Math // Probability

Tree diagrams

Generate theoretical sample spaces for single and combined events using listing, tables, and tree diagrams, and calculate theoretical probabilities as the number of favourable outcomes divided by the total number of equally likely outcomes

Ages
12 to 14
Type
Method
Builds on
3 topics
Leads to
0 topics
Standards
4 codes

Quick check

Can the learner draw a tree diagram to map out all possible outcomes of a two-step probability situation — like flipping a coin and then rolling a die — and calculate the probability of any particular combination?

Ask this to see whether the topic is secure.

Topic // Mastery

What mastery looks like

What a learner can do once this topic is secure.
  • List all 36 outcomes when rolling two dice and find P(total = 7)
  • Draw a tree diagram for two successive events and multiply along branches for combined probabilities
  • Calculate the probability of a combined event using a sample space diagram and simplify the fraction

Topic // Map

Where this fits

What to learn first and what this topic opens up, with the reason for each link. Required links come first.
  • RequiredMust come first
  • HelpfulHelps, but not essential

Before this

3 topics
  • Experimental probabilityRequired

    Probability // Ages 11 to 13

    Theoretical probability builds on experience with experimental probability

  • Sets & Venn DiagramsRequired

    Probability // Ages 12 to 14

    Sample spaces for combined events require systematic enumeration skills

  • Multiplying fractions (age 10+)Helpful

    Fractions // Ages 10 to 11

    Calculating probabilities for combined events often requires multiplying fractions

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Tree diagrams

Ages 12 to 14 // Method

Leads to

0 topics

Nothing in the catalogue builds on this topic yet. Browse Probability

Topic // Research-informed

Teaching notes

From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.

The approach

Students build complete, equally-likely sample spaces (ordered lists or grids) for single and combined events, then compute theoretical probability as favourable ÷ total. They draw probability trees, multiply along a single path (AND), add across separate, mutually exclusive complete paths (OR), use the complement for 'at least one', and recompute second-level branches for without-replacement (dependent) events. The central hazard is assuming aggregated outcomes — such as the eleven dice totals — are equally likely, when it is the 36 ordered outcomes that are equally likely.

Common misconceptions

  • Assuming aggregated outcomes are equally likely — treating the eleven totals of two dice as each 1/11 (equiprobability bias)
  • Ignoring order when listing — counting 3+4 and 4+3 as one outcome, undercounting the sample space
  • Adding probabilities along a branch instead of multiplying (misapplying AND/OR)
  • Reusing the first-event probabilities on the second branch of a without-replacement (dependent) event

Useful representations

  • Sample-space grid / two-way table (e.g. 6×6 for two dice)
  • Systematic ordered listing of outcomes
  • Probability tree with values on branches; multiply along, add across
  • Reduced-denominator second branches for dependent events

Teaching practices

  • Build the complete sample space first (grid or list), respecting ORDER, so favourable/total rests on genuinely equally-likely outcomes
  • Confront the two-dice trap head-on: the eleven totals 2–12 are NOT equally likely; there are 36 equally-likely ordered outcomes
  • Teach the AND rule (multiply along a branch) and OR rule (add across paths) explicitly, and the complement shortcut for 'at least one'
  • For without-replacement problems, recompute the second-level branch probabilities from the reduced situation (dependent events)
  • Leave fractions uncancelled until the end so common denominators are available for adding paths

Activities to try

  1. 1

    The two-dice grid: why totals are not equally likely

    Why it works: The 36 ordered (die1, die2) outcomes are equally likely; the eleven totals 2–12 are NOT, because different totals arise from different numbers of ordered outcomes.

    1. On paper or a 6×6 grid, write the total in every one of the 36 cells for (first die, second die), keeping the two dice as an ordered pair so 3+4 and 4+3 are separate cells.
    2. Shade all cells where the total is 7 and count them (6 cells: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1), then shade cells where the total is 2 (1 cell: 1+1).
    3. State P(total = 7) = 6/36 = 1/6 and P(total = 2) = 1/36.
    4. Explain in a sentence why P(7) is NOT 1/11, referring to the 36 equally-likely ordered outcomes.
  2. 2

    Trees: multiply along a path, add across paths

    Why it works: Along one path, combined events use AND and you multiply; across separate complete paths that each satisfy the question, you use OR and add. These are different operations for different situations.

    1. Draw a probability tree for a coin flip followed by a die roll, writing the probability on each branch and checking each set of branches sums to 1.
    2. Find P(tail and 6) by multiplying along the single path: 1/2 × 1/6 = 1/12.
    3. Find a two-path 'OR' probability by computing each separate complete path's product, then adding the path products together (the paths are mutually exclusive, so adding is valid).
    4. Leave fractions uncancelled until the final step so paths share common denominators when added.
  3. 3

    Without replacement: the second branch changes

    Why it works: For dependent (without-replacement) events, the second draw's probabilities depend on the first outcome, so they must be recomputed from the reduced contents rather than reused.

    1. Set up a bag problem, e.g. 3 red and 2 blue counters, and draw a tree for taking two counters without replacing the first.
    2. Write the first-draw probabilities (3/5 red, 2/5 blue), then recompute each second-draw branch from the reduced situation — one fewer counter in total, and one fewer of whatever was drawn (e.g. after a red: 2/4 red, 2/4 blue).
    3. If an optional tool offers a 'replace / don't replace' switch, toggle it and watch the second-level branch probabilities change (with no replacement) or stay the same (with replacement) to see the dependency.
    4. Find a combined probability, e.g. P(two reds) = 3/5 × 2/4 = 6/20 = 3/10, multiplying along the path with the updated second branch.

Topic // Standards

Curriculum standards

Codes for the curriculum statements this topic maps to. Each publisher owns its standard text.
Common Core State Standards for Mathematics
7.SP.87.SP.8a7.SP.8b
The national curriculum in England: Key stages 1 and 2 framework document
KS3.Maths.Prob.4

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