Statistics, probability and decisions for HSC Maths Standard 2: checking claims with relative frequency, expected frequency, independent events and absolute versus relative risk (Year 12 Relative frequency and probability, 2024 syllabus)

Syllabus dot point

“Examine how statistics and associated probabilities influence decisions made by media, governments and companies, using relative frequency, expected frequency and the multiplication of probabilities for independent events”

10 min readUpdated 7 Oct 2026

Quick answer

Relative frequency estimates probability and improves with more trials. Expected frequency $np$ lets companies and governments plan, independent events multiply, and "at least one" uses the complement. To judge a claim, check the sample size, how the sample was chosen, and whether a change is quoted as relative or absolute. Part of Year 12 Relative frequency and probability in the 2024 Mathematics Standard syllabus.

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What this dot point is asking
The answer
Practice questions

What this dot point is asking

In the Mathematics Standard 11-12 Syllabus (2024), first examined in the 2027 HSC, Year 12 Standard 2 Relative frequency and probability finishes with a content point that asks you to examine how statistics and associated probabilities influence decisions made by media, governments and companies. It sits beside the calculation points (relative frequency, expected frequency $np$ , and $P(A \text{ and } B) = P(A) \times P(B)$ for independent multistage events), so questions typically give you a claim and ask you to calculate with it and then comment.

Note

Numbers in a headline, a budget or an advertisement have usually been chosen to make a point. Your job is to ask three questions. Where did the numbers come from (how many, and how chosen)? What is the calculation behind them? What has been left out? The probability rules you already know let you check the first two, and common sense on the third.

The answer

Key fact

Relative frequency $= \frac{\text{frequency}}{\text{trials}}$ estimates a probability, and it gets closer to the true probability as the number of trials increases.
Expected frequency $= np$ is the long-run average count, used by companies and governments to plan.
Independent events: $P(A \text{ and } B) = P(A) \times P(B)$ .
Complement: $P(\text{not } A) = 1 - P(A)$ , the quick route to "at least one".
Absolute versus relative: a change from $0.04\%$ to $0.08\%$ is a doubling (relative) but only $0.04$ percentage points (absolute).

Judging a claim

Sample size. Small samples give unreliable relative frequencies. A spinner that lands on red $7$ times in $20$ spins ( $0.35$ ) can have a true probability near $0.26$ , as $2000$ spins ( $520$ reds) suggest.
How the sample was chosen. A survey in a company's own shop is likely to favour the company.
Absolute and relative change. Check the starting level. Doubling a tiny risk is still a tiny risk.
Independence. Multiplying probabilities is only correct when events are independent.
Expected values for decisions. Insurers, governments and businesses plan on expected values $np$ because the average over many cases is predictable even when single cases are not.

How companies use expected frequency

An insurer with $12\,000$ policies and a claim probability of $0.03$ expects $360$ claims. At $8000 per claim, the expected cost is $2 880 000, or $240 per policy. A premium has to be higher than that to cover running costs and profit (for example $15\%$ more is $276).

Worked examples

Two independent machines

The probabilities of failure are $0.04$ and $0.05$ . Both fail with probability $0.04 \times 0.05 = 0.002$ , so in $5000$ days we expect $5000 \times 0.002 = 10$ such days. At least one fails with probability $1 - 0.96 \times 0.95 = 0.088$ .

Marker's note: show the complement as a single line, since it replaces three cases.

"Doubles the risk"

Risk goes from $4$ in $10\,000$ to $8$ in $10\,000$ . The relative increase is $100\%$ , but the absolute increase is $0.04\%$ , so across $250\,000$ people, about $100$ extra cases. Both numbers belong in a fair report.

Marker's note: for "why might this mislead", say that the relative figure hides how small the absolute risk is.

Small, biased samples

" $2$ out of $3$ prefer our brand" from $8$ of $12$ people asked in the company's own shop is a small and biased sample. A fair survey of $400$ with a true probability of $0.5$ would be expected to give $200$ preferring the brand.

Marker's note: name the problem (size or bias) and say how it affects the claim.

Common traps

Adding instead of multiplying: For independent events "A and B" means multiply.
Treating relative frequency from a few trials as the true probability: More trials give a better estimate.
Quoting only the relative change: Always find the absolute figures.
Using $np$ for a single outcome: Expected frequency is an average over many trials, not a promise for one.
Forgetting the complement: "At least one" is almost always $1 - P(\text{none})$ .

