Interpreting results | Year 9

01 / LEARN

Understand the idea.

State what the evidence describes

State what the data actually measure. For 32 successes in 100 trials, the relative frequency is $\frac{32}{100} = 0.32$ . Using it to predict future results assumes similar conditions; it does not determine the next outcome or guarantee 32 successes in the next hundred.

Before combining groups, check that they repeated the same experiment and used the same success definition. Add successes and trials separately. A group with more trials contributes more evidence, so unequal sample sizes should not receive equal weight.

Give a conclusion and its limit

A sound conclusion links an estimate, a comparison and a limitation. An observed rate of 0.32 is close to a model probability of $\frac{1}{3}$ , but closeness does not prove the model correct. Check the procedure and collect more comparable trials.

For independent trials with unchanged probabilities, past results do not change the next trial's chance. A losing streak does not make a win due. Long-run proportions can settle near a model value without the device compensating for earlier results.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Unequal sample sizes

Group A has 8 successes in 20 trials; B has 54 in 100. Compare rates and combine them.

Worked solution

1
Calculate each group's success rate.
$\begin{aligned} Group A: & \frac{8}{20} = 0.40 \\ Group B: & \frac{54}{100} = 0.54 \end{aligned}$
Comparing proportions accounts for the groups' different numbers of trials; B has the higher observed success rate.
2
Pool the successes and trials.
$\begin{aligned} Successes: & 8 + 54 = 62 \\ Trials: & 20 + 100 = 120 \end{aligned}$
A combined estimate treats every trial equally, assuming both groups ran the same experiment under comparable conditions.
3
Calculate the combined relative frequency.
$\frac{62}{120} = \frac{31}{60} \approx 0.517$
Divide the pooled number of successes by the pooled number of trials.
4
Check why averaging the two rates would be unsuitable.
$\frac{0.40 + 0.54}{2} = 0.47$
This gives the 20-trial group the same weight as the 100-trial group, instead of weighting each trial equally.

Answer

Combined estimate $31 / 60 \approx 0.517$ .

EXAMPLE 02

A losing streak

A fair independent coin gives T five times. Sam says H is more likely next. Assess this.

Worked solution

1
State the fair coin's next-toss probability.
$P (H next) = \frac{1}{2}$
Fairness gives heads and tails equal probabilities on each toss.
2
Apply the independence assumption to the streak.
$P (H next after five T) = \frac{1}{2}$
Independence means the five previous tails do not change the probability of the next result.
3
Assess Sam's claim.
$P (H next) = P (T next)$
Heads remains just as likely as tails; the coin does not compensate for the earlier streak.

Answer

The claim is unsupported; P(H next) remains $1 / 2$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A model predicts 25 successes per 100 trials on average. A sample gives 23. Must the model be wrong?

Show answer & reasoning

No.

No. An average prediction of 25 successes is an expected count, not a compulsory result. A count of 23 can arise through random variation, so that sample alone does not establish that the model is wrong.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

A game has win probability 0.2. After four losses, what is the chance of the next independent win?
1 mark
A device gives 48 successes in 150 trials. State the observed relative frequency.
1 mark
Two reports give 18 and 25 successes but no trial totals. What is missing for a fair rate comparison?
2 marks
Two groups report 12 successes in 40 trials and 28 in 60. Find the combined estimate.
2 marks
A simulation replaces counters, but the real game keeps the first out. Can its result be used unchanged?
3 marks
A model probability is $\frac{1}{3}$ . An experiment gives 31 successes in 90 trials. Write a conclusion with a limitation.
3 marks

Read the worked solutions

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

A game has win probability 0.2. After four losses, what is the chance of the next independent win?
1 mark
The unchanged independent trial has the same chance.
$0.2$
A device gives 48 successes in 150 trials. State the observed relative frequency.
1 mark
$\frac{48}{150} = 0.32$ .
$0.32$
Two reports give 18 and 25 successes but no trial totals. What is missing for a fair rate comparison?
2 marks
Counts alone do not show proportions.
The number of trials in each experiment.
Two groups report 12 successes in 40 trials and 28 in 60. Find the combined estimate.
2 marks
$12 + 28 = 40$ successes in $40 + 60 = 100$ trials.
$0.40$
A simulation replaces counters, but the real game keeps the first out. Can its result be used unchanged?
3 marks
Without replacement, the remaining counts change.
Update the collection within each trial.
No; it models different second-draw probabilities.
A model probability is $\frac{1}{3}$ . An experiment gives 31 successes in 90 trials. Write a conclusion with a limitation.
3 marks
$\frac{31}{90} \approx 0.344$ .
Compare with $0.333 \dots$ and recommend more correctly conducted trials rather than claiming proof.
The estimate is about 0.344, close to $\frac{1}{3}$ ; finite-sample variation remains.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

Year 9 compound-event sample spaces, probabilities, experiments and simulations.

Official curriculum source ↗

In this lesson

Understand the idea.

State what the evidence describes

Give a conclusion and its limit

See the method in action.

Unequal sample sizes

Calculate each group's success rate.

Pool the successes and trials.

Calculate the combined relative frequency.

Check why averaging the two rates would be unsuitable.

A losing streak

State the fair coin's next-toss probability.

Apply the independence assumption to the streak.

Assess Sam's claim.

Try it yourself.

Check your understanding

Keep practising