Probability modelling: assumptions, simulation and validation

A probability model is a mathematical description of a random situation. It specifies possible outcomes or events and assigns probabilities to them. We use the model to calculate predictions, then compare those predictions with evidence.

A model is not the real process. It is a deliberately simplified representation:

$$\boxed{\text{real situation}\longrightarrow\text{assumptions}\longrightarrow\text{probability model}\longrightarrow\text{prediction}}$$

Good modelling therefore needs more than correct arithmetic. You must decide whether the assumptions are reasonable, interpret results in context, and recognise when evidence suggests that the model should be revised.

Prerequisites

You should be able to:

• use fractions, decimals and percentages;
• identify complements, unions and intersections using probability rules;
• multiply probabilities for independent events and add probabilities for mutually exclusive events;
• calculate a mean and interpret frequencies from data.

The probability modelling cycle

A useful modelling cycle has five stages.

1. Define the process. State what is observed and which event matters.
2. Make assumptions. For example, outcomes may be treated as equally likely or repeated trials as independent.
3. Construct and use the model. Assign probabilities and calculate the required prediction.
4. Compare with data. Decide whether discrepancies are plausible random variation or evidence of a weakness.
5. Refine or report. Change an assumption if necessary, or state the conclusion with its limitations.

The cycle matters because several different models can describe the same real situation. The best choice depends on purpose and on the available evidence.

Worked example 1: building a simple model

A cafe records whether its next customer buys tea, coffee or another drink. From recent data, the manager proposes

$$\operatorname{P}(T)=0.35, \qquad \operatorname{P}(C)=0.50.$$

The three categories are mutually exclusive and exhaustive, so their probabilities must sum to $1$. Therefore

$$\begin{aligned} \operatorname{P}(\text{other}) &=1-\operatorname{P}(T)-\operatorname{P}(C)\\ &=1-0.35-0.50\\ &=\boxed{0.15}. \end{aligned}$$

This model assumes that the recent data remain relevant. It may be poor on an unusually hot day, after a price change, or at a different time of day. The calculation can be flawless while the prediction is still unsuitable.

Assigning probabilities

Probabilities can come from theoretical reasoning, observed data, expert judgement, or a combination of these.

Equally likely outcomes

If a finite sample space contains $n$ equally likely outcomes and event $A$ contains $a$ of them, then

$$\boxed{\operatorname{P}(A)=\frac{a}{n}.}$$

The phrase equally likely is an assumption, not an automatic fact.

Worked example 2: when counting is justified

A fair six sided die is rolled. Find the probability of obtaining a prime number.

The sample space is

$$S=\{1,2,3,4,5,6\}.$$

The prime outcomes are $2,3,5$. Fairness makes all six faces equally likely, so

$$\operatorname{P}(\text{prime})=\frac{3}{6}=\boxed{\frac12}.$$

If the die were not known to be fair, merely counting three favourable faces would not justify the answer $1/2$.

Possible does not mean equally likely

The outcomes rain and no rain are two possibilities, but this does not make each probability $1/2$. Counting outcomes works only when the elementary outcomes are equally likely.

Experimental probability

If event $A$ occurs $r$ times in $n$ trials, its relative frequency is

$$\boxed{\widehat{p}=\frac{r}{n}.}$$

This observed value can be used as an estimate of the unknown probability $p$. The hat in $\widehat p$ indicates an estimate, not an exact population value.

Worked example 3: estimating from data

A drawing pin lands point upwards $138$ times in $200$ throws. Estimate the probability that it lands point upwards and predict the number of such results in the next $750$ throws.

The estimated probability is

$$\widehat p=\frac{138}{200}=0.69.$$

Using this estimate for future throws gives

$$750(0.69)=\boxed{517.5}.$$

The number of actual successes must be an integer, but $517.5$ is an expected frequency, not a promised outcome. It would be reasonable to report about $518$ point upwards results.

The prediction assumes that throwing conditions and the drawing pin do not change and that the $200$ recorded throws are representative.

Self-check 1

A machine produced $17$ faulty items among $400$ inspected items. Estimate the fault probability and predict the number of faulty items among the next $2400$ if conditions remain unchanged.

