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Tree Diagrams






When Probability Branches


Not all random situations happen in a single step.
In many problems, uncertainty unfolds sequentially, with each outcome opening the door to new possibilities.

Tree diagrams provide a visual way to represent this kind of staged randomness.
They arrange possible outcomes as branches, making it clear how one step leads to the next and how complete outcomes are formed along paths.

By laying out possibilities step by step, tree diagrams turn conditional probability into something visible.
They help track how probabilities change as information accumulates and make complex multi-stage situations easier to reason about.

Key Terms

Probability Tree— a branching diagram for sequential random processes
Conditional Probability— each branch label is a conditional probability
Event— each node in the tree represents an event
Mutual Exclusiveness— branches at each node are mutually exclusive

See All Probability Definitions →


What a Tree Diagram Represents

Some random processes happen in stages, and the later stages depend on the earlier ones.
A tree diagram represents a random process that unfolds in stages.
Say event AA happens with probability 0.60.6. Whether BB then happens depends on how the first stage went: probability 0.70.7 after AA, but only 0.30.3 when AA did not happen. A flat list of outcomes would lose that dependence.
stagepossible resultsprobabilities
1: does A happen?A, or not A0.6 and 0.4
2, after A: does B happen?B, or not B0.7 and 0.3
2, after not A: does B happen?B, or not B0.3 and 0.7
Drawing each stage as a fork, and multiplying along each route, gives every full outcome.
Start with a fork for stage one, AA or not AA. From each end, fork again for stage two with that stage's own probabilities. Each route from the start to an end is one complete outcome, and its probability is the product of the numbers along it.
pathmultiply along itprobability
A, then B0.6 × 0.70.42
A, then not B0.6 × 0.30.18
not A, then B0.4 × 0.30.12
not A, then not B0.4 × 0.70.28
all four paths1.00
BB happens on two of the paths, so P(B)=0.42+0.12=0.54P(B) = 0.42 + 0.12 = 0.54 — read straight off the routes.
P(A) = 0.600P(Aᶜ) = 0.400P(B|A) = 0.700P(Bᶜ|A) = 0.300P(B|Aᶜ) = 0.300P(Bᶜ|Aᶜ) = 0.700AP = 0.6000AᶜP = 0.4000A ∩ BP = 0.4200A ∩ BᶜP = 0.1800Aᶜ ∩ BP = 0.1200Aᶜ ∩ BᶜP = 0.2800
The same process as a tree: one fork for AA, then a fork for BB under each branch, with each end labelled by its path probability. The same state is on the conditional probability tree.
Nothing in those numbers was chosen: each route's probability was forced by the branches it follows.
Each stage corresponds to a point where several outcomes are possible.
From that point, the process branches, and each branch represents one possible result of the next step.
The second-stage numbers are conditional — 0.70.7 is the chance of BB given that AA already happened — so multiplying them along a route is exactly the multiplication rule. And because each fork's numbers add to 11, the four routes together add to 11 as well.
The definition
A tree diagram is a picture of a random process in stages: each fork shows the possible results of one stage, each branch carries the probability of its result given the path so far, and each route from the start to an end is one complete outcome.
Here: two stages, four routes, probabilities 0.420.42, 0.180.18, 0.120.12 and 0.280.28.
A tree diagram is a way of organizing a probability problem, not a new rule.
A complete path from the start of the tree to a final node represents one full sequence of outcomes.
The tree does not define randomness on its own — it organizes an already defined situation so that sequential outcomes and their relationships are explicit.
Forking and multiplying generalize; two things grow or go wrong as trees get bigger.
What generalizes: any number of stages and any number of results per stage. What was particular: two stages with two results each, so only four routes. Each extra stage multiplies the number of routes, so trees for long processes get large quickly. And every branch after the first carries a conditional probability; copying a first-stage number onto a later branch, as if the stages were independent, is the commonest error — the common mistakes section shows it.
The rest of the page covers the parts of a tree, how probability values are placed on it, its link to conditional probability and to the law of total probability, how to read paths, and the common mistakes.

Components of a Tree Diagram


    A tree diagram is built from a small number of structural elements, each with a specific role.

  • Root
    The starting point of the process, before any outcomes have occurred.

