S3 Probability — revision question pack

3 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section S3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

S3.1 · Understand and use mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions.

Explanation

Mutually exclusive events cannot occur together, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$ . Independent events satisfy $P(A\cap B)=P(A)P(B)$ , equivalently $P(A\mid B)=P(A)$ when $P(B)>0$ .
Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ for any two events, subtracting the overlap to avoid double-counting. Events of positive probability cannot be both mutually exclusive and independent; the same rules apply to events defined from discrete or continuous random variables.
Independence gives $0.18=P(A)P(B)=0.45P(B)$ , so $P(B)=0.18/0.45=0.40$ . Then $P(A\cup B)=0.45+0.40-0.18=0.67$ .
State explicitly which event property justifies each probability equation in the solution.
For a continuous random variable, probability is represented by area under its density curve, so interval probabilities are integrals and the total area is $1$ .

Worked example

Events $A$ and $B$ are independent. Given $P(A)=0.45$ and $P(A\cap B)=0.18$ , find $P(B)$ and $P(A\cup B)$ .

1.Independence gives $0.18=P(A)P(B)=0.45P(B)$ , so $P(B)=0.18/0.45=0.40$ .
2.Then $P(A\cup B)=0.45+0.40-0.18=0.67$ .

Answer: $P(B)=0.40$ .; $P(A\cup B)=0.67$ .

Common mistakes

Don't add $P(A)$ and $P(B)$ for overlapping events without subtracting $P(A\cap B)$ .
Don't treat mutually exclusive events as independent, making their intersection both zero and a product.

Exam tip

State whether the events are independent or mutually exclusive before choosing the intersection and union formulae.

Tier 1 · Easy

1.
Events $A$ and $B$ are mutually exclusive, with $P(A)=0.38$ and $P(B)=0.27$ . Find $P(A\cup B)$ .
(2)
(Total for Question 1 is 2 marks)
2.
Events $C$ and $D$ satisfy $P(C)=0.6$ , $P(D)=0.5$ and $P(C\cap D)=0.3$ . Determine whether they are independent and whether they are mutually exclusive.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Events $A$ and $B$ are mutually exclusive, with $P(A)=0.3$ and $P(B)=0.4$ . Find $P(A\cup B)$ and determine whether $A$ and $B$ are independent.
(3)
(Total for Question 1 is 3 marks)
2.
Events $C$ and $D$ satisfy $P(C)=0.6$ , $P(D\mid C)=0.25$ and $P(D\mid C')=0.50$ . Find $P(D)$ and determine whether $C$ and $D$ are independent.
(4)
(Total for Question 2 is 4 marks)
3.
Events $A$ and $B$ are independent, with $P(A)=0.35$ and $P(B)=0.60$ . Show that $A'$ and $B$ are independent, and find $P(A\cup B')$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A discrete random variable $X$ has $P(X=0)=0.2$ , $P(X=1)=0.5$ and $P(X=2)=0.3$ . Independently, $Y$ is uniformly distributed on $0\leq y\leq5$ . Let $A$ be the event $X\geq1$ and $B$ the event $Y<2$ . Find $P(A\cap B)$ and $P(A\cup B)$ .
(5)
(Total for Question 1 is 5 marks)
2.
Events $A$ and $B$ are independent. Given that $P(B)=2P(A)$ and $P(A\cup B)=\frac{21}{25}$ , find the exact values of $P(A)$ and $P(A\cap B)$ .
(5)
(Total for Question 2 is 5 marks)
3.
A fair coin is tossed twice, so $HH,HT,TH,TT$ are equally likely. Let $A$ be the event that the first toss is a head, $B$ the event that the second toss is a head, and $C$ the event that both tosses give the same result. Show that each pair of events is independent, but that the three events are not mutually independent.
(6)
(Total for Question 3 is 6 marks)
4.
A discrete random variable $X$ has probabilities $P(X=1)=k$ , $P(X=2)=0.2$ , $P(X=3)=0.3$ and $P(X=4)=0.5-k$ , where all probabilities are non-negative. Let $A$ be the event $X\leq2$ and $B$ the event that $X$ is even. Find all values of $k$ for which $A$ and $B$ are independent. For these values, find $P(A\cup B)$ and state whether $A$ and $B$ are mutually exclusive.
(6)
(Total for Question 4 is 6 marks)
5.
The continuous random variable $X$ is uniformly distributed on $-2\leq x\leq3$ . For $-1<t<1$ , let $A$ be the event $X>t$ and $B$ the event $|X|<1$ . Find the value of $t$ for which $A$ and $B$ are independent. For this value of $t$ , find $P(A\cup B)$ .
(5)
(Total for Question 5 is 5 marks)

S3.2 · Understand and use conditional probability, including the use of tree diagrams, Venn diagrams and two-way tables; understand and use the conditional probability formula P(A|B) = P(A∩B)/P(B).

Explanation

Conditional probability restricts the sample space: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ for $P(B)>0$ . On a tree diagram, multiply probabilities along a path and add the probabilities of mutually exclusive paths that satisfy the event.
Without replacement, later branch probabilities change because both the total and the relevant category count may have changed.
In a Venn diagram or two-way table, use the condition as the denominator; reversing $P(A\mid B)$ to $P(B\mid A)$ is a common error.
There are two valid tree paths.
Amber then blue has probability $\frac58\cdot\frac37=\frac{15}{56}$ .

Worked example

A bag contains $5$ amber counters and $3$ blue counters. Two counters are taken without replacement. Find the probability that exactly one counter of each colour is taken.

1.There are two valid tree paths.
2.Amber then blue has probability $\frac58\cdot\frac37=\frac{15}{56}$ .
3.Blue then amber has probability $\frac38\cdot\frac57=\frac{15}{56}$ .
4.Adding gives $\frac{30}{56}=\frac{15}{28}$ .

Answer: $\frac{15}{28}$

Common mistakes

Don't multiply probabilities from mutually exclusive branches and add probabilities along a single tree path.
Don't use replacement probabilities on a without-replacement tree, leaving second-stage denominators unchanged.

Exam tip

Label every branch with the conditional probability after the first outcome, then add the required disjoint paths.

Tier 1 · Easy

1.
Of $60$ students, $24$ travel by bus and $15$ of those bus travellers arrive before $8{:}30$ . Find the probability that a randomly chosen bus traveller arrives before $8{:}30$ .
(2)
(Total for Question 1 is 2 marks)
2.
For events $A$ and $B$ , $P(A\cap B)=0.18$ and $P(B)=0.45$ . Find $P(A\mid B)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
In a group of $60$ students, $35$ study Mathematics, $28$ study Physics and $18$ study both. Find the probability that a randomly chosen Physics student studies Mathematics, and the probability that a randomly chosen student studies neither subject.
(4)
(Total for Question 1 is 4 marks)
2.
In a shipment, $2\%$ of devices are defective, $4\%$ are flagged by an automated check, and $1.5\%$ are both defective and flagged. Find the probability that a flagged device is defective and the probability that a defective device is flagged.
(4)
(Total for Question 2 is 4 marks)
3.
Machine A produces $60\%$ of a factory's components and machine B produces the rest. Their defective rates are $2\%$ and $5\%$ respectively. After learning that a randomly selected component is defective, calculate the conditional probability that its source was machine B.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
For events $A$ and $B$ , $P(A)=0.60$ , $P(B)=0.50$ and $P(A\mid B)=0.70$ . Find $P(A'\mid B')$ .
(5)
(Total for Question 1 is 5 marks)
2.
Events $A$ and $B$ satisfy $P(A)=0.4$ , $P(B\mid A)=0.7$ and $P(A\mid B)=0.5$ . Given that $P(B\mid A')=q$ , find $q$ and $P(A\cup B)$ .
(5)
(Total for Question 2 is 5 marks)
3.
Box A is selected with probability $0.7$ and box B with probability $0.3$ . Box A contains $4$ red and $2$ blue counters; box B contains $3$ red and $1$ blue counter. Two counters are drawn without replacement from the selected box and both are red. Find the probability that box B was selected.
(5)
(Total for Question 3 is 5 marks)
4.
In a group of $100$ people, $40$ use service $A$ and $30$ use service $B$ . Let $x$ be the number who use both services. Complete the four cell counts of a two-way table in terms of $x$ . Given that using $A$ and using $B$ are independent, find $x$ and then calculate $P(A'\mid B')$ .
(5)
(Total for Question 4 is 5 marks)
5.
A customer is a premium member with probability $0.30$ and otherwise is a standard member. A premium member renews with probability $0.80$ , while a standard member renews with probability $q$ . The overall renewal probability is $0.59$ . Find $q$ . Given that a customer does not renew, find the probability that the customer is a standard member. Two customers are then selected independently from this population; find the probability that exactly one renews.
(6)
(Total for Question 5 is 6 marks)

