Year 7 Mathematics · AC9M7P01

Sample Spaces and Single-stage Probability

We are learning to identify sample spaces for single-stage chance events, assign probabilities and predict relative frequencies

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Learning goalsSay it simply

We are learning to identify sample spaces for single-stage chance events, assign probabilities and predict relative frequencies.

A sample space lists every possible elementary outcome of one chance action. When elementary outcomes are equally likely, an event's probability is its favourable outcome count divided by the total outcome count; when sectors, objects or rules have different weights, probabilities must come from those weights rather than from the number of labels.

An event is a subset of the sample space, so its probability is the sum of the probabilities of its mutually exclusive elementary outcomes. All elementary outcome probabilities must total 1, and the complement rule P(not A) = 1 − P(A) accounts for every outcome outside an event.

Expected frequency equals probability multiplied by the number of trials and predicts the long-run count for planning or comparison. It is not a guarantee for one run, because observed relative frequency can vary even when the probability model is correct.

Success criteria

  • I can list every possible elementary outcome without omissions or duplication.
  • I can assign and combine outcome probabilities, including for outcomes that are not equally likely.
  • I can calculate a complement and use probability to predict an expected frequency while recognising chance variation.
Key conceptTeach from the board

Use these models to connect the mathematical idea to values, diagrams, coordinates, graphs or structure before moving to symbolic calculation.

Follow each example from representation and setup through calculation/reasoning, interpretation and an independent check.

Example 1

Assigning probabilities to a weighted spinner.

  1. A spinner has red covering one half of its area, blue covering one quarter and green covering one quarter, so the colour sample space is {red, blue, green}.
  2. Assign P(red) = 1/2, P(blue) = 1/4 and P(green) = 1/4 from sector size rather than giving each label one-third.
  3. Check 1/2 + 1/4 + 1/4 = 1, and add outcomes to find P(red or blue) = 1/2 + 1/4 = 3/4.
  4. For 200 spins, calculate the expected red-or-blue frequency as 200 × 3/4 = 150.

Final answer: The weighted outcome probabilities are 1/2, 1/4 and 1/4, and red or blue has probability 3/4 with expected frequency 150 in 200 spins.

Check: The elementary probabilities total 1 and the expected colour counts 100, 50 and 50 total all 200 planned spins.

Example 2

Building probabilities from objects in a bag.

  1. A bag contains 3 red counters, 2 blue counters and 5 yellow counters, making 10 equally likely individual counters.
  2. Assign P(red) = 3/10, P(blue) = 2/10 and P(yellow) = 5/10.
  3. The event 'not yellow' contains red or blue, so P(not yellow) = 3/10 + 2/10 = 5/10 = 1/2, which also equals 1 − 5/10.
  4. In 40 draws with replacement, predict an expected not-yellow frequency of 40 × 1/2 = 20.

Final answer: The probability of not drawing yellow is 1/2, so its expected frequency in 40 replacement draws is 20.

Check: The red, blue and yellow probabilities total 10/10, and the complement and addition methods both give 1/2 for not yellow.

Example 3

Application problem 1

Problem: A spinner has 8 equal sectors: 3 red, 2 blue and 3 green. Find each colour probability and predict the expected number of blue results in 120 spins.

  1. Plan: Represent the information first, then calculate, interpret and independently check the result.
  2. Work: P(red)=3/8, P(blue)=2/8=1/4, P(green)=3/8. Expected blue results = 120×1/4 = 30.
  3. Interpret: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Final answer: P(red)=3/8, P(blue)=2/8=1/4, P(green)=3/8. Expected blue results = 120×1/4 = 30.

Check: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Example 4

Application problem 2

Problem: A game outcome has probability 3/8 and is played 160 times; calculate the expected frequency and explain why observing 55 outcomes would not by itself disprove the probability model.

  1. Plan: Find the predicted count first, then distinguish a difference of five from proof of an incorrect model.
  2. Work: The expected frequency is 160 × 3/8 = 60, and an observed count of 55 can occur through chance variation, so repeated evidence is needed before judging the model.
  3. Interpret: Expected frequency is a long-run prediction rather than an exact requirement for one run.

Final answer: The expected frequency is 160 × 3/8 = 60, and an observed count of 55 can occur through chance variation, so repeated evidence is needed before judging the model.

