A probability tree diagram shows all possible outcomes of two or more events and their probabilities. To find the probability of a combined outcome, multiply along the branches. To find the probability of two or more different combined outcomes, add the results.
The two golden rules
- Multiply along the branches — to find the probability of one specific sequence of outcomes.
- Add between branches — to find the probability of multiple acceptable outcomes.
These two rules handle almost every tree diagram question at KS3 and GCSE.
How to draw a probability tree diagram — step by step
Step 1 — identify the events and their outcomes
Each event becomes one "level" of the tree. List all possible outcomes at each level.
Step 2 — draw the first set of branches
Draw a dot (starting point), then draw one branch for each outcome of the first event. Label each branch with the outcome and its probability.
Step 3 — draw the second set of branches
From the end of each first-level branch, draw branches for each outcome of the second event. Label them with the correct probabilities.
Step 4 — check each set of branches sums to 1
The probabilities on each group of branches from one node must add to 1.
Step 5 — calculate the combined probabilities
Multiply along each complete path for its probability. The total of all path probabilities must equal 1.
Worked example 1 — two independent events (coins)
A fair coin is flipped twice. Draw a tree diagram and find P(two heads).
First flip: P(H) = 1/2, P(T) = 1/2
Second flip (independent — same probabilities each time): P(H) = 1/2, P(T) = 1/2
| Path | Calculation | Probability |
|---|---|---|
| H then H | 1/2 × 1/2 | 1/4 |
| H then T | 1/2 × 1/2 | 1/4 |
| T then H | 1/2 × 1/2 | 1/4 |
| T then T | 1/2 × 1/2 | 1/4 |
| Total | 4/4 = 1 ✓ |
P(two heads) = 1/4
Worked example 2 — two independent events (bag of balls, with replacement)
A bag contains 3 red balls and 7 blue balls. A ball is drawn, its colour noted, and it is replaced. A second ball is then drawn.
P(red) = 3/10, P(blue) = 7/10 — same on both draws (because the ball is replaced).
Find P(both the same colour):
P(red then red) = 3/10 × 3/10 = 9/100
P(blue then blue) = 7/10 × 7/10 = 49/100
P(both the same colour) = 9/100 + 49/100 = 58/100 = 29/50
Worked example 3 — without replacement
The same bag (3 red, 7 blue, 10 total). This time the ball is NOT replaced. Find P(one ball of each colour).
First draw: P(red) = 3/10, P(blue) = 7/10
Second draw — changes because one ball is removed:
| Path | Second draw | Calculation | Probability |
|---|---|---|---|
| Red then Blue | P(blue after red) = 7/9 | 3/10 × 7/9 | 21/90 |
| Blue then Red | P(red after blue) = 3/9 | 7/10 × 3/9 | 21/90 |
P(one of each colour) = 21/90 + 21/90 = 42/90 = 7/15
Notice: after drawing a red ball without replacement, only 9 balls remain and still 7 are blue. After drawing a blue ball, only 9 remain and still 3 are red.
With replacement vs without replacement
| Feature | With replacement | Without replacement |
|---|---|---|
| Total after first draw | Same (n) | Reduced by 1 (n − 1) |
| Second-draw probabilities | Same as first | Change depending on first outcome |
| Are events independent? | Yes | No |
Common mistakes
| Mistake | Consequence | Fix |
|---|---|---|
| Adding along branches instead of multiplying | Probabilities exceed 1 | Multiply first, then add between paths |
| Forgetting probabilities change without replacement | Wrong answer | Adjust the denominator for the second draw |
| Branch probabilities not summing to 1 at each node | Error in setup | Re-check each node sums to exactly 1 |
| Missing a path in the final addition | Incomplete answer | List ALL acceptable paths before adding |
Tree diagrams in the national curriculum
The DfE's KS3 mathematics programme of study requires pupils to enumerate sets using systematic methods including tree diagrams, and to calculate the probability of independent and combined events. The AQA GCSE Mathematics specification confirms that both with-replacement and without-replacement scenarios are assessed at KS3 and GCSE maths.
Frequently asked questions
When do I add probabilities on a tree diagram?
You add the probabilities of separate paths that all give the outcome you want. For example, if you want "at least one head" from two coin flips, there are three paths that achieve this (HH, HT, TH). Multiply along each path, then add the three results.
How do I know if two events are independent?
Two events are independent if the outcome of the first does not affect the probabilities of the second. Drawing a ball with replacement gives independent events (the bag resets). Drawing without replacement gives dependent events — what you drew first changes what remains.
What should all my tree diagram probabilities add to?
The probabilities of all end-of-tree paths (complete left-to-right routes) must add to exactly 1. This is a quick check: if your paths sum to something other than 1, you have made an arithmetic or setup error somewhere.
Can I use a tree diagram for three events?
Yes. Add a third level of branches, one for each outcome of the third event. Multiply all three branch probabilities along each complete path. The number of paths grows quickly (2 × 2 × 2 = 8 paths for three fair coins), so list them systematically to avoid missing any.
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