Probability tree diagrams, explained with examples

By The STEM-Tutor team · Updated October 2026

A tree diagram organises a problem with two or more events, so you can see every possible outcome and its probability. It is one of the most reliable ways to answer GCSE probability questions.

How a tree diagram works

Each set of branches represents one event, such as one coin flip or one counter taken from a bag. Each branch is labelled with an outcome and its probability. The probabilities on branches from the same point must add up to 1, because one of those outcomes has to happen.

Two rules do almost all the work:

• Multiply along the branches to find the probability of a combined outcome, such as red then blue.
• Add the end results when more than one combined outcome answers the question, such as “one of each colour”.

A quick check: the probabilities at the ends of all the paths must add to 1. If they do not, a branch is missing or a probability is wrong, so find the error before going on.

Independent events: with replacement

If the first event does not affect the second, the events are independent, and the second set of branches has the same probabilities each time. Taking a counter, putting it back, and taking another is an example.

Example 1: with replacement

A bag has 3 red and 2 blue counters. A counter is taken, replaced, and a second counter is taken. Find P(both red) and P(one of each colour).

• Each draw: P(red) = 3/5, P(blue) = 2/5.
• P(red, red) = 3/5 × 3/5 = 9/25.
• One of each: P(red, blue) + P(blue, red) = 6/25 + 6/25 = 12/25.

Dependent events: without replacement

If the counter is not replaced, the bag changes, so the second set of branches has different probabilities. Both the number of counters of that colour and the total go down by one.

Example 2: without replacement

Using the same bag, two counters are taken without replacement. Find P(both the same colour).

• P(red, red) = 3/5 × 2/4 = 6/20. P(blue, blue) = 2/5 × 1/4 = 2/20.
• P(same colour) = 6/20 + 2/20 = 8/20 = 2/5.

Example 3: at least one

Using the without-replacement tree, find P(at least one blue).

• The only outcome without blue is red, red, with probability 6/20.
• P(at least one blue) = 1 − 6/20 = 14/20 = 7/10. “At least one” questions are quickest using 1 minus the opposite.

Example 4: a real context

The probability it rains on Monday is 0.3. If it rains on Monday, the probability it rains on Tuesday is 0.6; otherwise it is 0.2. Find the probability it rains on exactly one of the two days.

• Rain then dry: 0.3 × 0.4 = 0.12. Dry then rain: 0.7 × 0.2 = 0.14.
• P(exactly one) = 0.12 + 0.14 = 0.26.

Example 5: three stages

A fair coin is flipped three times. Find P(exactly two heads).

• Each path has probability (1/2)³ = 1/8. Paths with exactly two heads: HHT, HTH, THH.
• P = 3 × 1/8 = 3/8.

Conditional probability (Higher)

Conditional probability is the probability of an event given that another has happened. On a tree, the second set of branches already shows conditional probabilities: in Example 4, 0.6 is P(rain Tuesday given rain Monday). Harder questions turn this round.

Example 6: working backwards

Using Example 4, given that it rained on Tuesday, what is the probability it also rained on Monday?

• P(rain Tuesday) = 0.3 × 0.6 + 0.7 × 0.2 = 0.18 + 0.14 = 0.32.
• P(rain Monday given rain Tuesday) = 0.18/0.32 = 9/16, about 0.56.

Example 7: forming an equation

A bag has n counters, of which 4 are red. Two are taken without replacement. P(both red) = 1/5. Show that n² − n − 60 = 0 and find n.

• P(both red) = (4/n) × (3/(n − 1)) = 12/(n(n − 1)) = 1/5.
• So n(n − 1) = 60, giving n² − n − 60 = 0, which factorises to (n − 8)(n + 7) = 0.
• n = 8, since n must be positive. This algebra-and-probability style is a common grade 8 or 9 question.

Frequency trees

A frequency tree looks like a probability tree but shows numbers of people or items instead of probabilities. They often appear on Foundation papers.

Example 8: a frequency tree

Of 80 students, 35 are in Year 10 and the rest in Year 11. 20 of the Year 10 students and 30 of the Year 11 students walk to school. A student is chosen at random. Find the probability they are in Year 11 and do not walk.

• Year 11: 80 − 35 = 45 students. Year 11 not walking: 45 − 30 = 15.
• P = 15/80 = 3/16. Fill in every number on the frequency tree first; the probability then comes straight from the counts.

Tips for drawing trees

• Use fractions unless the question uses decimals; they are exact and easier to check.
• Do not simplify branch fractions without replacement; keeping the same denominator makes adding easier.
• Label every branch with both the outcome and the probability.
• Write the result of each path at its end before answering the question.

Common mistakes

• Adding along branches instead of multiplying.
• Forgetting to reduce the total, and the count of the colour taken, when there is no replacement.
• Missing one of the paths, such as including red-blue but not blue-red.
• Branch probabilities from one point that do not add up to 1.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Are tree diagrams on the Foundation paper?

Yes. Tree diagrams with replacement are common on Foundation papers. Without-replacement and conditional probability questions are mostly Higher tier.

When should I use a Venn diagram instead?

Use a Venn diagram when events overlap at the same time, such as students who study French and Spanish. Use a tree diagram for events that happen in stages, one after the other.

Key takeaways

• Multiply along branches; add the end results you need.
• Branches from one point add to 1.
• Without replacement, change both the count and the total.
• Use 1 minus the opposite for “at least one”.
• Conditional probability: divide the joint probability by the probability of the given event.

Related guides

• GCSE Maths topics
• Factorising quadratics
• GCSE Maths revision