Check: Expected frequency is a long-run prediction rather than an exact requirement for one run.

Clean visual examplesOne-page board

Clean one-page examples

AC9M7P01 - Sample Spaces and Single-stage Probability
Example 1

Example 1 Assigning probabilities to a weighted spinner. A spinner has red covering one half of its area, blue covering one quarter and green covering one quarter, so the colour sample space is {red, blue, green}. Assign P(red) = 1/2, P(blue) = 1/4 and P(green) = 1/4 from sector size rather than giving each label one-third. Check 1/2 + 1/4 + 1/4 = 1, and add outcomes to find P(red or blue) = 1/2 + 1/4 = 3/4. For 200 spins, calculate the expected red-or-blue frequency as 200 × 3/4 = 150. Final answer: The weighted outcome probabilities are 1/2, 1/4 and 1/4, and red or blue has probability 3/4 with expected frequency 150 in 200 spins. Check: The elementary probabilities total 1 and the expected colour counts 100, 50 and 50 total all 200 planned spins.

Example 2

Example 2 Building probabilities from objects in a bag. A bag contains 3 red counters, 2 blue counters and 5 yellow counters, making 10 equally likely individual counters. Assign P(red) = 3/10, P(blue) = 2/10 and P(yellow) = 5/10. The event 'not yellow' contains red or blue, so P(not yellow) = 3/10 + 2/10 = 5/10 = 1/2, which also equals 1 − 5/10. In 40 draws with replacement, predict an expected not-yellow frequency of 40 × 1/2 = 20. Final answer: The probability of not drawing yellow is 1/2, so its expected frequency in 40 replacement draws is 20. Check: The red, blue and yellow probabilities total 10/10, and the complement and addition methods both give 1/2 for not yellow.

Example 3

Example 3 Application problem 1 Problem: A spinner has 8 equal sectors: 3 red, 2 blue and 3 green. Find each colour probability and predict the expected number of blue results in 120 spins. Plan: Represent the information first, then calculate, interpret and independently check the result. Work: P(red)=3/8, P(blue)=2/8=1/4, P(green)=3/8. Expected blue results = 120×1/4 = 30. Interpret: This synthesis problem combines the chapter's core representation, calculation and reasoning skills. Final answer: P(red)=3/8, P(blue)=2/8=1/4, P(green)=3/8. Expected blue results = 120×1/4 = 30. Check: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Example 4

Example 4 Application problem 2 Problem: A game outcome has probability 3/8 and is played 160 times; calculate the expected frequency and explain why observing 55 outcomes would not by itself disprove the probability model. Plan: Find the predicted count first, then distinguish a difference of five from proof of an incorrect model. Work: The expected frequency is 160 × 3/8 = 60, and an observed count of 55 can occur through chance variation, so repeated evidence is needed before judging the model. Interpret: Expected frequency is a long-run prediction rather than an exact requirement for one run. Final answer: The expected frequency is 160 × 3/8 = 60, and an observed count of 55 can occur through chance variation, so repeated evidence is needed before judging the model. Check: Expected frequency is a long-run prediction rather than an exact requirement for one run.

Curriculum examplesCopied content

Content description: identify the sample space for single-stage events; assign probabilities to the outcomes of these events and predict relative frequencies for related events.

Questions and answersWith answers

Attempt each problem before opening Check answer. The set moves from core understanding to application and synthesis.

  1. 1. List the sample space for one roll of a standard six-sided die.

    Check answer

    Answer: The sample space is {1, 2, 3, 4, 5, 6}.

    Hint: Write every face value that could appear after one roll.

    Why: The six faces list every possible elementary outcome exactly once.

  2. 2. A spinner has one half red, one quarter blue and one quarter green; find P(blue).

    Check answer

    Answer: P(blue) = 1/4.

    Hint: Use sector area rather than the number of colour labels.

    Why: Probability follows the blue sector's one-quarter share of the spinner.

  3. 3. If P(rain) = 0.37, find P(no rain).

    Check answer

    Answer: P(no rain) = 1 − 0.37 = 0.63.

    Hint: Subtract the stated event probability from 1.

    Why: Complementary events together have probability 1.

  4. 4. A bag contains 3 red, 2 blue and 5 yellow counters; find P(red or blue) for one draw.