Answer

$$\widehat p=\frac{17}{400}=0.0425.$$

The predicted frequency is

$$2400(0.0425)=\boxed{102}.$$

This is a model based prediction. It does not say that exactly $102$ items will be faulty.

Long run behaviour and random variation

For repeated trials under stable conditions, relative frequency tends to become more stable as the number of trials increases. This is the law of large numbers:

$$\frac{X_n}{n}\longrightarrow p \quad\text{as }n\text{ becomes large},$$

where $X_n$ is the number of successes in $n$ trials and $p$ is the success probability.

This does not mean that the relative frequency moves closer to $p$ after every trial. It can move away temporarily. Nor does it mean that a success becomes more likely because several failures have just occurred.

Worked example 4: interpreting a short run

A fair coin gives $7$ heads in $10$ tosses. Is this evidence that the model $\operatorname{P}(H)=0.5$ is false?

The observed relative frequency is

$$\frac7{10}=0.7,$$

which differs from $0.5$. However, short sequences naturally vary. In fact, under the fair coin model,

$$\begin{aligned} \operatorname{P}(\text{exactly }7\text{ heads}) &=\binom{10}{7}\left(\frac12\right)^7\left(\frac12\right)^3\\ &=\frac{120}{1024}\\ &\approx0.117. \end{aligned}$$

This particular result is not especially rare. Ten tosses provide weak evidence about fairness. More observations would make the estimate more informative.

What becomes more stable?

The difference between the number of heads and $n/2$ need not shrink as $n$ grows. It is the proportion of heads, $X_n/n$, that tends to approach $1/2$.

Expected frequency

If an event has modelled probability $p$ in each of $n$ trials, its expected frequency is

$$\boxed{np.}$$

It is the long run average count over many repetitions of the whole set of $n$ trials. It need not be an integer and need not occur in any one set of trials.

Worked example 5: an expectation that is not possible

For a fair die rolled $20$ times, the expected number of sixes is

$$20\left(\frac16\right)=\frac{10}{3}\approx3.33.$$

It is impossible to observe $3.33$ sixes. The value means that if the experiment of $20$ rolls were repeated many times, the mean number of sixes per experiment would approach $10/3$.

Self-check 2

A seed has probability $0.82$ of germinating. Give the expected number that germinate from $150$ seeds. Explain why the actual number need not equal your answer.

Answer

$$150(0.82)=\boxed{123}.$$

The probability model describes long run behaviour. Random variation means one batch can produce more or fewer than $123$ germinations.

Assumptions that models commonly make

An assumption should be specific enough to test or criticise. Saying only that a model is unrealistic earns little credit.

Assumption	Mathematical consequence	A possible failure
Outcomes are equally likely	Probability can be found by counting	A spinner has unequal sectors or an off-centre pivot
Trials are independent	Joint probabilities can be multiplied	Weather on consecutive days is related
Success probability is constant	One value of $p$ applies throughout	A player’s fatigue changes their scoring chance
Categories are exhaustive	Their probabilities sum to $1$	An overlooked response category exists
Data are representative	Relative frequency estimates the target probability	Data come from one untypical time or group
Conditions remain stable	Past data can predict future behaviour	A process, population or policy changes

Independence and constant probability are different assumptions. A machine can retain a constant long run fault rate while faults occur in clusters, which violates independence.

Worked example 6: criticising an independence assumption

A commuter is late on $20\%$ of working days. A simple model treats lateness on different days as independent. Find the modelled probability that the commuter is late on both Monday and Tuesday, then assess the model.

Under the stated model,

$$\operatorname{P}(L_M\cap L_T) =\operatorname{P}(L_M)\operatorname{P}(L_T) =(0.2)(0.2) =\boxed{0.04}.$$

The independence assumption may be doubtful. Severe weather, engineering work, illness, or a disrupted rail service can affect both days. If lateness clusters, the true probability of two late days could exceed $0.04$.

The useful criticism identifies the assumption, gives a contextual reason it may fail, and states how the prediction could be affected.

