What is P1: Frequency trees?

A frequency is a count of how many times an outcome occurs.

How many marks are P1 questions worth?

In verified past-paper appearances, P1 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has P1 appeared in?

P1 has appeared in 7 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Record, describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees

Frequency trees

Worked answers, methods and verified real exam appearances for P1 on Edexcel GCSE Maths 1MA1.

Explanation

A frequency is a count of how many times an outcome occurs.
In a frequency table, all the frequencies should add to the total number of trials.
A frequency tree splits a total into groups: the child branches at every split must add back to their parent.
Complete a missing branch by subtraction, then use the finished tree to compare counts or calculate relative frequencies.
When groups have different totals, compare proportions rather than raw frequencies.

Worked example

A frequency tree starts with $70$ people. $43$ choose tea and the rest choose coffee. Of the tea drinkers, $18$ add sugar. Complete the missing frequencies.

1.Coffee drinkers: $70-43=27$ .
2.Tea drinkers without sugar: $43-18=25$ .
3.Check the first split: $43+27=70$ .

Answer: $27$ choose coffee and $25$ choose tea without sugar.

Common mistakes

Don't fall into the trap of writing probabilities on a tree that asks for frequencies.
Don't fall into the trap of subtracting a branch from the overall total instead of its parent.

Exam tip

Check every split by adding its two branches back to the parent frequency.

Worked practice

Tier 1 · Easy

A spinner is used

30

times. It lands on red

18

times and on blue

12

times. Complete a frequency table for the two outcomes and work out the relative frequency of red.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Red: $18$ ; blue: $12$ . Relative frequency of red $=\dfrac{18}{30}=0.6$ .	2	Record the observed counts in the table and check that $18+12=30$ . Divide the red frequency by the total number of spins: $18\div30=0.6$ .

Tier 2 · Standard

A frequency tree describes

80

visits to a kiosk. On

52

visits a hot drink is chosen and on the remaining visits a cold drink is chosen. A snack is also bought on

31

hot-drink visits and on

7

cold-drink visits. Complete all terminal frequencies and work out the percentage of visits on which a snack is bought.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Cold drink: $28$ ; hot drink and no snack: $21$ ; cold drink and no snack: $21$ . Snack percentage $=47.5\%$ .	3	The cold-drink branch has frequency $80-52=28$ . The no-snack frequencies are $52-31=21$ and $28-7=21$ . Snacks are bought on $31+7=38$ visits, so the percentage is $\dfrac{38}{80}\times100=47.5\%$ .

Tier 3 · Hard

Two weeks of trials are combined. In week 1,

72

120

trials succeed;

45

successful trials and

18

unsuccessful trials are fast. In week 2,

99

180

trials succeed;

66

successful trials and

27

unsuccessful trials are fast. Construct a combined frequency table for success or failure against fast or not fast. Compare the proportions that are fast in the two outcome groups.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Success: $111$ fast and $60$ not fast; failure: $45$ fast and $84$ not fast. Proportion of successful trials that are fast $=\dfrac{111}{171}\approx64.9\%$ ; proportion of failed trials that are fast $=\dfrac{45}{129}\approx34.9\%$ . A successful trial is more likely to be fast.	5	Combine like branches: successful trials total $72+99=171$ , of which $45+66=111$ are fast, so $60$ are not fast. Failures total $(120-72)+(180-99)=129$ , of which $18+27=45$ are fast, so $84$ are not fast. Compare within each outcome group: $111/171\approx0.649$ and $45/129\approx0.349$ .

Tier 1 · Easy

A weather station records one result on each of

28

days. There are

11

dry days and

9

wet days. All the other days are windy. Work out the number of windy days.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$8$ windy days.	1	The three frequencies must total $28$ . The windy frequency is $28-11-9=8$ . Check that $11+9+8=28$ .

Tier 2 · Standard

A frequency tree represents

96

sports-club members. Of the junior members,

28

train on weekdays and

17

train at weekends. Of the adult members,

33

train on weekdays. Work out the three missing frequencies.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Junior: $45$ ; adult: $51$ ; adult and weekend: $18$ .	3	The junior frequency is $28+17=45$ . The adult frequency is $96-45=51$ . The adult weekend branch is $51-33=18$ .

