What is S3.3: Modelling with probability?

A probability model simplifies a real process by specifying possible outcomes and assigning probabilities to them.

How many marks are S3.3 questions worth?

S3.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has S3.3 appeared in?

We have not yet indexed a verified past-paper appearance for S3.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

S3.3

Modelling with probability, including critiquing assumptions made and the likely effect of more realistic assumptions.

Draft — not yet indexed

Modelling with probability

Worked answers and methods for S3.3 on Edexcel A-level Maths 9MA0.

Explanation

A probability model simplifies a real process by specifying possible outcomes and assigning probabilities to them.
State assumptions explicitly, such as independence, constant probabilities, identical trials or equally likely outcomes, and judge them in context.
More realistic dependence or changing probabilities can alter both central probabilities and tail risks, so identify the likely direction of the effect where possible.
Validate a model by comparing its predictions with observed data; a close fit in one sample does not prove that its assumptions are true.

Worked example

A factory model treats defects in two items from the same batch as independent, each with probability $0.03$ . It therefore predicts probability $0.03^2$ that both are defective. Explain how batch-to-batch variation is likely to affect this prediction.

1.The model gives $P(\text{both})=0.03^2=0.0009$ .
2.If an unobserved batch condition raises the defect probability for both items, learning that one is defective increases the chance that the other is defective.
3.This positive dependence makes the simple independent model likely to underestimate the joint probability.

Answer: Items from the same poor-quality batch are positively associated rather than independent.; Therefore two defects together are likely to occur more often than the modelled probability $0.0009$ .

Common mistakes

Don't assume repeated trials are independent although the first outcome changes the conditions for later trials.
Don't accept independence because it simplifies calculation without considering shared batch conditions.

Exam tip

Critique a probability model by naming the dependence mechanism and stating the likely direction of bias.

Worked practice

Tier 1 · Easy

A model assigns probability

1/200

to each ticket winning a draw. State one assumption behind this model and one reason it might fail.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Assumption: all $200$ tickets are equally likely to be selected. It could fail if the mixing or selection mechanism favours some tickets.	2
Notes
Equal probabilities require a fair randomising process. Any systematic difference in ticket placement, shape or handling would undermine that assumption.

(2 marks)

Tier 2 · Standard

A weather model assumes that rain on successive days is independent and that the probability of rain each day is

0.2

. Find its probability of five consecutive dry days, then give one reason the independence assumption may be unrealistic and one possible refinement.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$0.8^5=0.32768$ A valid reason is that weather systems, pressure or seasonal conditions persist, so nearby days may be dependent and the rain probability may change. Use conditional probabilities, separate probabilities for different conditions, or a transition model whose probability depends on the previous day's weather.	3
Notes
A dry day has probability $1-0.2=0.8$ , so independence gives $P(\text{five dry days})=0.8^5=0.32768$ . In reality, pressure systems can make one day's weather informative about the next. A transition model could use separate probabilities of rain following a wet day and following a dry day.

(3 marks)

Tier 3 · Hard

A transport model assumes that each of

5

commuters independently chooses route A with probability

0.4

. Calculate the modelled probability that all

5

choose route A. A closure on route B can influence every commuter on the same day. Critique the independence assumption and state the likely effect on the probability just calculated.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Modelled probability $=0.01024$ . A shared closure creates positive dependence, so the model is likely to underestimate the probability that all $5$ choose route A.	5
Notes
Under the model, $P(\text{all choose A})=0.4^5=0.01024$ . A route-B closure is a common cause affecting all five choices, so the choices are not independent on that day. It makes simultaneous choices of route A more likely, so a model including road conditions would usually give a larger probability for this event.

(5 marks)

Tier 1 · Easy

A model treats two prize draws as independent and uses the same winning probability for both, although a prize is removed after the first draw. State which model assumption fails and why.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	The independence and constant-probability assumptions fail. Removing a prize changes the composition before the second draw, so its winning probability depends on the first result.	2
Notes
Without replacement, the first outcome changes both the number of prizes and the total number of entries available. The second-trial probability is therefore conditional on the first outcome.

(2 marks)

Tier 2 · Standard

A delivery model assumes that each parcel has a constant probability

0.10

of being late and that parcels are independent. It therefore predicts

5

late parcels in a group of

50

. On one day,

12

50

parcels are late after the same road closure. Critique the model and suggest one refinement.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	The common road closure makes lateness probabilities depend on conditions and can make parcel outcomes dependent. The constant- $0.10$ independent model may therefore underestimate the number of late parcels on disrupted days. Use different conditional probabilities for normal and disrupted conditions, estimated from data for each condition.	3
Notes
The prediction $50(0.10)=5$ relies on one stable probability and independent outcomes. A shared closure affects many parcels simultaneously, so both assumptions are doubtful. A condition-dependent model can separate ordinary days from disruption days.