Worked example 7: sampling without replacement

A batch contains $1000$ components, of which $40$ are faulty. Two are selected without replacement. A quick model treats the selections as independent with fault probability $0.04$.

The approximate probability that both are faulty is

$$(0.04)^2=0.0016.$$

The exact probability is

$$\frac{40}{1000}\times\frac{39}{999} \approx0.0015616.$$

The selections are not exactly independent because the first selection changes the batch. The approximation is nevertheless close because the sample of $2$ is tiny relative to the population of $1000$.

This illustrates an important principle: an assumption can be false in a literal sense but still give a useful approximation.

Simulation

A simulation imitates a random process using random numbers. It is valuable when exact calculation is difficult, when a model has several stages, or when we want to study its long run behaviour.

A valid simulation must include:

• a clear mapping from random numbers to outcomes;
• probabilities that match the proposed model;
• one complete definition of a trial;
• a large number of repetitions;
• a recorded quantity that answers the question.

If an event has probability $0.37$, two digit random integers from $00$ to $99$ could represent it by assigning $00$ to $36$ to the event and $37$ to $99$ to its complement. This uses exactly $37$ of the $100$ equally likely values.

Worked example 8: designing a simulation

A basketball player scores each free throw with probability $0.72$. Design a simulation to estimate the probability that the player scores at least $4$ of $5$ throws, assuming independence.

Use two digit random integers from $00$ to $99$.

1. Let $00$ to $71$ represent a score and $72$ to $99$ represent a miss.
2. Generate five integers. These represent one set of five throws.
3. Record a success for the set if at least four integers represent scores.
4. Repeat the set many times, say $10\,000$ times.
5. Estimate the required probability by

$$\frac{\text{number of successful sets}}{10\,000}.$$

The simulation itself does not establish that $0.72$ is the correct probability or that throws are independent. It explores the consequences if those assumptions hold.

For comparison, the exact probability under this model is

$$\begin{aligned} \operatorname{P}(X\geq4) &=\binom54(0.72)^4(0.28)+\binom55(0.72)^5\\ &\approx\boxed{0.577}. \end{aligned}$$

Simulation estimates fluctuate around this value. More repetitions usually reduce, but do not eliminate, simulation error.

Self-check 3

A customer buys a warranty with probability $0.13$. Describe a simulation using random digits $0$ to $9$ to estimate the probability that at least one of the next four customers buys one.

Answer

One digit cannot represent probability $0.13$ exactly, so combine digits in pairs to form equally likely integers $00$ to $99$.

• Assign $00$ to $12$ to a warranty purchase and $13$ to $99$ to no purchase.
• Generate four pairs of digits for one trial.
• Record whether at least one pair is from $00$ to $12$.
• Repeat many times.
• Divide the number of recorded successes by the number of trials.

The method assumes customers make independent decisions with constant purchase probability $0.13$.

Validating a model against data

Suppose a model assigns probabilities $p_1,p_2,\ldots,p_k$ to $k$ categories. In $n$ observations, the expected frequency in category $i$ is

$$E_i=np_i.$$

Compare expected and observed frequencies. Some difference is inevitable because of random variation. A close match supports the model’s usefulness but does not prove it true. A large or systematic discrepancy prompts questions about assumptions, data quality, or changing conditions.

Worked example 9: assessing a die model

A die is rolled $120$ times, giving the following results.

Score	$1$	$2$	$3$	$4$	$5$	$6$
Observed frequency	$14$	$18$	$17$	$20$	$23$	$28$

Under a fair die model, every expected frequency is

$$120\left(\frac16\right)=20.$$

The differences, observed minus expected, are

$$-6,\ -2,\ -3,\ 0,\ 3,\ 8.$$

The six appears more often and the one less often than expected, but discrepancies alone do not prove bias. A careful conclusion is:

The data show some departure from the fair die model, particularly for scores $1$ and $6$. This could be random variation, so more data or a formal goodness of fit procedure would be needed before concluding that the die is biased.

At A level, the key skill here is measured interpretation. Do not say that observed and expected frequencies should be identical.