Tier 3 · Hard

A factory checks containers made by two machines. For Machine A,

21

containers are faulty and

259

are not faulty. For Machine B,

30

containers are faulty and

470

are not faulty. Work out the relative frequency of faulty containers for each machine and compare the results.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Machine A: $0.075$ ; Machine B: $0.06$ . Machine A has the greater relative frequency of faulty containers, by $0.015$ (or $1.5$ percentage points).	3	For Machine A, $21\div280=0.075$ . For Machine B, $30\div500=0.06$ . Since $0.075>0.06$ , Machine A has the greater relative frequency; the difference is $0.075-0.06=0.015$ .

Tier 2 · Standard

A wildlife centre records

180

animal sightings. There are

48

morning sightings. The afternoon frequency is

24

more than the morning frequency. All the other sightings are in the evening. Two thirds of the evening sightings are birds and the rest are mammals. Work out the missing frequencies and the relative frequency of an evening mammal sighting among all

180

sightings.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Afternoon: $72$ ; evening: $60$ ; evening birds: $40$ ; evening mammals: $20$ . Relative frequency of an evening mammal sighting $=\dfrac{20}{180}=\dfrac{1}{9}$ .	4	The afternoon frequency is $48+24=72$ , so the evening frequency is $180-48-72=60$ . Two thirds of $60$ is $40$ , leaving $20$ evening mammal sightings. Divide this frequency by all $180$ sightings to get $20/180=1/9$ .

Tier 3 · Hard

A travel survey records

250

journeys by car, bus or walking. There are

140

morning journeys and the rest are evening journeys. In total,

92

journeys are by car and

101

are by bus. In the morning,

56

journeys are by car and the bus frequency is twice the walking frequency. Complete a two-way frequency table for time of day against method of travel.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	Morning: car $56$ , bus $56$ , walking $28$ , total $140$ . Evening: car $36$ , bus $45$ , walking $29$ , total $110$ . Totals: car $92$ , bus $101$ , walking $57$ , all journeys $250$ .	5	Let the morning walking frequency be $x$ , so the morning bus frequency is $2x$ . Then $56+2x+x=140$ , giving $x=28$ and a morning bus frequency of $56$ . Subtract the morning entries from the column totals: evening car is $92-56=36$ and evening bus is $101-56=45$ . The evening total is $250-140=110$ , so evening walking is $110-36-45=29$ .

Tier 3 · Hard

A frequency tree records whether parcels are domestic or international and whether they are delayed. There are

42

international parcels. Of the domestic parcels,

18

are delayed. Of the international parcels,

9

are delayed. The relative frequency of a delayed parcel among all the parcels is

0.15

. Work out the total number of parcels and all the missing frequencies in the tree.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Total parcels: $180$ ; domestic parcels: $138$ . Domestic and not delayed: $120$ ; international and not delayed: $33$ .	4	There are $18+9=27$ delayed parcels. If $N$ is the total number, then $27/N=0.15$ , so $N=180$ . There are $180-42=138$ domestic parcels. The remaining terminal frequencies are $138-18=120$ and $42-9=33$ .

Q10

Tier 3 · Hard

A workshop tests

120

components and

78

pass. Components for a new order will be made in full trays of

80

. The workshop wants the expected number that pass to be at least

500

. Using the test result, decide the minimum number of full trays the workshop should make. Show that one fewer tray is not enough.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$10$ full trays (or $800$ components). $9$ trays give an expected $468$ passes, whereas $10$ trays give an expected $520$ passes.	4	The test gives a pass relative frequency of $78/120=13/20$ . One tray therefore gives an expected $80\times13/20=52$ passes. Nine trays give $9\times52=468$ , which is below $500$ , while ten trays give $10\times52=520$ . Therefore the minimum is $10$ full trays.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-11	1H	Q10	4	Non-calculator	Higher	QP MS
2019-06	1F	Q17	2	Non-calculator	Foundation	QP MS
2019-11	1F	Q15	4	Non-calculator	Foundation	QP MS
2023-06	2F	Q19	5	Allowed	Foundation	QP MS
2023-11	3F	Q15	5	Allowed	Foundation	QP MS
2021-11	2F	Q17	5	Allowed	Foundation	QP MS
2022-06	1F	Q17	5	Non-calculator	Foundation	QP MS

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