S3.3 · Modelling with probability, including critiquing assumptions made and the likely effect of more realistic assumptions.

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.

Tier 1 · Easy

1.
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)
2.
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 2 is 2 marks)

Tier 2 · Standard

1.
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 1 is 3 marks)
2.
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, $12$ of $50$ parcels are late after the same road closure. Critique the model and suggest one refinement.
(3)
(Total for Question 2 is 3 marks)
3.
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 3 is 4 marks)

Tier 3 · Hard

1.
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 1 is 5 marks)
2.
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 2 is 5 marks)
3.
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 3 is 6 marks)
4.
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 4 is 6 marks)
5.
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 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

S3.1 · Understand and use mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions.

Tier 1 · Easy

Mark scheme for S3.1 Tier 1 · Easy
Question	Scheme	Marks
1	$0.65$	2
(2 marks)		2
Notes
Mutually exclusive events have no overlap, so $P(A\cup B)=P(A)+P(B)=0.38+0.27=0.65$ .
2	$C$ and $D$ are independent because $P(C)P(D)=0.3=P(C\cap D)$ . $C$ and $D$ are not mutually exclusive because $P(C\cap D)\neq0$ .	2
(2 marks)		2
Notes
Independence requires the intersection probability to equal the product $0.6(0.5)=0.3$ , which it does. Mutual exclusivity requires a zero intersection, which it does not have.

Tier 2 · Standard

Mark scheme for S3.1 Tier 2 · Standard
Question	Scheme	Marks
1	$P(A\cup B)=0.7$ $A$ and $B$ are not independent.	3
(3 marks)		3
Notes
Mutual exclusivity gives $P(A\cap B)=0$ , so $P(A\cup B)=0.3+0.4=0.7$ . Independence would require $P(A\cap B)=P(A)P(B)=0.12$ . Since $0\neq0.12$ , the events are not independent.
2	$P(D)=0.35$ $C$ and $D$ are not independent.	4
(4 marks)		4
Notes
Split according to $C$ : $P(D)=0.6(0.25)+0.4(0.50)=0.35$ . If $C$ and $D$ were independent, $P(D\mid C)$ would equal $P(D)$ , but $0.25\neq0.35$ . Therefore they are not independent.
3	$P(A'\cap B)=0.39=P(A')P(B)$ , so $A'$ and $B$ are independent. $P(A\cup B')=0.61$	4
(4 marks)		4
Notes
Independence gives $P(A\cap B)=0.35(0.60)=0.21$ . Therefore $P(A'\cap B)=P(B)-P(A\cap B)=0.60-0.21=0.39$ . Also $P(A')P(B)=0.65(0.60)=0.39$ , proving independence. The complement of $A\cup B'$ is $A'\cap B$ , so $P(A\cup B')=1-0.39=0.61$ .

