MR13 - Probability modelling and assumptions

MR13 - Probability modelling and assumptions

Build models for clearly defined populations, compare dependence and group assumptions, and judge when binomial or Normal distributions are unsuitable. MR13 asks you to connect the numerical prediction with the situation it is meant to describe, including the health records in the large data set.

Build a model for a defined target

Can the success rate stay at 0.70.7 for each of two contacts while the chance of two failures more than doubles? Yes. A pair of equal success rates leaves room for very different patterns of joint outcomes. We will compare complete models rather than trusting the headline rate alone.

A record of 42 successes in 60 contacts gives proportion 42/60=0.742/60=0.7. Define success, for example a reply within an agreed time, and classify every attempt once. Using the proportion for future contacts assumes the recorded people and conditions are relevant. Existing customers may reply differently from people who have never used the service.

You already know

Observed proportions depend on their denominator and selection rule. Independent probabilities multiply; a conditional branch uses the information from its earlier outcome. An expected frequency is not a guaranteed future count.

A model needs a target population, not just a decimal probability.

Empirical probability model

An empirical probability model assigns probabilities using observed proportions in a defined group. A uniform selection from that finite group has exactly those probabilities. Applying the proportions to a wider or future population requires evidence that the group represents it.

For the June 2027 course, OCR's LDS 9 contains 200 individuals aged 16 or over from the 2003–04 US NHANES health survey. Age is recorded for all 200. There are 57 individuals aged at least 60 and 43 aged at least 65. These counts describe the archived records; they are not current UK population estimates.

A recorded event proportion

For kk qualifying records out of n>0n>0 relevant records,

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

Define the event, the relevant records and the intended target before using the proportion. Missing observations cannot automatically be counted as “no”.

Not in the formula sheet: learn it.

We can use the LDS exactly for a selection experiment on its own rows, while keeping wider-population claims separate.

Worked example 1: select from the LDS records

Question. Select a record uniformly from the 200 LDS 9 individuals, replace it and select independently again. Calculate the probability that both recorded ages are at least 60.

Plan. Find the record proportion, then multiply under the stated independent replacement rule.

p=57/200qualifying records out of all recordsP(both)=(57/200)2independent selections=3249/40000=0.081225retain the exact value\begin{aligned} p&=57/200 &&\text{qualifying records out of all records}\\ \mathrm P(\text{both})&=(57/200)^2 &&\text{independent selections}\\ &=3249/40000=0.081225 &&\text{retain the exact value} \end{aligned}

Why this step? Every row is equally likely, and replacement allows the same row to be selected again without changing the probability.

Answer. The probability is 3249/400003249/40000, approximately 0.0810.081 to 2 significant figures.

Check. It is smaller than the single-record probability 0.2850.285. The denominator includes all 200 because none of these ages is missing.

Change only the threshold from 60 to 65. Keep the dataset and selection design fixed.

Worked example 2: a more restrictive age event

Question. Under the same design, calculate the probability that both recorded ages are at least 65.

Plan. Use the favourable count for the narrower age event.

p=43/20043 qualifying recordsP(both)=(43/200)2=1849/40000=0.046225evaluate the product\begin{aligned} p&=43/200 &&\text{43 qualifying records}\\ \mathrm P(\text{both})&=(43/200)^2\\ &=1849/40000=0.046225 &&\text{evaluate the product} \end{aligned}

Why this step? The population of selectable rows is unchanged; only the event definition changes.

Answer. The probability is 1849/400001849/40000, approximately 0.0460.046 to 2 significant figures.

Check. The event is a subset of the previous event, so its probability is smaller.

Representativeness must be argued again for a different target. These records come from an older US survey and form a selected educational subset; they do not by themselves describe UK adults today. A larger sample with the same mismatch would not remove that mismatch. For another variable, restrict to relevant non-missing records before estimating its proportion, and consider whether missingness could bias the result.

Watch out

Wrong: “The probability that any UK adult is at least 65 is 43/20043/200.” Right: this is the proportion in the 200 archived LDS records. Generalising needs a suitable population, time period and sampling design.

Use one more actual LDS count with the same selection method.

Compare a conditional model with independence

A dependence assumption changes complete paths even when the individual success probabilities stay fixed.

Let SiS_i denote a successful contact and Fi=Si′F_i=S_i' a failure. Two contacts might share a household or encounter the same system outage. Such a link can make an independence assumption doubtful, but numerical conditional probabilities still need relevant evidence.

A conditional refinement

A conditional refinement replaces a simplified probability rule by probabilities that depend on an earlier event. Its value is judged by the assumptions it addresses and the predictions it changes, not simply by calling it “more realistic”.

We will compare two models in which each contact has marginal success probability 0.70.7. The only structural change is how the two outcomes are linked.

Worked example 3: independent contacts

Question. Two independent contacts each succeed with probability 0.70.7. Calculate the four path probabilities and the probabilities of at least one failure and two failures.

Plan. Use unchanged success and failure probabilities on the second branches.

P(SS)=0.7(0.7)=0.49two successesP(SF)=0.7(0.3)=0.21second failsP(FS)=0.3(0.7)=0.21first failsP(FF)=0.3(0.3)=0.09two failures\begin{aligned} \mathrm P(SS)&=0.7(0.7)=0.49 &&\text{two successes}\\ \mathrm P(SF)&=0.7(0.3)=0.21 &&\text{second fails}\\ \mathrm P(FS)&=0.3(0.7)=0.21 &&\text{first fails}\\ \mathrm P(FF)&=0.3(0.3)=0.09 &&\text{two failures} \end{aligned}

Why this step? Independence allows the same second-stage probability regardless of the first result.

Answer. At least one failure has probability 1−0.49=0.511-0.49=0.51; two failures has probability 0.090.09.

Check. The four paths sum to 1, and the second-contact success probability is 0.49+0.21=0.70.49+0.21=0.7.

Change the dependence rule: use success probability 0.850.85 after success and 0.350.35 after failure, keeping the first probability 0.70.7. These branches preserve the second marginal, so the comparison isolates clustering rather than a changed headline success rate.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account