Exam technique

For a "comment on the claim" question, give one calculation that tests the claim and then one sentence of judgement that mentions sample size, bias or the starting level. Show every probability as a decimal or fraction and keep the full value until the end.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A spinner is spun

20

times and lands on red

7

times. It is then spun

2000

times and lands on red

520

times. Find the relative frequency of red for each set of trials, and say which is the better estimate of the probability of red.

Show worked solution →

Relative frequencies. $\frac{7}{20} = 0.35$ and $\frac{520}{2000} = 0.26$ .

Better estimate. $0.26$ , from $2000$ trials, because relative frequencies move closer to the true probability as the number of trials increases.

Marker's note: one mark for both values and one for the choice with the reason "more trials".

foundation2 marks

A machine fails on any given day with probability

0.04

. A second, independent machine fails with probability

0.05

. Find the probability that both fail on the same day.

Show worked solution →

Independent events. $P(\text{both}) = P(A) \times P(B) = 0.04 \times 0.05 = 0.002$ .

Marker's note: one mark for multiplying and one for $0.002$ (or $\frac{1}{500}$ ). Adding the probabilities is the usual error.

core3 marks

The machines in the previous question run for

5000

days. Find the expected number of days on which both fail, and the probability that at least one fails on a given day.

Show worked solution →

Expected days. Expected frequency $= np = 5000 \times 0.002 = 10$ days.

At least one fails. Use the complement: neither fails with probability $0.96 \times 0.95 = 0.912$ , so

P(\text{at least one}) = 1 - 0.912 = 0.088.

Marker's note: one mark for $10$ , one for $0.912$ , one for $0.088$ . The complement is quicker than adding three cases.

core3 marks

An insurance company has

12\,000

policies. Past records show

3\%

of policies result in a claim each year, and the average claim is $8000. (a) Find the expected number of claims. (b) Find the expected total cost of claims. (c) Find the minimum premium per policy that covers the expected cost, and the premium if the company adds

15\%

for running costs and profit.

Show worked solution →

(a) Expected claims: $np = 12\,000 \times 0.03 = 360$ .
(b) Expected cost: $360 \times 8000 = \$2\,880\,000$ .
(c) Premium: Break-even premium $= \frac{2\,880\,000}{12\,000} = \$240$ per policy. Adding $15\%$ : $240 \times 1.15 = \$276$ .

Marker's note: one mark for each part. The company prices on the expected value because individual claims are unpredictable but the average over many policies is stable.

exam4 marks

A news headline reads "New medication DOUBLES the risk of a rare side effect!" The study showed that the side effect occurred in

4

per

10\,000

people not taking the medication and

8

per

10\,000

people taking it. (a) Show the claim of "doubles" is correct as a relative change. (b) Find the increase in absolute risk as a percentage. (c) If

250\,000

people take the medication, how many extra cases would be expected? (d) Why might the headline mislead?

Show worked solution →

(a) Relative change: $\frac{8}{4} = 2$ , so the risk is multiplied by $2$ , an increase of $100\%$ .
(b) Absolute change: $\frac{8}{10\,000} - \frac{4}{10\,000} = 0.0008 - 0.0004 = 0.0004 = 0.04\%$ .
(c) Extra cases: $250\,000 \times 0.0004 = 100$ extra cases.
(d) Misleading: "Doubles" sounds alarming, but the risk goes from $0.04\%$ to $0.08\%$ , so the chance remains very small for any individual. Quoting the relative change without the absolute risk exaggerates the effect.

Marker's note: one mark for each part. Decisions by a government or a patient need both numbers.

exam4 marks

An advertisement says "

2

out of

3

people prefer our brand." The company asked

12

people in its own shop and

8

chose its brand. (a) Show the claim matches the sample. (b) A competitor says the claim is not reliable. Give two reasons. (c) If the true proportion who prefer the brand is

0.5

and

400

people are surveyed fairly, how many would you expect to prefer it?

Show worked solution →

(a) Matches: $\frac{8}{12} = \frac{2}{3}$ .
(b) Two reasons: (i) The sample is small: $12$ people gives a relative frequency that can differ a lot from the true probability. (ii) The sample is biased: it was taken in the company's own shop, so those asked are likely to favour the brand.
(c) Expected number: $np = 400 \times 0.5 = 200$ .

Marker's note: one mark for (a), one mark for each valid reason, one for (c).

Practise this

Sources & how we know this

Mathematics Standard 11-12 Syllabus (2024): Year 12 Standard 2 Relative frequency and probability — NESA (2024)
Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA