Tree Diagrams — Multiply Along, Add Across

A step-by-step guide to with/without replacement and multi-step probability at GCSE.

GCSE Probability
Tree Diagrams — Multiply Along, Add Across – cover
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Introduction

Tree diagrams are a visual way to organise outcomes in probability questions. They are especially useful when events happen in sequence (first pick, second pick) and when we need to consider with replacement vs without replacement. At GCSE level, tree diagrams appear frequently in questions about picking balls from a bag, probabilities involving coins/dice drawn more than once, or choosing people/items from groups.

This tutorial will teach you how to build clear and accurate tree diagrams, read probabilities off branches, and calculate combined probabilities by multiplying along branches and adding across different paths.

Key Vocabulary

  • Outcome: A possible result (e.g. “Red”, “Heads”, “Even”).
  • Event: A set of outcomes (e.g. “pick a red ball”).
  • Independent events: One event does not affect the next (e.g. coin tosses).
  • Dependent events: One event affects the next (e.g. picking cards without replacement).
  • With replacement: After the first pick, you put the item back; the probabilities stay the same.
  • Without replacement: You do not put the item back; probabilities change on the next draw.
  • Branch probability: Probability written on a branch of the tree.
  • Path probability: Multiply along branches; add across different paths.
  • Complement: The “opposite” event (e.g. not red).

Core Ideas

  • Sum to 1 at each split: At every branching point, probabilities add to 1.
  • Multiply along a path: To find the probability of a sequence (e.g. Red then Blue), multiply branch probabilities along that path.
  • Add across different paths: If the desired outcome can happen in several ways, add the path probabilities.
  • Replacement changes independence: With replacement ⇒ probabilities stay the same ⇒ events are typically independent. Without replacement ⇒ probabilities change ⇒ events are dependent.
Exam Tip: Label branches clearly and keep fractions in simplest form. Check each split sums to 1.

Step-by-Step Method

  1. Decide splits: What are the outcomes for the first event? For the second?
  2. Draw the tree: Start from a point, draw branches for the first event with their probabilities.
  3. Extend branches: From each first branch, draw branches for the second event, with updated probabilities (especially without replacement).
  4. Mark target paths: Highlight the routes that represent the outcome the question asks for.
  5. Multiply along, add across: Multiply along each target path, then add those path probabilities if there are multiple routes.
  6. Check totals: All final outcome probabilities should add to 1.

Worked Examples

Example 1 (With Replacement, Foundation)

A bag contains 3 red and 2 blue counters. You pick one, replace it, then pick again. Find the probability of getting two reds.

On the first pick: \(P(R)=\tfrac{3}{5}\), \(P(B)=\tfrac{2}{5}\). With replacement, the second pick has the same probabilities. The path “R then R” has probability \(\tfrac{3}{5}\times \tfrac{3}{5}=\tfrac{9}{25}\). Answer: \(\tfrac{9}{25}\).

Example 2 (Without Replacement, Foundation)

A bag has 4 green and 1 yellow sweet. Two are taken without replacement. Find the probability of getting two greens.

First pick: \(P(G)=\tfrac{4}{5}\). If the first is green, remaining are 3 green out of 4 total, so \(P(G\text{ on 2nd} \mid G\text{ on 1st})=\tfrac{3}{4}\). Multiply along the path: \(\tfrac{4}{5}\times \tfrac{3}{4}=\tfrac{12}{20}=\tfrac{3}{5}\). Answer: \(\tfrac{3}{5}\).

Example 3 (Either Order, Add Across, Foundation)

A box has 2 red and 3 blue pens. Two pens are selected without replacement. Find the probability of one red and one blue.

Two paths: R then B, or B then R.

  • \(P(\text{R then B})=\tfrac{2}{5}\times \tfrac{3}{4}=\tfrac{6}{20}=\tfrac{3}{10}\).
  • \(P(\text{B then R})=\tfrac{3}{5}\times \tfrac{2}{4}=\tfrac{6}{20}=\tfrac{3}{10}\).

