Digital simulations and explaining differences from predictions | Year 7

01 / LEARN

Understand the idea.

Match the model to the chances

A simulation uses a random rule to stand in for a physical trial. To model a fair die, generate an integer from $1$ to $6$ with each equally likely. In a spreadsheet, enter =RANDBETWEEN(1,6) in A2 and fill down to A61 for sixty rolls.

Check both the possible outputs and their chances. Integers from $0$ to $6$ give seven possibilities, so they do not model a six-faced die. To model red probability $\frac{3}{10}$ , use ten equally likely integers and assign exactly three to red.

Preserve and interpret one run

For results in A2:A61, =COUNTIF(A2:A61,6) counts sixes; divide the count by $60$ for their relative frequency. Random cells may recalculate when the sheet changes. Copy a completed run and use Paste values when you need to preserve that exact record.

Report the random rule, trial count, observed proportion and theoretical comparison. The supplied table contains separate example runs. Larger samples tend to have more stable proportions, but a correct simulation can still differ from its prediction and need not improve with every added trial.

Supplied example die-simulation records; each row is a separate run
Trials	1	2	3	4	5	6
60	7	12	9	13	8	11
600	101	98	95	107	103	96

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Compare two runs

Use the supplied records to find each even-result relative frequency and compare with the theoretical fair-die chance.

Supplied example die-simulation records; each row is a separate run
Trials	1	2	3	4	5	6
60	7	12	9	13	8	11
600	101	98	95	107	103	96

Worked solution

1
Find the theoretical even-result chance.
$E = {2, 4, 6}, P (E) = \frac{3}{6} = 0.5$
A fair die has three even faces among six equally likely faces.
2
Calculate the observed share in the sixty-trial run.
$\begin{aligned} Even count & = 12 + 13 + 11 = 36 \\ Relative frequency & = \frac{36}{60} = 0.6 \end{aligned}$
The counts in the columns for two, four and six are the favourable observations.
3
Calculate the observed share in the six-hundred-trial run.
$\begin{aligned} Even count & = 98 + 107 + 96 = 301 \\ Relative frequency & = \frac{301}{600} \approx 0.502 \end{aligned}$
Use this run's event count and its own trial total; the decimal is rounded to three decimal places.
4
Compare each estimate with one half.
$\begin{aligned} \frac{36}{60} - \frac{1}{2} & = \frac{1}{10} \\ \frac{301}{600} - \frac{1}{2} & = \frac{1}{600} \end{aligned}$
Both estimates are above one half, but the larger run's difference is smaller. This shows greater closeness in these records, not a guarantee for every pair of runs.

Answer

Estimates 0.6 and about 0.502; theoretical chance 0.5.

EXAMPLE 02

A model for unequal colour chances

Model a bag with 3 red and 7 blue equal-size counters, with replacement.

Worked solution

1
Choose ten equally likely random integers.
$S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}$
The bag contains ten equally selectable counters; =RANDBETWEEN(1,10) supplies ten equally likely labels.
2
Assign three labels to red and the remaining seven to blue.
$\begin{aligned} R & = {1, 2, 3} \\ B & = {4, 5, 6, 7, 8, 9, 10} \end{aligned}$
Each label has one colour, and the label counts match the bag's three red and seven blue counters.
3
Check the model probabilities and repeat the same rule.
$\begin{aligned} P (R) & = \frac{3}{10} \\ P (B) & = \frac{7}{10} \end{aligned}$
Keeping all ten labels available for every new random draw models replacement, which restores the bag's original chances.

Answer

Choose 1–10 equally: 1–3 means red, 4–10 means blue.

03 / YOUR TURN

Try it yourself.

Check your understanding

A model generates 1 to 8 equally and calls 1 or 2 a success. Does it model success probability $1 / 4$ ?

Show answer & reasoning

Yes.

Exactly two of the eight equally likely integers are successes. Their probability is $\frac{2}{8} = \frac{1}{4}$ , so the rule models the required success chance.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Which rule models a fair die: =RANDBETWEEN(0,6) or =RANDBETWEEN(1,6)? Explain.
2 marks
Describe a random-integer model for a fair coin using integers 1 and 2.
2 marks
Use the supplied table. Find the relative frequency of 6 in the 60-trial run.
2 marks
Supplied example die-simulation records; each row is a separate run
Trials 1 2 3 4 5 6
60 7 12 9 13 8 11
600 101 98 95 107 103 96
For the 600-trial run, state observed and expected six counts, and their difference.
3 marks
Supplied example die-simulation records; each row is a separate run
Trials 1 2 3 4 5 6
60 7 12 9 13 8 11
600 101 98 95 107 103 96
A model chooses 1 to 10 equally and labels 1 to 4 as green. Find green chance and expected greens in 50 trials.
3 marks
A simulation estimate is 0.6 for an event with theoretical chance 0.5. Must its coding be wrong? State one check and one alternative explanation.
3 marks

Read the worked solutions

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

Which rule models a fair die: =RANDBETWEEN(0,6) or =RANDBETWEEN(1,6)? Explain.
2 marks
The required outcomes are exactly 1 to 6.
The other rule adds a seventh outcome, 0.
=RANDBETWEEN(1,6).
Describe a random-integer model for a fair coin using integers 1 and 2.
2 marks
Each integer has chance $\frac{1}{2}$ .
Each recorded outcome matches one integer.
Choose 1 or 2 equally; use 1 for heads and 2 for tails.
Use the supplied table. Find the relative frequency of 6 in the 60-trial run.
2 marks
Supplied example die-simulation records; each row is a separate run
Trials 1 2 3 4 5 6
60 7 12 9 13 8 11
600 101 98 95 107 103 96
The six frequency is 11.
Divide by 60; the decimal is rounded to three decimal places.
$\frac{11}{60} \approx 0.183$ .
For the 600-trial run, state observed and expected six counts, and their difference.
3 marks
Supplied example die-simulation records; each row is a separate run
Trials 1 2 3 4 5 6
60 7 12 9 13 8 11
600 101 98 95 107 103 96
Expected six count: $600 \times \frac{1}{6} = 100$ .
Read the actual count 96.
$100 - 96 = 4$ .
96 observed; 100 expected; 4 fewer than expected.
A model chooses 1 to 10 equally and labels 1 to 4 as green. Find green chance and expected greens in 50 trials.
3 marks
Four of ten integers give green.
$50 \times \frac{2}{5} = 20$ .
Chance $\frac{4}{10} = \frac{2}{5}$ ; expected greens 20.
A simulation estimate is 0.6 for an event with theoretical chance 0.5. Must its coding be wrong? State one check and one alternative explanation.
3 marks
A correct model can give an unbalanced finite record.
Check that outcomes are assigned in the intended proportions.
Do not adjust recorded data to force an expected fraction.
Not necessarily. Check the random-outcome mapping; random variation can explain the difference.

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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.

Match the model to the chances

Preserve and interpret one run

See the method in action.

Compare two runs

Find the theoretical even-result chance.

Calculate the observed share in the sixty-trial run.

Calculate the observed share in the six-hundred-trial run.

Compare each estimate with one half.

A model for unequal colour chances

Choose ten equally likely random integers.

Assign three labels to red and the remaining seven to blue.

Check the model probabilities and repeat the same rule.

Try it yourself.

Check your understanding

Keep practising