Expected frequencies | Year 7

01 / LEARN

Understand the idea.

From chance to predicted count

An expected frequency turns a probability into a predicted count: multiply the number of trials by the event probability. With blue chance $\frac{1}{4}$ , the expected count in $80$ spins is $80 \times \frac{1}{4} = 20$ blue results.

Use a model that applies throughout the experiment. Replacing and remixing a counter restores the original bag chances. An expected share of one quarter does not prescribe the order: four consecutive trials need not contain exactly one success.

expected frequency = trials \times probability

Expected is not guaranteed

Expected describes an average prediction, not a compulsory result. Random counts may lie above or below it. With more trials, proportions tend to settle nearer the model probability, but they need not improve after every trial and count differences can still grow.

A predicted average can lie between whole numbers. With probability $\frac{1}{4}$ over six trials, the expectation is $1.5$ . Each actual run records a whole count; averages across many such runs can be close to one and a half.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Predict multiples on a die

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

Worked solution

1
Find the chance of a multiple of three.
$E = {3, 6}, P (E) = \frac{2}{6} = \frac{1}{3}$
Two of the six equally likely die faces are multiples of three.
2
Multiply the trial count by that probability.
$120 \times \frac{1}{3} = 40$
The same fair-die probability applies to each of the one hundred and twenty rolls.
3
State what the prediction means.
Forty multiples of three are expected, but a particular run may contain more or fewer.

Answer

40 results are expected.

EXAMPLE 02

A prediction between whole numbers

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

Worked solution

1
Calculate the expected blue count.
$6 \times \frac{1}{4} = \frac{6}{4} = 1.5$
The expected count is the number of spins multiplied by blue's probability.
2
Distinguish the prediction from a possible recorded count.
$Possible blue counts: 0, 1, 2, 3, 4, 5, 6$
Each spin either contributes one blue result or contributes none, so an actual count must be whole.
3
Interpret the non-whole expectation.
Across many comparable six-spin runs, the average blue count is expected to be near one and a half; no individual spin produces half a blue result.

Answer

Expected count 1.5; it is an average prediction, not a possible half-result.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

25 heads.

A fair coin has heads probability $\frac{1}{2}$ , so $50 \times \frac{1}{2} = 25$ heads are expected. This is the average prediction, while an actual fifty-toss count can differ.

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6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

An event has probability $\frac{1}{5}$ . Find the expected frequency in 40 trials.
2 marks
An ideal spinner is half red. Find the expected red count in 90 spins.
2 marks
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
A fair die is rolled 60 times. How many results greater than 4 are expected?
2 marks
An event has chance $\frac{1}{4}$ and is repeated 10 times. Is an expected frequency of 2.5 a mistake? Explain.
2 marks
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.

An event has probability $\frac{1}{5}$ . Find the expected frequency in 40 trials.
2 marks
$40 \times \frac{1}{5} = 8$ .
8.
An ideal spinner is half red. Find the expected red count in 90 spins.
2 marks
Red chance is $\frac{1}{2}$ .
$90 \times \frac{1}{2} = 45$ .
45.
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 $\frac{3}{5}$ .
$50 \times \frac{3}{5} = 30$ .
30.
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 $\frac{1}{3}$ .
$60 \times \frac{1}{3} = 20$ .
20.
An event has chance $\frac{1}{4}$ and is repeated 10 times. Is an expected frequency of 2.5 a mistake? Explain.
2 marks
$10 \times \frac{1}{4} = 2.5$ .
A predicted average count can be fractional although actual counts are whole.
No.
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 $50 \times \frac{1}{2} = 25$ .
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 ↗

In this lesson

Understand the idea.

From chance to predicted count

Expected is not guaranteed

See the method in action.

Predict multiples on a die

Find the chance of a multiple of three.

Multiply the trial count by that probability.

State what the prediction means.

A prediction between whole numbers

Calculate the expected blue count.

Distinguish the prediction from a possible recorded count.

Interpret the non-whole expectation.

Try it yourself.

Check your understanding

Keep practising