Comparing candidate models

Model choice depends on the variable and mechanism, not only on the appearance of data.

Worked example 10: two models for demand

A bakery wants to model the number of loaves sold each weekday. It considers:

• Model A: every weekday has the same demand distribution;
• Model B: Mondays, Fridays and other weekdays have separate distributions.

Model A is simpler and needs fewer data. Model B can represent a genuine day effect but needs enough observations for each category. If Friday demand is consistently higher, Model A may systematically underpredict Fridays and overpredict quieter days.

A sensible process is to fit both models using earlier data, test their predictions on later data, and prefer the simpler model unless the added complexity produces a worthwhile improvement.

A strong modelling conclusion

State what the model predicts, name the key assumptions, compare with evidence, and qualify the conclusion. Phrases such as “under this model”, “approximately”, and “if conditions remain unchanged” are mathematically meaningful.

Common misconceptions

”There are two outcomes, so each has probability $1/2$”

False unless the two outcomes are equally likely. Winning the lottery and not winning are two outcomes with very different probabilities.

”The expected frequency is what will happen”

Expected frequency is a long run mean. Actual frequencies vary randomly.

”A larger sample removes all uncertainty”

A larger representative sample usually reduces random sampling error. It does not repair bias, poor measurement, dependence, or changing conditions.

”After five tails, a head is due”

For independent fair coin tosses,

$$\operatorname{P}(H\mid TTTTT)=\frac12.$$

Past tosses do not compensate for an imbalance. Believing that they must is the gambler’s fallacy.

”Agreement proves the model”

Different models can make similar predictions, especially with limited data. Agreement means the model may be adequate for a stated purpose, not that every assumption is literally true.

Mixed exam style example

A company claims that $90\%$ of parcels arrive the next day. In a sample of $80$ parcels, $68$ arrive the next day.

a. Find the observed relative frequency.

b. Find the expected number of next day arrivals under the company’s model.

c. State two assumptions needed when applying the model to future parcels.

d. Explain why the sample does not by itself prove the claim false.

Solution

a.

$$\widehat p=\frac{68}{80}=\boxed{0.85}.$$

b.

$$80(0.9)=\boxed{72}.$$

c. Suitable assumptions include:

• the sampled parcels are representative of the parcels covered by the claim;
• delivery conditions remain stable;
• one parcel’s arrival does not materially affect another’s;
• the meaning and recording of “next day” are consistent.

d. The observed count is $4$ below the expected count, but samples vary randomly. A formal test would measure how unusual $68$ or fewer is under the $p=0.9$ model. This leads to binomial hypothesis testing.

Final self-check

A spinner is claimed to land on red with probability $0.4$. It lands on red $93$ times in $250$ spins.

1. Find the experimental probability of red.
2. Find the expected red frequency under the proposed model.
3. State one assumption behind using the model for another spinner.
4. Explain briefly whether the result disproves the model.

Answer

1. The experimental probability is

$$\frac{93}{250}=\boxed{0.372}.$$

2. The expected frequency is

$$250(0.4)=\boxed{100}.$$

3. For example, the other spinner must be manufactured and operated under sufficiently similar conditions. Alternatively, the spins should be independent and the probability should remain constant.

4. No. The observed frequency differs from the expectation by $7$, but random variation is expected. The result is evidence to assess, not automatic disproof. More data or a formal probability calculation would be needed for a stronger conclusion.

Exam-ready checklist

You should now be able to:

• distinguish a probability model from the real process;
• assign probabilities from equally likely outcomes or relative frequencies;
• calculate and interpret expected frequencies;
• explain long run relative frequency without invoking the gambler’s fallacy;
• identify independence, constant probability, representativeness and stability assumptions;
• design a random number simulation with a correct outcome mapping;
• compare observed and expected results without overclaiming;
• state contextual limitations and suggest sensible refinements.

Next steps

• Learn how new information changes probabilities in conditional probability.
• Turn outcomes into numerical quantities in discrete random variables.
• Study repeated independent trials in the binomial distribution.
• Compare standard probability models in choosing a distribution.
• Use probability to judge evidence in hypothesis testing language.