  • Branches
    Lines extending from a node, each representing a possible outcome at a given stage.

  • Nodes
    Points where branches meet or split, representing intermediate states of the process.

  • Paths
    Sequences of connected branches from the root to a terminal node, representing complete outcome sequences.

  • These components work together to make the order of events and the structure of sequential randomness explicit.

    Fact: A best-of-seven tree has 70 leaves
    A path stops as soon as one side has 4 wins, so the tree has 70 leaves, not 27=1282^7 = 128. With evenly matched sides, ending in game 6 and in game 7 are equally likely, 516\frac{5}{16} each: at 33–22, game 6 either ends it or makes it 33–33.

    Not every path has to reach the same depth. When a process stops as soon as its result is settled, a node can be terminal at an early stage while other branches keep splitting. A match played as a best-of-seven series is a typical case: some paths end after four games, others run to the seventh. The leaves of such a tree sit at different stages, and each one is still a complete outcome sequence.

Probability Values on a Tree Diagram


Probabilities in a tree diagram are assigned to the branches, not to the paths directly.

Each branch probability represents the likelihood of an outcome given that the process has reached the corresponding node. In this way, branch probabilities are conditional by nature.

The probability of a complete path is obtained by following the path from the root and combining the probabilities along its branches. This reflects how uncertainty accumulates across successive stages of the process.

Pitfall: Adding along a path
Adding the branch values on a path, for two draws without replacement from 3 red and 2 blue balls, gives P(R1∩R2)=35+24=1110P(R_1 \cap R_2) = \frac{3}{5} + \frac{2}{4} = \frac{11}{10}, more than 1. Along a path the values multiply: 35⋅24=310\frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}.

Tree diagrams therefore make it explicit how local, step-by-step probabilities combine to produce probabilities of full outcome sequences.

Tree Diagrams and Conditional Probability


Tree diagrams make conditional probability explicit by construction.

Each branch represents the probability of an outcome given that the process has reached a certain stage. Moving along a branch means accepting the condition imposed by all previous outcomes on the path.

Pitfall: Treating later branches as unconditional
Copying P(R1)=35P(R_1) = \frac{3}{5} onto the second-draw branch, for draws without replacement from 3 red and 2 blue balls, gives P(R1∩R2)=35⋅35=925P(R_1 \cap R_2) = \frac{3}{5} \cdot \frac{3}{5} = \frac{9}{25}. Once a red is gone the branch is P(R2∣R1)=24P(R_2 \mid R_1) = \frac{2}{4}, so the path is 310\frac{3}{10}.

Because of this, conditional probability is not an extra concept added to the diagram — it is already built into how the diagram is read. Probabilities are interpreted step by step, with each stage conditioning on what has happened before.

This is why tree diagrams are especially useful for reasoning about sequential experiments, updating information, and situations where later outcomes depend on earlier ones.
P(A) = 0.600P(Aᶜ) = 0.400P(B|A) = 0.700P(Bᶜ|A) = 0.300P(B|Aᶜ) = 0.300P(Bᶜ|Aᶜ) = 0.700AP = 0.6000AᶜP = 0.4000A ∩ BP = 0.4200A ∩ BᶜP = 0.1800Aᶜ ∩ BP = 0.1200Aᶜ ∩ BᶜP = 0.2800
The subtree below A: branch weights are conditional probabilities

Every branch beyond the root carries a conditional probability: the two branches below A are P(B given A) and its complement, and they add to 1 because they exhaust what can happen once A has occurred. The tree builds conditioning into its structure. Change P(B given A) and watch only the subtree respond on the conditional probability tree diagram.

The same structure carries the law of total probability.

Tree Diagrams and the Law of Total Probability


Tree diagrams provide a direct visual interpretation of the Law of Total Probability.

Each first-level branch of a tree represents a distinct case that cannot occur together with the others. These branches form a partition of all possible outcomes at that stage of the process.

RULELaw of Total Probability
When a later event can occur through several different branches, its overall probability is obtained by accounting for all paths that lead to it. Each path contributes according to the probability values along that path, and the total probability is obtained by combining these contributions.

