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Year 7 · Probability · 15 min

Expected frequencies

Predict an expected count from a probability, and explain why actual counts need not equal the prediction.

01 / LEARN

The idea, step by step

From chance to predicted count

For event probability p, the expected frequency in n repetitions is n⁢p. If a quarter of an ideal spinner is blue, about a quarter of many spins should be blue. For 80 spins, 80×1⁢/4=20 blue results are expected.

The model probability must apply throughout the repeated experiment. Replacing a drawn counter can preserve the bag’s colour chances. The prediction does not mean that every group of four spins contains one blue result.

expected frequency=trials×probability

Expected is not guaranteed

Random variation means actual counts may be above or below their expectation. A fair coin can produce more heads than tails in a short run. Larger runs often make proportions more stable, but the raw difference in counts need not decrease.

An expected count can be fractional. Six spins with blue chance 1⁢/4 give expected count 1.5. Each individual run has a whole blue count; 1.5 describes an average across many comparable runs, not half of a recorded outcome.

A common trap

The event must occur soon to make up for earlier misses.

Instead: An expected count does not force a sequence or a catch-up outcome. Future trials still follow the stated chance model.

02 / WORKED EXAMPLES

See the thinking

Example 1

Predict multiples on a die

A fair die is rolled 120 times. How many multiples of 3 are expected?

  1. The favourable faces are 3 and 6: chance 2⁢/6=1⁢/3.
  2. Multiply 120×1⁢/3=40.
  3. State this as a prediction rather than a required exact count.
40 results are expected.
Example 2

A prediction between whole numbers

An ideal spinner has blue chance 1⁢/4. Interpret the expected blue count in 6 spins.

  1. Calculate 6×1⁢/4=1.5.
  2. An individual run only has a whole number of blue results.
  3. Across many six-spin runs, their average count is expected to be near 1.5.
Expected count 1.5; it is an average prediction, not a possible half-result.

Check your understanding

A fair coin is tossed 50 times. How many heads are expected?

Show answer

25 heads.

50×1⁢/2=25; actual counts need not be exactly 25.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Read the questions online
  1. An event has probability 1⁢/5. Find the expected frequency in 40 trials.
    2 marks
  2. An ideal spinner is half red. Find the expected red count in 90 spins.
    2 marks
  3. A bag has 3 green and 2 white equal-size counters. Each randomly chosen counter is returned and remixed. How many green results are expected in 50 draws?
    3 marks
  4. A fair die is rolled 60 times. How many results greater than 4 are expected?
    2 marks
  5. An event has chance 1⁢/4 and is repeated 10 times. Is an expected frequency of 2.5 a mistake? Explain.
    2 marks
  6. A fair coin gave 31 heads in 50 tosses instead of the expected 25. Does that alone show the prediction calculation was wrong?
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. An event has probability 1⁢/5. Find the expected frequency in 40 trials.
    2 marks

    40×1⁢/5=8.

    8.
  2. An ideal spinner is half red. Find the expected red count in 90 spins.
    2 marks

    Red chance is 1⁢/2.

    90×1⁢/2=45.

    45.
  3. A bag has 3 green and 2 white equal-size counters. Each randomly chosen counter is returned and remixed. How many green results are expected in 50 draws?
    3 marks

    Returning maintains chance 3⁢/5.

    50×3⁢/5=30.

    30.
  4. A fair die is rolled 60 times. How many results greater than 4 are expected?
    2 marks

    The favourable faces are 5 and 6: chance 1⁢/3.

    60×1⁢/3=20.

    20.
  5. An event has chance 1⁢/4 and is repeated 10 times. Is an expected frequency of 2.5 a mistake? Explain.
    2 marks

    10×1⁢/4=2.5.

    A predicted average count can be fractional although actual counts are whole.

    No.
  6. A fair coin gave 31 heads in 50 tosses instead of the expected 25. Does that alone show the prediction calculation was wrong?
    3 marks

    The expected count remains 50×1⁢/2=25.

    31 is one random observed count; expected does not mean guaranteed.

    No.
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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Year 7 single-stage sample spaces, equally likely outcomes, probabilities, expected frequencies, chance experiments, simulations and relative frequencies.

Official curriculum source ↗
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