What is P7: Possibility spaces?

A possibility space lists every permitted outcome once, often as ordered pairs in a grid.

How many marks are P7 questions worth?

In verified past-paper appearances, P7 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has P7 appeared in?

P7 has appeared in 13 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Construct theoretical possibility spaces for single and combined experiments with equally likely outcomes and use these to calculate theoretical probabilities

Possibility spaces

Worked answers, methods and verified real exam appearances for P7 on Edexcel GCSE Maths 1MA1.

Explanation

A possibility space lists every permitted outcome once, often as ordered pairs in a grid.
If two experiments are independent, with $m$ and $n$ equally likely outcomes, the complete space has $mn$ equally likely ordered pairs.
Probability is then favourable pairs divided by all pairs.
Order matters when there is a first and second result: $(a,b)$ and $(b,a)$ are different.
Do not use favourable outcomes over total outcomes unless the combined outcomes are equally likely.

Worked example

Two fair six-sided dice are rolled. Use the possibility space to find the probability that the total is $9$ .

1.There are $6\times6=36$ equally likely ordered pairs.
2.The favourable pairs are $(3,6),(4,5),(5,4),(6,3)$ .
3.Probability $=\dfrac{4}{36}=\dfrac{1}{9}$ .

Answer: $\dfrac{1}{9}$ .

Common mistakes

Don't fall into the trap of counting $(3,6)$ and $(6,3)$ as one outcome.
Don't fall into the trap of using favourable outcomes over an incomplete possibility space.

Exam tip

Label both axes of a grid and check that it contains the expected $mn$ outcomes.

Worked practice

Tier 1 · Easy

A fair coin is tossed and a fair spinner labelled

1

2

3

is spun. Write the six outcomes as ordered pairs and find the probability of getting a head and an even number.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$(H,1),(H,2),(H,3),(T,1),(T,2),(T,3)$ . $P(\text{head and even})=\dfrac{1}{6}$ .	2	Pair each coin result with each spinner result. Only $(H,2)$ satisfies both conditions, so one of the six equally likely outcomes is favourable.

Tier 2 · Standard

Spinner A has equal sectors labelled

1

2

3

. Spinner B has equal sectors labelled

1

2

3

4

. Construct a possibility space and find the probability that the two scores have a sum greater than

5

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Possibility space: $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)$ . The favourable ordered pairs are $(2,4),(3,3),(3,4)$ . $P(\text{sum greater than }5)=\dfrac{3}{12}=\dfrac{1}{4}$ .	3	A $3$ by $4$ grid contains $12$ equally likely ordered pairs. Mark the entries whose sums exceed $5$ : $6,6,7$ at $(2,4),(3,3),(3,4)$ . Therefore the probability is $3/12=1/4$ .

Tier 3 · Hard

Two different cards are chosen in order from cards labelled

1

2

3

4

and

6

. The first card is the tens digit and the second is the units digit. Construct the theoretical possibility space and find the probability that the two-digit number is divisible by

3

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Possibility space: $12,13,14,16,21,23,24,26,31,32,34,36,41,42,43,46,61,62,63,64$ . There are $5\times4=20$ equally likely ordered outcomes. The favourable numbers are $12,21,24,42,36,63$ . Probability $=\dfrac{6}{20}=\dfrac{3}{10}$ .	4	Make a $5$ by $5$ grid and exclude its diagonal because the cards must be different. A number is divisible by $3$ when its digit sum is divisible by $3$ . Testing the $20$ allowed ordered pairs gives the six listed numbers, so the probability is $6/20=3/10$ .

Tier 1 · Easy

A fair five-sided spinner is labelled

2

3

5

8

and

10

. Write down its possibility space and work out the probability that it lands on a prime number.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	Possibility space: $\{2,3,5,8,10\}$ ; probability $=\dfrac{3}{5}$ .	2	The five equally likely outcomes are $2,3,5,8,10$ . Three of them, $2,3,5$ , are prime, so the probability is $3/5$ .

Tier 2 · Standard

Fair spinner A is labelled

1

4

and

7

. Fair spinner B is labelled

2

and

5

. Write down the possibility space of the total score and work out the probability that the total is even.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Totals: for A= $1$ : $3,6$ ; for A= $4$ : $6,9$ ; for A= $7$ : $9,12$ . $P(\text{even total})=\dfrac{3}{6}=\dfrac{1}{2}$ .	3	The six equally likely totals are $3,6,6,9,9,12$ . Three of these totals are even, so the probability is $3/6=1/2$ .

Tier 3 · Hard

Fair spinner A is labelled

1

2

and

5

. Fair spinner B is labelled

2

3

4

and

6

. The two scores are multiplied together. Write down the possibility space and work out the probability that the result is a multiple of

4

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Products: for A= $1$ : $2,3,4,6$ ; for A= $2$ : $4,6,8,12$ ; for A= $5$ : $10,15,20,30$ . $P(\text{multiple of }4)=\dfrac{5}{12}$ .	4	The $3$ by $4$ possibility space has $12$ equally likely products. The multiples of $4$ are $4$ in the first row, $4,8,12$ in the second row and $20$ in the third row. There are $5$ favourable outcomes, so the probability is $5/12$ .