In this way, the law of total probability is not an abstract rule added afterward.
It is read directly from the structure of the tree: split the process into disjoint cases, follow the branches, and combine their contributions — the multiply-along-a-path, add-across-paths reading that total probability notation writes as marks.
P(A1) = 0.330P(A2) = 0.330P(A3) = 0.340P(B1|A1) = 0.400P(B2|A1) = 0.300P(B3|A1) = 0.300P(B1|A2) = 0.300P(B2|A2) = 0.500P(B3|A2) = 0.200P(B1|A3) = 0.200P(B2|A3) = 0.300P(B3|A3) = 0.500A1P = 0.330A1 ∩ B1 = 0.132A1 ∩ B2 = 0.099A1 ∩ B3 = 0.099A2P = 0.330A2 ∩ B1 = 0.099A2 ∩ B2 = 0.165A2 ∩ B3 = 0.066A3P = 0.340A3 ∩ B1 = 0.068A3 ∩ B2 = 0.102A3 ∩ B3 = 0.170
One outcome highlighted across every branch

The first-level branches are the disjoint cases of the law of total probability, and the highlighted outcome appears once in each; its overall probability is the sum of the path products leading to it. The tree makes the partition and the sum visible at once. Highlight an outcome and add its paths on the law of total probability visualizer.

Computing with paths is the subject of the next section.

Using Tree Diagrams to Compute Probabilities


Tree diagrams provide a clear method for computing probabilities in multi-stage situations.

Each complete path through the diagram represents one possible sequence of outcomes. The probability of such a sequence is obtained by following the path and combining the probability values along its branches.

Pitfall: Counting only one order
Taking only the path red then blue for one ball of each colour, drawn from 3 red and 2 blue, gives 35⋅24=310\frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}. Blue then red is a second path, so the probability is 310+25⋅34=35\frac{3}{10} + \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{5}.

When a question involves several possible sequences, the corresponding path probabilities are combined to obtain the final result. In this way, tree diagrams turn complex probability questions into structured path-based calculations.

This approach is especially helpful when outcomes depend on earlier stages and when direct formulas are difficult to apply.
P(A) = 0.600P(Aᶜ) = 0.400P(B|A) = 0.700P(Bᶜ|A) = 0.300P(B|Aᶜ) = 0.300P(Bᶜ|Aᶜ) = 0.700AP = 0.6000AᶜP = 0.4000A ∩ BP = 0.4200A ∩ BᶜP = 0.1800Aᶜ ∩ BP = 0.1200Aᶜ ∩ BᶜP = 0.2800
One complete path: multiply along the branches

A complete path is one sequence of outcomes, and its probability is the product of the branch weights along it, here P(A) times P(B given A). Adding the products of several paths gives the probability of any event made of those sequences. Follow other paths and multiply on the conditional probability tree diagram.

Bayes theorem reads the same tree backwards.
Multiply along a path for one sequence, add across paths for an event: every tree calculation is these two moves.

Tree Diagrams and Bayes' Theorem


Tree diagrams provide a natural visual setting for understanding Bayes' theorem.

A tree can be read in one direction to represent how probabilities are assigned before any information is observed. Once an outcome at a later stage is known, the same tree can be used to reason backward, focusing only on the paths consistent with the observed information.

Pitfall: Reading the branch in the wrong direction
Answering P(R1∣B2)P(R_1 \mid B_2), first ball red given second blue, with the branch P(B2∣R1)=24P(B_2 \mid R_1) = \frac{2}{4} gives 12\frac{1}{2} (3 red, 2 blue, no replacement). The paths ending in blue carry 310\frac{3}{10} and 110\frac{1}{10}, so P(R1∣B2)=310÷410=34P(R_1 \mid B_2) = \frac{3}{10} \div \frac{4}{10} = \frac{3}{4}.

By restricting attention to these paths and renormalizing their probability values, the diagram shows how probabilities are updated in light of new evidence. This makes the logic of Bayes' theorem visible without relying solely on algebraic formulas.

Tree diagrams therefore serve as an intuitive bridge between conditional probability and Bayesian updating.

Tree Diagrams vs Other Representations


Tree diagrams are one of several ways to represent probabilistic situations.