Add across: \(\tfrac{3}{10}+\tfrac{3}{10}=\tfrac{6}{10}=\tfrac{3}{5}\). Answer: \(\tfrac{3}{5}\).

Example 4 (Independent Events: Coins, Higher)

A biased coin has \(P(H)=0.6\). It is tossed twice. Find the probability of exactly one head.

Paths: H then T, or T then H.

  • \(0.6\times 0.4=0.24\)
  • \(0.4\times 0.6=0.24\)

Add: \(0.24+0.24=0.48\). Answer: 0.48.

Example 5 (Conditional Branch Update, Higher)

A drawer contains 5 black socks and 3 white socks. Two are taken at random without replacement. Find \(P(\text{both white})\).

First white: \(\tfrac{3}{8}\). If white first, remaining whites = 2 of total 7 ⇒ \(\tfrac{2}{7}\). Multiply: \(\tfrac{3}{8}\times \tfrac{2}{7}=\tfrac{6}{56}=\tfrac{3}{28}\). Answer: \(\tfrac{3}{28}\).

Method Reminder: Without replacement ⇒ update both the numerator (fewer of that colour) and the denominator (one fewer total).

Common Mistakes & Fixes

  • Forgetting to update without replacement: Keep track of both successful items and total items after the first pick.
  • Adding instead of multiplying along a path: Always multiply along branches for a sequence of events.
  • Forgetting both orders: “One of each” usually has two paths; add them.
  • Not checking splits sum to 1: At each node, probabilities must add to 1.
  • Switching between decimals and fractions mid-solution: Stick to one format or convert consistently to avoid errors.

Practice Questions — Foundation

  1. A bag has 3 red and 2 blue counters. Two counters are drawn with replacement. Find \(P(\text{two blue})\).
  2. A box has 5 green and 1 yellow bead. Two are drawn without replacement. Find \(P(\text{green then yellow})\).
  3. A spinner has 4 equal sections: A, B, C, D. It is spun twice. Find \(P(\text{A then C})\).
  4. A drawer has 2 black and 3 white socks. Two are drawn without replacement. Find \(P(\text{two white})\).
  5. A bag has 2 red, 1 blue. Two are drawn with replacement. Find \(P(\text{exactly one red})\).

Practice Questions — Higher

  1. A jar has 4 red, 3 blue, 1 green bead. Two are drawn without replacement. Find \(P(\text{one red and one blue in any order})\).
  2. A biased coin has \(P(H)=0.7\). It is tossed twice. Find \(P(\text{at least one head})\).
  3. A bag has 6 black and 4 white balls. Two are drawn without replacement. Find \(P(\text{black then white})\) and \(P(\text{white then black})\), then the probability of “one black and one white”.
  4. Box X has 3 red, 2 blue; Box Y has 1 red, 4 blue. You choose a box at random (equal chance) and take one ball. Find \(P(\text{red})\).
  5. From a class with 12 girls and 8 boys, two students are chosen without replacement. Find \(P(\text{both girls})\).

Challenge Questions

  1. A bag contains 5 red and 3 blue counters. Two are drawn without replacement. Find \(P(\text{at least one blue})\).
  2. Urn A: 2 red, 3 green. Urn B: 4 red, 1 green. You flip a fair coin: Heads ⇒ choose from A; Tails ⇒ choose from B. Then pick one counter. Find \(P(\text{green})\).
  3. A box has 3 defective and 7 good bulbs. Two bulbs are selected without replacement. Find \(P(\text{both good})\) and \(P(\text{exactly one defective})\).
  4. A fair die is rolled twice. Use a tree or reasoning to find \(P(\text{sum is 7})\).
  5. A bag has 4 red, 4 blue, 2 yellow counters. Two are drawn without replacement. Find \(P(\text{two counters of the same colour})\).

Quick Revision Sheet

  • Multiply along a path.
  • Add across paths that lead to the same overall outcome.
  • With replacement: probabilities stay the same.
  • Without replacement: update both numerator and denominator.
  • Check splits sum to 1.