Tier 2 · Standard

A fair coin is tossed and a fair spinner labelled

2

4

and

6

is spun. For a head, the score is one more than the spinner number. For a tail, the score is one less than the spinner number. Construct the possibility space for the score and work out the probability that the score is

3

5

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Scores: for H, $3,5,7$ ; for T, $1,3,5$ . $P(\text{score is }3\text{ or }5)=\dfrac{4}{6}=\dfrac{2}{3}$ .	3	Pair each coin result with each spinner result. The six equally likely scores are $3,5,7,1,3,5$ . Four outcomes give $3$ or $5$ , so the probability is $4/6=2/3$ .

Tier 3 · Hard

A numerator is chosen at random from

\{2,5,8\}

and a denominator is chosen independently at random from

\{3,6,10,12\}

. Each choice in a set is equally likely. Construct the possibility space of fractions. Work out the probability that the fraction is greater than

\dfrac{1}{2}

and the probability that its simplest form has denominator

3

. Which event is more likely?

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Possibility space: $\dfrac23,\dfrac26,\dfrac2{10},\dfrac2{12},\dfrac53,\dfrac56,\dfrac5{10},\dfrac5{12},\dfrac83,\dfrac86,\dfrac8{10},\dfrac8{12}$ . $P(\text{fraction}>\dfrac12)=\dfrac{7}{12}$ . $P(\text{simplest denominator is }3)=\dfrac{6}{12}=\dfrac{1}{2}$ . A fraction greater than $\dfrac12$ is more likely.	5	A $3$ by $4$ grid gives $12$ equally likely fractions. The seven fractions greater than $1/2$ are $2/3,5/3,5/6,8/3,8/6,8/10,8/12$ . After simplifying, six outcomes have denominator $3$ : $2/3,2/6,5/3,8/3,8/6,8/12$ . Compare $7/12$ with $6/12$ .

Tier 3 · Hard

A fair six-sided die is rolled once. If the die score is

n

, a game awards

n(8-n)

points. Construct the theoretical possibility space for the points awarded. Work out the probability that the points awarded are a multiple of

3

. Explain why the four different possible point scores are not equally likely.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	For $n=1,2,3,4,5,6$ , the point scores are $7,12,15,16,15,12$ . $P(\text{multiple of }3)=\dfrac{4}{6}=\dfrac{2}{3}$ . The scores $12$ and $15$ each come from two die results, while $7$ and $16$ each come from one.	4	Substitute each equally likely die result into $n(8-n)$ . Four of the six results give a multiple of $3$ : $12,15,15,12$ . Therefore the probability is $4/6=2/3$ . Repeated point scores have twice the probability of scores produced by only one face.

Q10

Tier 3 · Hard

A fair coin is tossed. After a head, a fair two-sided spinner labelled

2

and

4

is spun. After a tail, a fair three-sided spinner labelled

1

3

and

5

is spun. Draw a possibility tree and work out the probability that the spinner score is greater than

2

. Explain why counting three favourable scores out of the five listed outcomes would not give the correct probability.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Tree outcomes: H₂ and H₄ each have probability $\dfrac{1}{4}$ ; T₁, T₃ and T₅ each have probability $\dfrac{1}{6}$ . $P(\text{spinner score}>2)=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{7}{12}$ . The five outcomes are not equally likely because the head branch splits into two outcomes while the tail branch splits into three.	5	The coin gives probability $1/2$ to each first branch. Therefore H₂ and H₄ each have probability $1/2\times1/2=1/4$ , while T₁, T₃ and T₅ each have probability $1/2\times1/3=1/6$ . The favourable routes are H₄, T₃ and T₅, so add $1/4+1/6+1/6=7/12$ rather than using $3/5$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-06	2F	Q6	2	Allowed	Foundation	QP MS
2019-06	1H	Q22	5	Non-calculator	Higher	QP MS
2024-06	1F	Q13	3	Non-calculator	Foundation	QP MS
2022-11	1F	Q7	3	Non-calculator	Foundation	QP MS
2022-06	3F	Q10	4	Allowed	Foundation	QP MS
2024-11	2F	Q16	3	Allowed	Foundation	QP MS
2024-11	1F	Q11	5	Non-calculator	Foundation	QP MS
2019-06	3H	Q1	5	Allowed	Higher	QP MS
2019-06	3F	Q24	5	Allowed	Foundation	QP MS
2023-11	3H	Q16	3	Allowed	Higher	QP MS
2023-06	2F	Q15	3	Allowed	Foundation	QP MS
2024-06	1F	Q23	3	Non-calculator	Foundation	QP MS
2019-11	1F	Q11	2	Non-calculator	Foundation	QP MS

Other points in P Probability

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