Fact: A poker hand sits on 120 paths
Dealing 5 cards one at a time gives a tree with 52⋅51⋅50⋅49⋅48=311,875,20052 \cdot 51 \cdot 50 \cdot 49 \cdot 48 = 311{,}875{,}200 paths, yet there are only 2,598,9602{,}598{,}960 different hands. The tree records order, so each hand appears on 5!=1205! = 120 paths.

Compared to tables, tree diagrams emphasize order and sequence, making them better suited for problems where outcomes occur in stages. Compared to formulas, they highlight structure and dependencies rather than algebraic relationships.

However, tree diagrams are not always the best choice. As the number of stages or possible outcomes grows, diagrams can become large and difficult to read. In such cases, tabular or formula-based methods may be more efficient.

Tree diagrams are most effective when the process is sequential and the number of stages is small enough to be visualized clearly.
Aspect Tree diagrams Tables Formulas
Emphasis order and sequence parallel outcomes and joint counts algebraic relationships
Strength makes conditional reasoning explicit compact view of joint distributions works at any scale, including abstract problems
Best for sequential processes, few stages, dependencies between stages counting outcomes, joint events, two-variable summaries large problems, abstract reasoning, general rules
Limitation clutters with many stages or outcomes hides order; obscures conditional structure less visual; harder to spot structure at a glance

Common Mistakes and Misinterpretations


Tree diagrams are simple in structure, but they are often used incorrectly.

Typical mistakes include:
• treating branch probabilities as unconditional rather than conditional
• assigning probabilities to paths directly instead of to branches
• forgetting that branch probabilities at a node must sum to 1
• double-counting outcomes by summing overlapping paths
• mixing up stages of the process with events of interest

Another common error is reading the tree in the wrong direction after information is observed. Once an outcome at a later stage is known, only the paths consistent with that outcome should be considered.

Being careful about what each branch represents, and what information is being conditioned on at each stage, prevents most misunderstandings.

Adding along a path

The probability of a complete path is found by adding the branch values on it instead of multiplying them. Each later branch is a fraction of what is left after the earlier ones, and adding ignores that.

Wrong working: two balls are drawn without replacement from a bag of 3 red and 2 blue; the path is red, then red.
P(R1∩R2)=35+24=1110P(R_1 \cap R_2) = \frac{3}{5} + \frac{2}{4} = \frac{11}{10}


Right working: the branch values along the path multiply.
P(R1∩R2)=P(R1)⋅P(R2∣R1)=35⋅24=620=310P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{3}{5} \cdot \frac{2}{4} = \frac{6}{20} = \frac{3}{10}


Why it happens: a tree uses two operations, multiplying along a path and adding across paths, and the two are easily swapped.

How to check: a path probability is never larger than any branch value on it, since each step keeps only a fraction of what came before; 1110\frac{11}{10} is not even a probability. The four paths of this tree must also add to 11: 310+310+310+110=1\frac{3}{10} + \frac{3}{10} + \frac{3}{10} + \frac{1}{10} = 1.

Counting only one order

An event that can happen in more than one order is given the probability of a single path. "One red and one blue" is reached by red then blue and by blue then red, which are two different paths of the tree.

Wrong working: from the same bag of 3 red and 2 blue, two draws without replacement,
P(one of each)=P(R1∩B2)=35⋅24=310P(\text{one of each}) = P(R_1 \cap B_2) = \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}


Right working: both paths are added.
P(one of each)=35⋅24+25⋅34=310+310=35P(\text{one of each}) = \frac{3}{5} \cdot \frac{2}{4} + \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}


Why it happens: the question names colours, not an order, so only one path comes to mind; the tree records the draws in order and keeps the two orders apart.

How to check: list every leaf that matches the event before computing. The three events "both red", "both blue" and "one of each" must add to 11: 310+110+35=1\frac{3}{10} + \frac{1}{10} + \frac{3}{5} = 1.
Mistake What is actually correct
Treating branch probabilities as unconditional each branch label is conditional, given the path that reaches it
Assigning probabilities to paths directly probabilities live on branches; a path's probability is the product of its branch values
Forgetting branches at a node sum to 1 the outcomes leaving any single node form a partition and must sum to 1
Double-counting overlapping paths distinct paths are mutually exclusive; sum their probabilities without overlap
Mixing up stages with events of interest a stage is structural (where the tree splits); an event is the question being asked about outcomes
Reading the tree forward after observation once a later outcome is observed, restrict to the paths consistent with it and renormalize

When Tree Diagrams Are Most Useful


Tree diagrams are most effective when a random situation unfolds in a small number of clearly ordered stages.