Answers

Foundation

  1. \(\big(\tfrac{2}{5}\big)\big(\tfrac{2}{5}\big)=\tfrac{4}{25}\)
  2. \(\tfrac{5}{6}\times \tfrac{1}{5}=\tfrac{1}{6}\)
  3. \(\tfrac{1}{4}\times \tfrac{1}{4}=\tfrac{1}{16}\)
  4. \(\tfrac{3}{5}\times \tfrac{2}{4}=\tfrac{6}{20}=\tfrac{3}{10}\)
  5. Exactly one red with replacement: R then not R: \(\tfrac{2}{3}\times \tfrac{1}{3}=\tfrac{2}{9}\); not R then R: \(\tfrac{1}{3}\times \tfrac{2}{3}=\tfrac{2}{9}\); total \(=\tfrac{4}{9}\).

Higher

  1. One red & one blue (no replacement): R then B: \(\tfrac{4}{8}\times \tfrac{3}{7}=\tfrac{12}{56}=\tfrac{3}{14}\); B then R: \(\tfrac{3}{8}\times \tfrac{4}{7}=\tfrac{12}{56}=\tfrac{3}{14}\); total \(=\tfrac{3}{7}\).
  2. At least one head: \(1-P(\text{no heads})=1-(0.3\times 0.3)=1-0.09=0.91\).
  3. BW: \(\tfrac{6}{10}\times \tfrac{4}{9}=\tfrac{24}{90}=\tfrac{4}{15}\); WB: \(\tfrac{4}{10}\times \tfrac{6}{9}=\tfrac{24}{90}=\tfrac{4}{15}\); one of each \(=\tfrac{8}{15}\).
  4. Choose box: \(\tfrac{1}{2}\) each. From X red: \(\tfrac{3}{5}\); from Y red: \(\tfrac{1}{5}\). Total: \(\tfrac{1}{2}\cdot\tfrac{3}{5}+\tfrac{1}{2}\cdot\tfrac{1}{5}=\tfrac{3}{10}+\tfrac{1}{10}=\tfrac{4}{10}=\tfrac{2}{5}\).
  5. \(\tfrac{12}{20}\times \tfrac{11}{19}=\tfrac{132}{380}=\tfrac{33}{95}\).

Challenge

  1. At least one blue = \(1-P(\text{no blue})\). No blue = both red: \(\tfrac{5}{8}\times \tfrac{4}{7}=\tfrac{20}{56}=\tfrac{5}{14}\). So \(1-\tfrac{5}{14}=\tfrac{9}{14}\).
  2. \(\tfrac{1}{2}\cdot\tfrac{3}{5}+\tfrac{1}{2}\cdot\tfrac{1}{5}=\tfrac{3}{10}+\tfrac{1}{10}=\tfrac{2}{5}\).
  3. Both good: \(\tfrac{7}{10}\times \tfrac{6}{9}=\tfrac{42}{90}=\tfrac{7}{15}\). Exactly one defective = (D then G) + (G then D) \(=\tfrac{3}{10}\times \tfrac{7}{9}+\tfrac{7}{10}\times \tfrac{3}{9} =\tfrac{21}{90}+\tfrac{21}{90}=\tfrac{42}{90}=\tfrac{7}{15}\).
  4. Sum 7 on two fair rolls: favourable: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) ⇒ 6/36 = 1/6.
  5. Same colour (no replacement): RR: \(\tfrac{4}{10}\times \tfrac{3}{9}=\tfrac{12}{90}\); BB: \(\tfrac{4}{10}\times \tfrac{3}{9}=\tfrac{12}{90}\); YY: \(\tfrac{2}{10}\times \tfrac{1}{9}=\tfrac{2}{90}\); total \(=\tfrac{26}{90}=\tfrac{13}{45}\).

Conclusion & Next Steps

Tree diagrams turn multi-step probability into a clear, repeatable process: multiply along paths and add across paths. Next, study Venn diagrams (overlaps, union, intersection) and independent vs mutually exclusive events, so you can choose the best model for any probability question.

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