    Check answer

    Answer: P(red or blue) = (3 + 2)/10 = 5/10 = 1/2.

    Hint: Add the red and blue counts before dividing by the total count.

    Why: The event combines five favourable counters among ten equally likely counters.

  5. 5. A fair die is rolled once; find the probability of an outcome greater than 4.

    Check answer

    Answer: The favourable outcomes are {5, 6}, so the probability is 2/6 = 1/3.

    Hint: List values in the sample space that are strictly greater than 4.

    Why: Two of the six equally likely die outcomes satisfy the event.

  6. 6. An event has probability 0.35; predict its expected frequency in 80 trials.

    Check answer

    Answer: The expected frequency is 80 × 0.35 = 28.

    Hint: Calculate probability multiplied by the number of planned trials.

    Why: Multiplying trial count by event probability gives the predicted count.

  7. 7. A spinner has three labels A, B and C, but A covers half the spinner while B and C each cover one quarter; correct the claim that every label has probability one-third.

    Check answer

    Answer: The correct probabilities are P(A) = 1/2, P(B) = 1/4 and P(C) = 1/4 because likelihood follows sector size rather than label count.

    Hint: Compare the fraction of the whole spinner occupied by each label.

    Why: Named outcomes are not equally likely when their areas differ.

  8. 8. Design a fair six-sector spinner with events A, B and C having probabilities 1/2, 1/3 and 1/6 respectively, and justify the design.

    Check answer

    Answer: Label 3 equal sectors A, 2 equal sectors B and 1 equal sector C; the probabilities are 3/6 = 1/2, 2/6 = 1/3 and 1/6, and they total 1.

    Hint: Express every target probability with denominator 6 and assign that many equal sectors.

    Why: Allocating six equal sectors in the ratio 3:2:1 realises the required probabilities exactly.

  9. 9. A game outcome has probability 3/8 and is played 160 times; calculate the expected frequency and explain why observing 55 outcomes would not by itself disprove the probability model.

    Check answer

    Answer: The expected frequency is 160 × 3/8 = 60, and an observed count of 55 can occur through chance variation, so repeated evidence is needed before judging the model.

    Hint: Find the predicted count first, then distinguish a difference of five from proof of an incorrect model.

    Why: Expected frequency is a long-run prediction rather than an exact requirement for one run.

  10. 10. A spinner has 8 equal sectors: 3 red, 2 blue and 3 green. Find each colour probability and predict the expected number of blue results in 120 spins.

    Check answer

    Answer: P(red)=3/8, P(blue)=2/8=1/4, P(green)=3/8. Expected blue results = 120×1/4 = 30.

    Hint: Represent the information first, then calculate, interpret and independently check the result.

    Why: This synthesis problem combines the chapter's core representation, calculation and reasoning skills.

Practice and reviewReady for practice

Common mistake: Three colour labels on a spinner must each have probability one-third.

Correction: Probabilities depend on sector size, so labels are equally likely only when their total sectors are equal in size.

Common mistake: A multi-outcome event can be counted as though it were one elementary outcome.

Correction: The probabilities of all mutually exclusive elementary outcomes in the event must be added.

Common mistake: An expected frequency is the exact number that must occur in the next set of trials.

Correction: Expected frequency is a long-run prediction, and an observed count can differ through chance variation.

Curriculum alignmentStart here

We are learning to identify sample spaces for single-stage chance events, assign probabilities and predict relative frequencies.

A sample space lists every possible elementary outcome of one chance action. When elementary outcomes are equally likely, an event's probability is its favourable outcome count divided by the total outcome count; when sectors, objects or rules have different weights, probabilities must come from those weights rather than from the number of labels.

An event is a subset of the sample space, so its probability is the sum of the probabilities of its mutually exclusive elementary outcomes. All elementary outcome probabilities must total 1, and the complement rule P(not A) = 1 − P(A) accounts for every outcome outside an event.

Expected frequency equals probability multiplied by the number of trials and predicts the long-run count for planning or comparison. It is not a guarantee for one run, because observed relative frequency can vary even when the probability model is correct.

Success criteria

  • I can list every possible elementary outcome without omissions or duplication.
  • I can assign and combine outcome probabilities, including for outcomes that are not equally likely.
  • I can calculate a complement and use probability to predict an expected frequency while recognising chance variation.
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