The idea, step by step
Separate a prediction from a guarantee
A probability quantifies uncertainty under assumptions. An expected count gives a long-run average; a tail probability describes how often a specified threshold is crossed. State the time period, unit and event so that the numerical result answers the actual question.
For X counting faults in a sample, E(X) = 2 does not mean every sample contains two faults. If a decision depends on whether any fault occurs, P(X ≥ 1), not E(X), is the relevant number. If a decision depends on exceeding capacity, calculate the corresponding tail.
Keep the model open to challenge
Compare results with what is plausible in context. Check that the trial count, success probability and independence assumptions remain suitable. A small observed probability is evidence that an event is unusual under a model, not proof that the event cannot happen or that a different model must be true.
Across m comparable repetitions, the expected number of times an event A happens is mP(A). This expected-frequency rule does not make the actual count exact. Report rounding honestly and do not interpret a rounded displayed zero as a mathematically impossible event.
A common trap
A tail probability of 0.02 means the model is wrong if the event happens.
Instead: The model itself assigns that event a 2% chance. Investigate assumptions and accumulating evidence rather than declaring a single possible event impossible.
See the thinking
Zero is possible even when the mean is positive
The daily count X is modelled by Bin(10, 0.2). A manager claims there must be at least one success because E(X) = 2. Assess the claim.
- The expected count is 10 × 0.2 = 2, but it is an average.
- P(X = 0) = 0.8¹⁰ = 0.1073741824.
- A day with zero successes is possible and has probability about 10.7%; the claim is false.
Expected number of threshold crossings
Each of 50 independent groups has 5 independent trials with success probability 0.2. How many groups are expected to have at least one success?
- Within a group, P(at least one) = 1 − 0.8⁵ = 0.67232.
- Multiply this event probability by 50 groups.
- The expected number is 50 × 0.67232 = 33.616; the observed number must be an integer.
Check your understanding
An event has model probability 0.12. In 200 repetitions, how many occurrences are expected?
Show answer
24.
Expected count = 200 × 0.12; the actual count can differ.
Your turn
6 questions · 13 marks · Work at your own pace.
Read the questions online
- A sample fault count has mean 1.8. Explain whether any sample can contain exactly 1.8 faults.2 marks
- A model gives probability 0.04 that demand exceeds capacity. Find the expected number of exceedances in 250 comparable days.2 marks
- . Is observing X = 0 inconsistent with the model? Quantify your answer.3 marks
- Two systems have equal expected daily cost. Explain why that alone does not establish which one a risk-averse manager should choose.2 marks
- A binomial probability is displayed as 0.0000 after rounding. Must the event be impossible?2 marks
- A return rate estimated last year is used for this year’s changed product range. Give a specific limitation of a binomial forecast using the old rate.2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- A sample fault count has mean 1.8. Explain whether any sample can contain exactly 1.8 faults.2 marks
Expectation is a weighted average across possible counts.
It is not an individual sample outcome.
No; the count is an integer, while its mean need not be. - A model gives probability 0.04 that demand exceeds capacity. Find the expected number of exceedances in 250 comparable days.2 marks
Expected count = 250 × 0.04 = 10.
This is an average over repetitions, not a fixed schedule.
10 days. - . Is observing X = 0 inconsistent with the model? Quantify your answer.3 marks
Zero is in the support.
Its probability is about 59%, so it is not even unusual under the model.
No; P(X = 0) = 0.9⁵ = 0.59049. - Two systems have equal expected daily cost. Explain why that alone does not establish which one a risk-averse manager should choose.2 marks
Equal means match only one summary.
The manager needs the risk criteria and further distribution information.
Their variability and high-cost tail probabilities may differ. - A binomial probability is displayed as 0.0000 after rounding. Must the event be impossible?2 marks
A positive probability below 0.00005 rounds to 0.0000 at four decimal places.
Use more precision or scientific notation when the distinction matters.
No. - A return rate estimated last year is used for this year’s changed product range. Give a specific limitation of a binomial forecast using the old rate.2 marks
The binomial calculation is conditional on its p value.
Changed products can alter the return probability, so investigate updated data before trusting the forecast.
The success probability may no longer be appropriate or constant.
Curriculum reference & lesson scope
SACE Board of South Australia
Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.
Official curriculum source ↗