Tier 3 · Hard

Mark scheme for S3.1 Tier 3 · Hard
Question	Scheme	Marks
1	$P(A\cap B)=0.32$ . $P(A\cup B)=0.88$ .	5
(5 marks)		5
Notes
$P(A)=0.5+0.3=0.8$ . Since $Y$ is uniform, $P(B)=2/5=0.4$ . The variables are independent, so $P(A\cap B)=0.8(0.4)=0.32$ . Therefore $P(A\cup B)=0.8+0.4-0.32=0.88$ .
2	$P(A)=\dfrac{15-\sqrt{57}}{20}$ $P(A\cap B)=\dfrac{141-15\sqrt{57}}{100}$	5
(5 marks)		5
Notes
Let $P(A)=x$ , so $P(B)=2x$ and independence gives $P(A\cap B)=2x^2$ . The union formula gives $3x-2x^2=21/25$ , hence $50x^2-75x+21=0$ . Therefore $x=(15\pm\sqrt{57})/20$ . The plus sign gives $P(B)>1$ , so $P(A)=(15-\sqrt{57})/20$ . Substituting into $2x^2$ , or using $3x-21/25$ , gives $(141-15\sqrt{57})/100$ .
3	$P(A)=P(B)=P(C)=\frac12$ . $P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac14$ , equal to the relevant products, so every pair is independent. $P(A\cap B\cap C)=\frac14$ but $P(A)P(B)P(C)=\frac18$ . Therefore $A$ , $B$ and $C$ are pairwise independent but not mutually independent.	6
(6 marks)		6
Notes
Each event contains two of the four equally likely outcomes. Every pair intersects only in $HH$ , so each pair has intersection probability $1/4=(1/2)(1/2)$ . The intersection of all three events is also $\{HH\}$ and has probability $1/4$ , whereas mutual independence would require $(1/2)^3=1/8$ . Pairwise checks alone therefore do not establish mutual independence.
4	$k=0.2$ or $k=0.3$ $P(A\cup B)=0.7$ for either value $A$ and $B$ are not mutually exclusive because $P(A\cap B)=0.2>0$ .	6
(6 marks)		6
Notes
$P(A)=k+0.2$ , $P(B)=0.7-k$ and $P(A\cap B)=P(X=2)=0.2$ . Independence requires $(k+0.2)(0.7-k)=0.2$ , giving $k^2-0.5k+0.06=0=(k-0.2)(k-0.3)$ . Both roots make every listed probability non-negative. The union probability is $(k+0.2)+(0.7-k)-0.2=0.7$ . The positive intersection rules out mutual exclusivity.
5	$t=-\dfrac13$ $P(A\cup B)=\dfrac45$	5
(5 marks)		5
Notes
The uniform interval has length $5$ . For $-1<t<1$ , $P(A)=(3-t)/5$ , $P(B)=2/5$ and $P(A\cap B)=(1-t)/5$ . Independence gives $(1-t)/5=[(3-t)/5](2/5)$ , so $5-5t=6-2t$ and $t=-1/3$ . Then $P(A)=2/3$ , $P(B)=2/5$ and $P(A\cap B)=4/15$ , hence $P(A\cup B)=2/3+2/5-4/15=4/5$ .

S3.2 · Understand and use conditional probability, including the use of tree diagrams, Venn diagrams and two-way tables; understand and use the conditional probability formula P(A|B) = P(A∩B)/P(B).

Tier 1 · Easy

Mark scheme for S3.2 Tier 1 · Easy
Question	Scheme	Marks
1	$\frac{15}{24}=\frac58$	2
(2 marks)		2
Notes
The condition restricts the denominator to the $24$ bus travellers. Of these, $15$ arrived before $8{:}30$ , so the probability is $15/24=5/8$ .
2	$0.4$	2
(2 marks)		2
Notes
Use the conditional probability formula: $P(A\mid B)=P(A\cap B)/P(B)=0.18/0.45=0.4$ .

Tier 2 · Standard

Mark scheme for S3.2 Tier 2 · Standard
Question	Scheme	Marks
1	$P(M\mid P)=\dfrac{9}{14}$ $P(\text{neither})=\dfrac14$	4
(4 marks)		4
Notes
Among the $28$ Physics students, $18$ also study Mathematics, so $P(M\mid P)=18/28=9/14$ . The number studying at least one subject is $35+28-18=45$ , leaving $60-45=15$ studying neither. Thus $P(\text{neither})=15/60=1/4$ .
2	$P(D\mid F)=0.375$ $P(F\mid D)=0.75$	4
(4 marks)		4
Notes
$P(D\mid F)=P(D\cap F)/P(F)=0.015/0.04=0.375$ . Reversing the condition changes the denominator: $P(F\mid D)=0.015/0.02=0.75$ .
3	$P(B\mid D)=0.625$	4
(4 marks)		4
Notes
The joint probabilities are $P(A\cap D)=0.60(0.02)=0.012$ and $P(B\cap D)=0.40(0.05)=0.020$ . Hence $P(D)=0.012+0.020=0.032$ . Restricting the sample space to defective components gives $P(B\mid D)=0.020/0.032=0.625$ .

