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Probability: replacement, tree diagrams and checking outcomes

Probability starts with a clearly defined experiment. Decide whether outcomes are equally likely and whether a selection changes what is available for the next selection. The words ‘without replacement’ change the second denominator.

Choose the method

For equally likely outcomes, probability = favourable outcomes/total outcomes. The answer must lie between 0 and 1.

Multiply conditional probabilities along a tree branch. Add the probabilities of mutually exclusive branches that produce the event you want.

Use a complement for ‘at least one’ when it is easier: P(at least one) = 1 - P(none). Independence must be given or justified; separate events are not automatically independent.

Worked examples

Example 1

A bag contains 3 red and 2 blue counters. Two are drawn without replacement. Find P(both red).

  1. The first draw is red with probability 3/5.
  2. After a red draw, 2 red counters remain among 4 counters. The second probability is 2/4.
  3. P(both red) = 3/5 × 2/4 = 3/10.

Example 2

For the same bag, find P(one red and one blue) without replacement.

  1. Red then blue: 3/5 × 2/4 = 3/10.
  2. Blue then red: 2/5 × 3/4 = 3/10.
  3. These branches cannot happen together, so add: 3/10 + 3/10 = 3/5.

Try it yourself

A fair die is rolled twice independently. Find P(at least one six).

Show the worked answer
  1. The complement is no six on either roll: (5/6)² = 25/36.
  2. Required probability = 1 - 25/36 = 11/36.

Common mistakes

Sources and review

SEAB 2026 syllabus: O-Level Mathematics 4052: probability.

Original Rae practice, prepared with AI assistance. Selected numerical results and their displayed working are automatically checked at publication; this does not verify every explanation. Curriculum references checked on 5 September 2026. No teacher review or SEAB endorsement is claimed.

What the syllabus expects

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More O-Level Elementary Maths (E-Maths) topics

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