2.03e - Probability modelling and assumptions

2.03e - Probability modelling and assumptions

Build probability models for clearly defined selections, compare their predictions and explain what their assumptions leave out. You will use real census data, conditional paths and weighted groups to practise the modelling and criticism required by 2.03e.

An observed proportion can inform a model

Can two deliveries each have a one-in-five chance of being late, yet the chance of both being late triple? Yes: equal individual probabilities do not tell you whether failures cluster. We will build two complete models and see exactly where their predictions separate.

First ask what the model describes. A record with 80 on-time deliveries out of 100 suggests p=0.8p=0.8 for a comparable future delivery. It does not guarantee eight on-time arrivals in the next ten. If the record covers only short urban journeys, applying it to long rural journeys needs a new justification.

You already know

An observed proportion is a count divided by its relevant total. Independent probabilities multiply; conditional paths use the probability appropriate to the earlier outcome. Define the event and the population before calculating.

A useful model tells another person what is selected, how it is selected and what “success” means.

Probability model

A probability model specifies possible outcomes and their probabilities for a defined situation. An empirical probability uses observed relative frequencies. It is exact for a uniform selection from the recorded finite group, but using it to predict a different or future population involves an assumption that the record is relevant.

OCR's large data set contains 348 aligned local-authority areas in England and Wales. In its 2011 age-structure sheet, 229 areas have recorded median age at least 40, and 59 have median age at least 45. These are counts of areas, not counts of residents. Selecting an area uniformly gives each row the same weight even though populations differ.

An empirical proportion

For kk recorded outcomes satisfying an event among n>0n>0 relevant records,

p^=kn.\widehat p=\frac{k}{n}.

The event, denominator and target selection must match. Applying this as a probability for future cases requires representative, comparable evidence.

Not in the formula sheet: learn it.

The next pair uses actual 2011 census counts. The selections are a mathematical experiment on those rows, rather than a claim about present-day people.

Worked example 1: a uniform area selection

Question. Select an area uniformly from the 348 large-data-set rows, replace it, and select independently again. Find the probability that both areas have recorded 2011 median age at least 40.

Plan. Find the single-selection proportion, then use the specified independent replacement model.

p=229348229 qualifying area rowsP(both)=(229348)2independent selections=52441121104≈0.433evaluate at the end\begin{aligned} p&=\frac{229}{348} &&\text{229 qualifying area rows}\\ \mathrm P(\text{both})&=\left(\frac{229}{348}\right)^2 &&\text{independent selections}\\ &=\frac{52441}{121104}\approx0.433 &&\text{evaluate at the end} \end{aligned}

Why this step? Uniform area selection uses 348 as the denominator; replacing the selected row keeps that probability unchanged.

Answer. The probability is 0.4330.433 to 3 significant figures.

Check. The single-selection probability is approximately 0.6580.658. Requiring both gives a smaller probability, as expected.

Change only the event threshold from 40 to 45. Keep the same archived data and independent selection rule.

Worked example 2: a narrower event

Question. Under the same selection rule, find the probability that both areas have recorded 2011 median age at least 45.

Plan. Replace the favourable count by the count for the narrower event.

p=5934859 qualifying areasP(both)=(59348)2=3481121104≈0.02873 significant figures\begin{aligned} p&=\frac{59}{348} &&\text{59 qualifying areas}\\ \mathrm P(\text{both})&=\left(\frac{59}{348}\right)^2=\frac{3481}{121104}\\ &\approx0.0287 &&\text{3 significant figures} \end{aligned}

Why this step? The denominator describes the same population; only which rows count as favourable has changed.

Answer. The probability is 0.02870.0287.

Check. This event is a subset of the previous one, so its probability cannot be larger. Both probabilities refer to area-level medians in 2011.

Extending these proportions to today's authorities or to individual residents is a different model. Time, boundary definitions and unequal area populations matter. Collecting more similarly selected area records does not, by itself, turn an area proportion into a resident proportion.

Watch out

Wrong: “229/348229/348 of residents were at least 40.” Right: 229/348229/348 of the recorded areas had median age at least 40 in 2011. The statistical unit is an area and the variable is its median age.

Use one more actual threshold, with the same selection design.

A refined assumption changes the paths

Refining a model means replacing a simplifying assumption with a better-supported rule and checking what changes.

For deliveries, define on time as arrival by the agreed deadline, with every delivery classified once as on time or late. Shared traffic or weather might link two arrivals. Those possible causes motivate investigation; they do not supply numerical branch probabilities without evidence.

Marginal probability and dependence

A marginal probability concerns one event without conditioning on another. Equal marginal probabilities do not imply independence. A dependence model specifies how the chance on one branch changes when another event has occurred.

Let OiO_i mean delivery ii is on time and Li=Oi′L_i=O_i'. We compare models that both give each delivery marginal on-time probability 0.80.8.

Worked example 3: the independent delivery model

Question. Each of two deliveries is on time with probability 0.80.8, independently. Find all four path probabilities and the probabilities of at least one late arrival and both late.

Plan. Multiply each first-stage probability by the unchanged second-stage probability.

P(OO)=0.8(0.8)=0.64both on timeP(OL)=0.8(0.2)=0.16only second lateP(LO)=0.2(0.8)=0.16only first lateP(LL)=0.2(0.2)=0.04both late\begin{aligned} \mathrm P(OO)&=0.8(0.8)=0.64 &&\text{both on time}\\ \mathrm P(OL)&=0.8(0.2)=0.16 &&\text{only second late}\\ \mathrm P(LO)&=0.2(0.8)=0.16 &&\text{only first late}\\ \mathrm P(LL)&=0.2(0.2)=0.04 &&\text{both late} \end{aligned}

Why this step? Independence is what allows the same second-stage probabilities after either first result.

Answer. At least one is late with probability 1−0.64=0.361-0.64=0.36; both are late with probability 0.040.04.

Check. The four paths sum to 1, and the second on-time marginal is 0.64+0.16=0.80.64+0.16=0.8.

Change only the dependence rule. Keep the first on-time probability 0.80.8, but use conditional on-time probabilities 0.90.9 after on time and 0.40.4 after late. This alternative is constructed to preserve the second marginal.

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