They work especially well when:
• outcomes occur sequentially rather than all at once
• later possibilities depend on earlier outcomes
• conditional probabilities are central to the problem
• the number of branches remains manageable

In contrast, tree diagrams become less practical as the number of stages or possible outcomes grows. In such cases, the visual structure can become cluttered, and alternative representations such as tables or formulas may be more efficient.

Tree diagrams are therefore best viewed as a tool for structured reasoning, not a universal solution for all probability problems.

Interactive Tools


• Probability Tree Visualizer
• Conditional Probability Explorer
• Bayes Tree Simulator

Summary


Tree diagrams provide a clear way to organize probability problems that unfold in stages.

They represent sequential randomness through branches and paths, making conditional relationships explicit and traceable. By following paths through the diagram, probabilities of complex outcomes can be computed in a structured and transparent way.

Tree diagrams connect naturally to conditional probability, the law of total probability, and Bayes' theorem, serving as a visual bridge between models, rules, and calculations.
Concept How the tree represents it
Joint probability of a sequence multiply the branch probabilities along the path from root to terminal node
Marginal probability of a later event sum the path probabilities over every path that ends in that event
Conditional probability each branch label is itself a conditional probability, given everything that came before
Law of total probability sum contributions across the disjoint first-level branches that partition the process
Bayes' theorem read backward — keep only the paths consistent with the observation and renormalize their probabilities
Best for sequential outcomes, few stages, dependencies between stages, central conditional probability
Less suitable when many stages, many outcomes per stage, parallel events without natural order

Tree Diagrams FAQ

Why do you multiply along a path but add across paths?

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Two different rules are at work. Multiplying along a path applies the chain rule: each branch is conditional on everything before it, so the product gives the joint probability of that whole sequence. Adding across paths applies the law of total probability: distinct complete paths are disjoint outcomes, so their probabilities simply sum.Read more →

Are the numbers on a tree's branches conditional probabilities?

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Yes, from the second stage onward. A branch at stage two carries the probability of that outcome given everything upstream of it, not its standalone probability. Only the first stage's branches are unconditional. Reading later branches as plain probabilities is the most common tree error, and it quietly breaks every path product that follows.Read more →

Should the branches at a node add up to 1?

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Yes — every set of branches leaving a single node must sum to 1, because they enumerate all the ways that stage can turn out given how you got there. It is the fastest check on a tree you have just drawn. This applies node by node; the complete path probabilities also sum to 1, but that is a separate check.Read more →

How do you read a tree diagram after you learn the outcome?

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Keep only the paths consistent with what you observed, then compare them against each other. The branch probabilities were written for the forward direction, so you cannot read an answer straight off a branch; you divide the matching path's probability by the total of all surviving paths. That division is Bayes' theorem, done on the diagram.Read more →

Does every path in a tree diagram have to be the same length?

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No. A path can stop as soon as the result is settled while other branches keep splitting. In a best-of-seven series a path ends when one side reaches 4 wins, so the tree has 70 leaves rather than 27=1282^7 = 128, sitting after game 4, 5, 6 or 7. Each leaf is still one complete outcome; with evenly matched sides, the series ends in game 6 or in game 7 with the same probability, 516\frac{5}{16}.Read more →

Does a tree diagram count ordered or unordered outcomes?

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Ordered. Each stage is one step of the process, so a path records the order in which things happened, and different orders are different paths. Dealing 5 cards one at a time gives 52⋅51⋅50⋅49⋅48=311,875,20052 \cdot 51 \cdot 50 \cdot 49 \cdot 48 = 311{,}875{,}200 paths, yet there are only 2,598,9602{,}598{,}960 different hands: each hand sits on 5!=1205! = 120 paths. An unordered event, such as one red ball and one blue, adds every path that leads to it.Read more →