GCSE Maths Practice: tree-diagrams

Question 9 of 11

A box contains 6 red, 5 green and 4 yellow prize tokens. Three tokens are drawn one after another without replacement.

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\( \begin{array}{l}\textbf{A box contains 6 red, 5 green and 4 yellow prize tokens.}\\ \text{Three tokens are drawn one after another without replacement.}\\ \text{Find the probability that the first token is yellow, the second is red and the third is green.}\\ \text{(You may use a tree diagram.)} \end{array} \)

Diagram

Choose one option:

For Higher questions, extend the tree to three stages. Multiply along the Yellow → Red → Green path and reduce totals each time.

Higher Tree Diagrams: Three-Stage Probability Without Replacement

This question is GCSE Higher because it involves three dependent stages. Unlike simpler two-step questions, you must update the total number of items twice, and you must keep track of which colours have changed after each draw. A tree diagram helps organise this clearly, but it must be built carefully.

Why the Probabilities Change

The draws are without replacement. That means after each draw:

  • The total number of tokens decreases by 1.
  • The number of the colour drawn decreases by 1.
  • All other colours stay the same.

These are dependent events because the second and third probabilities depend on what happened earlier.

Tree Diagram Structure

For Higher tier, it is often necessary to show more branches than at Foundation. A suitable approach is:

  • Stage 1: split into Yellow / Not Yellow.
  • From Yellow: Stage 2 split into Red / Not Red.
  • From Yellow then Red: Stage 3 split into Green / Not Green.

This creates a clear three-level tree focused on the required path, without drawing every possible outcome.

Multiplying Along the Required Path

Once the tree is labelled, the probability of a specific sequence is found by multiplying the probabilities along that path. For example, the path “Yellow then Red then Green” is:

P(Y then R then G) = P(Y) × P(R|Y) × P(G|Y and R)

Worked Example (Different Numbers)

A jar contains 3 red, 2 green and 1 yellow counter. Three counters are drawn without replacement. Find the probability of drawing Yellow then Red then Green.

  • P(Y) = 1/6
  • After Y: P(R|Y) = 3/5
  • After Y then R: P(G|Y,R) = 2/4
  • Multiply: 1/6 × 3/5 × 2/4

Common Higher-Tier Mistakes

  • Forgetting to change the denominator twice: the totals should go 15, 14, 13.
  • Reducing the wrong numerator: only the drawn colour reduces, others stay the same.
  • Stopping after two stages: Higher questions often require an extra step.
  • Arithmetic slips when simplifying: multiply first, then simplify carefully.

Exam Tip

If a question involves three dependent selections, a tree diagram is one of the safest methods. Label the totals at each stage before you write any fractions. This prevents the most common errors.

Study tip: For three-stage trees, write the totals under each level (15 → 14 → 13) to keep your denominators correct.

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