Worked Example: Drawing Two Different Tokens
Suppose a hypothetical bag contains three tokens labelled A, B, and C. Two tokens are drawn one after another without replacement. Draw the tree, list the outcomes, and identify the event that the two-token sequence contains C.
At the first stage, any of A, B, or C can be drawn. At the second stage, the first token is no longer available:
- First A
- Then B: (A, B)
- Then C: (A, C)
- First B
- Then A: (B, A)
- Then C: (B, C)
- First C
- Then A: (C, A)
- Then B: (C, B)
There are six terminal paths and six ordered outcomes. The order matters: (A, C) and (C, A) describe different sequences, even though both sequences contain the same two token labels. The event “the two-token sequence contains C” consists of (A, C), (B, C), (C, A), and (C, B). It has four outcomes.
The tree also checks that no impossible path has been included: for example, there is no path (A, A), because the first A is not returned to the bag before the second draw. Branches must match the stated experiment, including how objects are handled between stages.
Counting Paths and Selecting Events
Once a tree is complete, counting its terminal paths gives the number of outcomes in the sample space. To count the outcomes in an event, use its condition to decide which terminal paths to include. This is a direct consequence of how the tree is constructed: every complete path corresponds to one outcome, and the full set of complete paths covers the possible outcomes.
Here is a short justification. Start with any possible outcome of the experiment. Its first-stage result identifies a branch from the start. Given that result, its second-stage result identifies the next branch, and the same reasoning applies at each later stage. Following those results produces a terminal path for that outcome. Conversely, a terminal path specifies one result at each stage, so it describes one complete outcome. Because the branches at each stage are labelled with distinct possible results, two different terminal paths differ at some stage and therefore describe different outcomes. Thus the paths and outcomes correspond one-to-one.
Worked Example: Three Successive Coin Tosses
Consider a hypothetical experiment consisting of three coin tosses. Write H for heads and T for tails. Draw the tree, count the outcomes, and find the outcomes in the event “exactly one toss is tails.”
The tree has two branches at each stage, labelled H and T. Its branches can be listed compactly by following each first result through the next two stages:
- First H
- Then H
- Then H: (H, H, H)
- Then T: (H, H, T)
- Then T
- Then H: (H, T, H)
- Then T: (H, T, T)
- Then H
- First T
- Then H
- Then H: (T, H, H)
- Then T: (T, H, T)
- Then T
- Then H: (T, T, H)
- Then T: (T, T, T)
- Then H
The eight terminal paths give the sample space: \(\Omega=\{(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)\}\). For exactly one T, select (H, H, T), (H, T, H), and (T, H, H). There are three outcomes in this event. The count of all outcomes is \(2\cdot2\cdot2=8\), consistent with the tree.
This is an outcome-counting exercise; the tree does not assign probabilities to its paths. If probabilities are needed, information about the experiment’s probability model must also be supplied. A tree with equally many branches at each stage does not, by itself, establish that all terminal outcomes are equally likely.
Common Mistakes and Why Trees Matter
Trees are useful not only for finding a total count but also for checking whether an outcome list is complete and whether an event has been interpreted correctly. They make the order of stages explicit and show where possible results change as the experiment proceeds.
- Counting branches instead of complete paths. A branch represents one stage’s result, not a complete outcome. Count the endpoints reached after all stages.
- Ignoring order. In a sequence of draws, (A, C) and (C, A) are different outcomes when the first and second stages are distinct. Keep the stage order in each outcome.
- Using the same branches at every point. If the experiment is without replacement, some second-stage options disappear. Label the branches that are actually possible after each first result.
- Leaving out a branch or adding an impossible one. Check the description of the experiment at every stage. Include all allowed results and exclude results that cannot occur.
- Assuming a count gives probabilities. A tree lists possible outcomes. To treat outcomes as equally likely or calculate probabilities, the experiment must provide the relevant probability information.
A common mistaken argument is: “There are eight paths in the three-toss tree, so every path has probability \(1/8\).” The tree justifies the count of eight, but not that probability claim. Equal probabilities require additional assumptions about the tosses, such as each toss being fair and the tosses being independent. Keeping outcome listing separate from probability assignment prevents a correct tree from being used to support an unsupported conclusion.
Check Your Understanding
For each question, focus on the branches and complete paths rather than on probabilities.
- A student chooses one of three notebooks and then one of two pens. How many terminal paths should a complete tree have? What does one path represent?
- A bag contains tokens X, Y, and Z. Two are drawn without replacement. Why should the second-stage branches after drawing X differ from those after drawing Y?
- For the three-toss experiment, list the outcomes in the event “the first toss is T.”
- Explain why (B, A) and (A, B) are distinct outcomes when two labelled objects are drawn in sequence.
- Does a tree with four terminal paths prove that each path has probability \(1/4\)? Explain what additional information would be needed.