GCSE Maths Practice: tree-diagrams

Higher Tree Diagrams: Multi-Stage Events and Changing Probabilities

Tree diagrams become especially useful at Higher GCSE when a probability problem involves several stages and the probabilities change as you move through the experiment. The big difference from simpler questions is that you cannot keep using the same fraction at each step. Instead, you must update the numerator and denominator to match what is still possible after earlier outcomes.

A common Higher situation is “without replacement”. This means once an outcome happens, it cannot happen again, so the sample space becomes smaller. In a tree diagram, this is shown by the denominators decreasing step by step (for example, 6 then 5 then 4…), while the numerators change only when you remove something that affects the outcome you are trying to get. This is exactly why tree diagrams help: each branch represents a different story, so you always know what has been removed and what remains.

Another Higher feature is that problems may combine different experiments in one question (for example, a coin flip followed by a second process). Even if the first event is independent (like a fair coin), once you move into a “without replacement” stage, the later events become dependent. The tree keeps this organised: you place the coin outcomes first, then extend only the branches that lead into the next stage. In many exam solutions, you will see students draw the full tree lightly, then focus only on the relevant route(s) needed for the event.

When answering a tree-diagram question, there are two main skills to practise:

• Multiplying along a path: If the question asks for one specific sequence (A then B then C), you multiply the probabilities on that route.
• Adding different paths: If the question allows more than one sequence (for example, “a 5 and a 6 in any order”), you calculate each valid route and add them.

Worked Example (Different to the Question)

A bag contains number cards 1 to 5. Two cards are taken without replacement. Find the probability of getting an even number then an odd number.

First draw: P(even) = 2/5 (cards 2 and 4). If an even is removed, 4 cards remain. The number of odd cards could be 3 or 2 depending on which even was taken, so a full solution would either split into cases or list the specific cards. This shows why Higher questions sometimes require careful structure: not every “even/odd” situation stays simple after removal.

Common Mistakes to Avoid

• Forgetting to reduce the denominator after a draw without replacement (e.g., writing 1/6 then 1/6 again).
• Changing the numerator when it should stay the same (e.g., if you removed a 5, the chance of getting 6 may still be 1 out of the remaining total).
• Mixing up “then” with “in any order” (the second usually needs multiple paths added together).

Tree diagrams are not just drawings — they are a way to keep track of changing sample spaces. If you write the correct fractions at each stage, the final probability is often just straightforward multiplying (and sometimes adding) at the end.