Tier 3 · Hard

Mark scheme for S3.2 Tier 3 · Hard
Question	Scheme	Marks
1	$0.50$	5
(5 marks)		5
Notes
From the conditional probability formula, $P(A\cap B)=P(A\mid B)P(B)=0.70(0.50)=0.35$ . Hence $P(A\cup B)=0.60+0.50-0.35=0.75$ , so $P(A'\cap B')=1-0.75=0.25$ . Also $P(B')=0.50$ . Therefore $P(A'\mid B')=0.25/0.50=0.50$ .
2	$q=\dfrac{7}{15}$ $P(A\cup B)=0.68$	5
(5 marks)		5
Notes
$P(A\cap B)=P(A)P(B\mid A)=0.4(0.7)=0.28$ . Since $P(A\mid B)=0.5$ , $P(B)=0.28/0.5=0.56$ . Total probability gives $0.56=0.4(0.7)+0.6q$ , so $q=0.28/0.6=7/15$ . Finally $P(A\cup B)=0.4+0.56-0.28=0.68$ .
3	$P(B\mid RR)=\dfrac{15}{43}$	5
(5 marks)		5
Notes
For box A, $P(RR\mid A)=\frac46\frac35=\frac25$ , so $P(A\cap RR)=\frac7{10}\frac25=\frac7{25}$ . For box B, $P(RR\mid B)=\frac34\frac23=\frac12$ , so $P(B\cap RR)=\frac3{10}\frac12=\frac3{20}$ . Therefore $P(RR)=\frac7{25}+\frac3{20}=\frac{43}{100}$ , and $P(B\mid RR)=(3/20)/(43/100)=15/43$ .
4	The cell counts for $(A\cap B),(A\cap B'),(A'\cap B),(A'\cap B')$ are $x,40-x,30-x,30+x$ . $x=12$ $P(A'\mid B')=\dfrac{42}{70}=0.6$	5
(5 marks)		5
Notes
The margins give the remaining cells by subtraction, and the neither cell is $100-[x+(40-x)+(30-x)]=30+x$ . Independence requires $x/100=(40/100)(30/100)$ , so $x=12$ . There are $70$ people in $B'$ , of whom $30+12=42$ are also in $A'$ , giving $42/70=0.6$ .
5	$q=0.50$ $P(\text{standard}\mid\text{not renew})=\dfrac{35}{41}$ $P(\text{exactly one renewal})=0.4838$	6
(6 marks)		6
Notes
Total probability gives $0.59=0.30(0.80)+0.70q$ , so $q=0.50$ . The joint probability of being standard and not renewing is $0.70(0.50)=0.35$ , while $P(\text{not renew})=0.41$ , so the conditional probability is $35/41$ . For two independently selected customers, exactly one renewal has probability $2(0.59)(0.41)=0.4838$ .

S3.3 · Modelling with probability, including critiquing assumptions made and the likely effect of more realistic assumptions.

Tier 1 · Easy

Mark scheme for S3.3 Tier 1 · Easy
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
(2 marks)		2
Notes
Equal probabilities require a fair randomising process. Any systematic difference in ticket placement, shape or handling would undermine that assumption.
2	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
(2 marks)		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.

Tier 2 · Standard

Mark scheme for S3.3 Tier 2 · Standard
Question	Scheme	Marks
1	$0.8^5=0.32768$ 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
(3 marks)		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.
2	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
(3 marks)		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	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
(4 marks)		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.

Tier 3 · Hard

Mark scheme for S3.3 Tier 3 · Hard
Question	Scheme	Marks
1	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
(5 marks)		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.
2	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
(5 marks)		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.
3	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
(6 marks)		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.
4	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
(6 marks)		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.
5	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
(6 marks)		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.