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How do you measure probability on the 0 to 1 scale, use sample spaces and handle mutually exclusive and combined events?

Use the probability scale from 0 to 1, calculate probabilities of equally likely outcomes using sample spaces and listings, and apply the rules for mutually exclusive and exhaustive events including combined events in two-way tables.

A focused answer to the WJEC GCSE Mathematics probability content on the basics, covering the 0 to 1 probability scale, equally likely outcomes, sample space diagrams, mutually exclusive and exhaustive events and combined events using two-way tables and listings.

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  1. What this dot point is asking
  2. The probability scale
  3. The complement and exhaustive events
  4. Mutually exclusive events
  5. Sample spaces and combined events
  6. Why this matters

What this dot point is asking

This is the foundation of WJEC probability. You are asked to use the probability scale from 00 to 11, to calculate the probability of equally likely outcomes by counting (using sample space diagrams and systematic listing), and to apply the rules for mutually exclusive and exhaustive events, including combined events shown in two-way tables. The central idea is that probability is a fraction of favourable outcomes over total outcomes, and that the probabilities of all outcomes sum to 11. It is examined on both components at every tier.

The probability scale

Every probability sits between 00 and 11.

So drawing a particular card from a pack of 5252 has probability 152\tfrac{1}{52}, and drawing any heart has probability 1352=14\tfrac{13}{52} = \tfrac{1}{4}.

The complement and exhaustive events

The probabilities of all possible outcomes add to 11.

Because something must happen, the probabilities of all the possible outcomes sum to 11. Events that together cover every possibility are exhaustive. The complement of event A (the event "A does not happen") therefore has probability P(not A)=1P(A)P(\text{not A}) = 1 - P(A). This is the fastest route to "find the probability it is not ..." questions: subtract from 11 rather than adding many cases.

Mutually exclusive events

Some events cannot occur at the same time.

So the probability of rolling a 11 or a 22 on a fair die is 16+16=13\tfrac{1}{6} + \tfrac{1}{6} = \tfrac{1}{3}. The "add for or" rule only applies when the events cannot overlap.

Sample spaces and combined events

Listing all outcomes makes combined events countable.

A two-way table organises combined data so probabilities can be read off directly from the totals.

Why this matters

Probability basics underpin every later topic in the strand: tree diagrams, Venn diagrams and expected outcomes all build on the scale, the complement rule and systematic counting. The marks reward writing probabilities correctly (as fractions, not "66 out of 3636" left unsimplified), using 1P(A)1 - P(A) efficiently, and listing combined outcomes without missing any. Recognising mutually exclusive events and applying the "add for or" rule correctly is a common discriminator, and a sample space or two-way table is the reliable tool for combined events.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20182 marksA bag contains 55 red, 33 blue and 22 green counters. One counter is taken at random. Work out the probability that it is not blue. (Foundation and Higher, Unit 1, non-calculator.)
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There are 5+3+2=105 + 3 + 2 = 10 counters in total, and 33 are blue.

The probability of blue is 310\dfrac{3}{10}, so the probability of not blue is 1310=7101 - \dfrac{3}{10} = \dfrac{7}{10}.

Markers award a mark for the probability of blue (or for counting the 77 non-blue counters) and a mark for 710\dfrac{7}{10}. Using 1P(blue)1 - P(\text{blue}) is the efficient method, since the complement of an event has probability 11 minus the event's probability.

WJEC 20213 marksTwo fair six-sided dice are rolled and their scores added. Use a sample space to find the probability that the total is 77. (Foundation and Higher, Unit 2, calculator.)
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A sample space of two dice has 6×6=366 \times 6 = 36 equally likely outcomes.

The totals giving 77 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)(1,6), (2,5), (3,4), (4,3), (5,2), (6,1): six outcomes.

So the probability is 636=16\dfrac{6}{36} = \dfrac{1}{6}.

Markers give a mark for the 3636 outcomes, a mark for the six ways of making 77 and a mark for 16\dfrac{1}{6}. Forgetting that (3,4)(3,4) and (4,3)(4,3) are different outcomes undercounts and is the usual error.

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