How to Draw and Use Probability Tree Diagrams
A probability tree diagram is a clear way to map out two or more events and the chance of each outcome. This page shows you how to build a tree, label every branch with its probability, multiply along the branches to find combined outcomes, and add separate paths together. We cover both independent events (where the first outcome does not change the second) and dependent events (where it does), then you can test yourself in the practice room with unlimited randomised questions.
What is a tree diagram?
A tree diagram shows two or more events one after another. Each event splits into branches, and every branch is labelled with the probability of that outcome. The branches that leave any single point must add up to \(1\).
Tree diagrams make combined events easy to handle because they turn a wordy problem into a picture you can read off systematically.
The two rules
- Multiply along the branches. To find the probability of a full path (one outcome followed by another), multiply the probabilities on each branch you travel along.
- Add between the paths. If more than one path gives the result you want, work out each path then add the results together.
The key formulae
Independent and dependent events
Whether the branch probabilities change on the second event depends on the type of problem.
🔑 Independent events
The first outcome does not affect the second. This happens when an item is replaced, or when you use separate objects such as a coin and a die.
The second set of branches has the same probabilities as the first.
⚠ Dependent events
The first outcome does affect the second. This happens when an item is taken and not replaced.
Both the numerator and denominator usually change on the second set of branches, because there are fewer items left.
Reading a tree diagram
The diagram below shows a bag with counters, where one counter is drawn, then a second. Follow the top route for the first outcome, then either branch for the second.
Worked examples
💡 Example 1: independent events (with replacement)
A bag holds \(3\) red and \(2\) yellow counters. A counter is drawn, its colour noted, then it is replaced. A second counter is then drawn. Find the probability that both counters are red.
\(P(\text{red}) = \dfrac{3}{5}\) on each draw, since the counter is replaced.
\[P(\text{red, red}) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\]
What's happening?
Because the counter is replaced, the bag is identical on the second draw, so both branches read \(\tfrac{3}{5}\).
This is one path, so we multiply along it.
💡 Example 2: one of each, in any order
Using the same bag (with replacement), find the probability of drawing one red and one yellow counter, in any order.
\(P(\text{red, yellow}) = \dfrac{3}{5} \times \dfrac{2}{5} = \dfrac{6}{25}\)
\(P(\text{yellow, red}) = \dfrac{2}{5} \times \dfrac{3}{5} = \dfrac{6}{25}\)
\[\frac{6}{25} + \frac{6}{25} = \frac{12}{25}\]
What's happening?
"In any order" means two different paths give what we want.
We multiply along each path, then add the two paths together.
💡 Example 3: dependent events (without replacement)
A box holds \(4\) blue and \(6\) green pens. Two pens are taken at random without replacement. Find the probability that both are blue.
First draw: \(P(\text{blue}) = \dfrac{4}{10}\)
Second draw: one blue gone, one fewer pen, so \(P(\text{blue}) = \dfrac{3}{9}\)
\[\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\]
What's happening?
Without replacement, the second branch changes: both the top number (\(4 \to 3\)) and the bottom number (\(10 \to 9\)) drop by one.
💡 Example 4: the "at least one" shortcut
A fair coin is flipped twice. Find the probability of getting at least one head.
\(P(\text{no heads}) = P(\text{tail, tail}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\)
\[P(\text{at least one head}) = 1 - \frac{1}{4} = \frac{3}{4}\]
What's happening?
Listing every winning path is slow. The opposite of "at least one head" is "no heads at all", which is a single path.
Find that, then subtract from \(1\).
🔑 Key points
- Branches from one point add up to \(1\).
- Multiply along a path; add between paths.
- Replacement means independent: branches stay the same.
- No replacement means dependent: the second branches change.
- For "at least one", use \(1 - P(\text{none})\).
⚠ Common pitfalls
- Forgetting to reduce the denominator on the second draw when there is no replacement.
- Adding along a single path instead of multiplying.
- Missing a path in "either order" questions.
- Rounding probabilities early: keep them as fractions for accuracy.