(3 marks)

Tier 3 · Hard

A ferry is delayed with probability

0.10

on a calm day and

0.50

on a stormy day. A day is stormy with probability

0.20

, and the same weather applies to two crossings made that day. A simple model uses the overall delay probability for each crossing and treats the crossings as independent. Calculate this model's probability that both crossings are delayed. Then calculate the probability using the shared-weather model and comment on the independence assumption.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Overall delay probability $=0.18$ , so the independent model gives $0.18^2=0.0324$ . The shared-weather model gives $0.8(0.10^2)+0.2(0.50^2)=0.058$ . Shared weather creates positive dependence, so the independent model underestimates the probability of two delays.	5
Notes
First average over weather: $P(D)=0.8(0.10)+0.2(0.50)=0.18$ . Squaring gives $0.0324$ only if crossings are independent. With one weather state shared by both crossings, condition first: $P(D_1\cap D_2)=0.8(0.10)^2+0.2(0.50)^2=0.058$ . The larger value shows the effect of positive association from the common condition.

(5 marks)

Tier 2 · Standard

A spinner has four sectors with equal angles. A student claims that this proves each outcome has probability

0.25

, so the model does not need to be checked. Critique the claim and describe how repeated spins could be used to validate the model, including how close agreement should be interpreted.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Equal angles support an equal-probability model but do not prove it, because the pointer, pivot, construction or spinning method could favour some sectors. Carry out a large number of spins under consistent conditions and record the frequency of every outcome. Compare each observed relative frequency with $0.25$ and look for a persistent pattern of disagreement, repeating the investigation if necessary. Close agreement supports the model for those conditions but does not prove that its assumptions are true.	4
Notes
Equal sector angles are only one part of the random mechanism: friction, balance and how the spinner is released can alter outcome probabilities. Validation therefore needs repeated observations of all four outcomes under controlled conditions. Relative frequencies close to $0.25$ provide evidence that the model is useful in those conditions, but sampling variation and untested conditions prevent a claim of proof.

(4 marks)

Tier 3 · Hard

A parcel-damage model assigns probability

0.08

to every parcel, regardless of route length or loading shift. A validation report gives observed damage proportions of

0.07

for short-route day parcels,

0.14

for short-route night parcels,

0.16

for long-route day parcels and

0.27

for long-route night parcels; each group contains at least

200

parcels. Without carrying out further probability calculations, assess the model. Identify two features it omits, explain why checking only the overall damage proportion could be misleading, and describe how to refine and validate the model.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The model is reasonably close for short-route day parcels but underestimates damage in the other three groups, especially for long-route night parcels. It omits route length and loading shift; in the reported proportions each factor is separately associated with higher damage. An overall proportion depends on the mixture of the four groups, so agreement after pooling could conceal systematic underprediction within particular groups. Use separate condition-dependent damage probabilities for the route-and-shift groups, estimated from suitable data. Validate the refined model on fresh parcels by comparing predicted and observed proportions within every group, while retaining the group sample sizes and checking whether the discrepancies persist.	6
Notes
Compare the fixed model probability with the supplied subgroup results rather than averaging them immediately. The increasing proportions across night loading and longer routes suggest structured lack of fit rather than merely random disagreement around one common value. Pooling can mask that structure when the group mix changes. A condition-dependent model should be fitted using one data set and tested on fresh data for the same four groups, because rechecking only the fitting data would give weak validation.

(6 marks)

Tier 3 · Hard

A group contains

20

animals, of which exactly

5

carry a parasite. Four animals are selected without replacement. A simple model treats the selections as independent, each with parasite probability

0.25

. Calculate the modelled probability of selecting at least one carrier. Calculate the exact probability without replacement, and explain the direction of the model's error.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Independent model: $P(\text{at least one})=1-0.75^4=0.6836$ to $4$ decimal places. Exact probability: $1-\dfrac{15}{20}\dfrac{14}{19}\dfrac{13}{18}\dfrac{12}{17}=\dfrac{232}{323}=0.7183$ to $4$ decimal places. The independent model underestimates the probability of at least one carrier. After a non-carrier is selected, the carrier proportion among those remaining increases, so the constant-probability assumption is not valid.	6
Notes
The replacement-style model gives complement $0.75^4$ . Without replacement, the probability that all four selected animals are non-carriers is $(15/20)(14/19)(13/18)(12/17)=91/323$ . Its complement is $232/323$ . Removing non-carriers while following the no-carrier path makes a later carrier increasingly likely, so the independent calculation is too small.

(6 marks)

Q10

Tier 3 · Hard

10.

A model treats four attempts as independent with one constant success probability equal to the average

0.60

. It therefore models the probability of four successes as

0.60^4

. A fatigue-based refinement instead uses successive success probabilities

0.90

0.75

0.50

and

0.25

, while retaining independence conditional on these stated probabilities. Calculate both probabilities of four successes, compare them, and explain which assumption of the simple model the refinement changes and what further data would be needed to validate it.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Simple model: $0.60^4=0.1296$ . Refined model: $0.90(0.75)(0.50)(0.25)=0.084375$ . The simple model overestimates the four-success probability by $0.045225$ . The refinement removes the constant-probability assumption by allowing success probability to fall with attempt number; it does not by itself establish that attempts are independent. Estimate attempt-specific probabilities from repeated sequences and validate predicted sequence frequencies on fresh data, also checking for dependence between outcomes.	6
Notes
Under one constant probability, multiply four factors of $0.60$ . Under the refined specification, multiply the four different conditional success probabilities. Their difference is $0.1296-0.084375=0.045225$ . Evidence for fatigue requires replicated ordered attempts, not just an overall average, and validation should compare predictions with new sequences.

(6 marks)

Verified exam appearances

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Other points in S3 Probability

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