What is P2: Expected outcomes?

A random experiment has an uncertain individual result, even though its long-run behaviour can be predicted.

How many marks are P2 questions worth?

In verified past-paper appearances, P2 questions on Edexcel GCSE Maths 1MA1 have been worth 1–5 marks.

Which past exam papers has P2 appeared in?

P2 has appeared in 9 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Apply ideas of randomness, fairness and equally likely events to calculate expected outcomes of multiple future experiments

Expected outcomes

Worked answers, methods and verified real exam appearances for P2 on Edexcel GCSE Maths 1MA1.

Explanation

A random experiment has an uncertain individual result, even though its long-run behaviour can be predicted.
If outcomes are equally likely, probability is the number of favourable outcomes divided by the total number of outcomes.
For $n$ repeated trials, expected frequency is $n\times p$ .
This is a prediction, not a guarantee.
To judge whether a game is fair, compare each player's chances or calculate the expected winnings after any cost; equal-looking sectors or prizes do not by themselves prove fairness.

Worked example

A fair spinner has four equal sectors labelled A, A, B and C. Player 1 wins on A; Player 2 wins on B or C. The spinner is used $360$ times. Decide whether the game is fair and find each player's expected number of wins.

1. $P(\text{Player 1 wins})=\dfrac{2}{4}=\dfrac{1}{2}$ .
2. $P(\text{Player 2 wins})=\dfrac{2}{4}=\dfrac{1}{2}$ , so the game is fair.
3.Expected wins for each player $=360\times\dfrac{1}{2}=180$ .

Answer: The game is fair; each player is expected to win $180$ times.

Common mistakes

Don't fall into the trap of treating an expected frequency as the exact result that must occur.
Don't fall into the trap of calling a game fair without considering both probabilities and rewards.

Exam tip

Write the probability first, then multiply it by the number of future trials.

Worked practice

Tier 1 · Easy

A fair six-sided die is rolled

240

times. Work out the expected number of rolls on which the score is greater than

4

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$80$ rolls	2	The favourable scores are $5$ and $6$ , so the probability is $2/6=1/3$ . The expected frequency is $240\times\dfrac{1}{3}=80$ .

Tier 2 · Standard

A fair spinner has four equal sectors labelled

1

1

2

and

3

. It is spun

600

times. Work out the expected total of all the scores.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$1050$	3	The expected score per spin is $\dfrac{1+1+2+3}{4}=\dfrac{7}{4}$ . Over $600$ spins, the expected total is $600\times\dfrac{7}{4}=1050$ .

Tier 3 · Hard

A box contains

3

green,

2

yellow and

1

red token. A token is chosen at random and replaced. A player receives

\pounds4

for green,

\pounds1

for yellow and nothing for red. The entry fee is

\pounds2.20

. Work out the organiser's expected profit from

900

plays and decide whether the game is fair to the player.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Expected payout per play $=\pounds\dfrac{7}{3}$ . Expected organiser profit from $900$ plays $=-\pounds120$ . The game is not fair; the player has an expected gain of about $13.3$ pence per play.	5	The expected payout is $4\times\dfrac{3}{6}+1\times\dfrac{2}{6}=2+\dfrac{1}{3}=\dfrac{7}{3}$ pounds. The organiser receives $\pounds2.20=\pounds\dfrac{11}{5}$ per play, so expected profit per play is $\dfrac{11}{5}-\dfrac{7}{3}=-\dfrac{2}{15}$ pounds. Over $900$ plays this is $900\times(-2/15)=-120$ pounds. A fair entry fee would equal the expected payout, so this game favours the player.

Tier 1 · Easy

A fair coin is tossed

90

times. Work out an estimate of how many times the coin will land on tails. Give a reason why the actual number of tails may be different.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$45$ tails. An expected frequency is a prediction, so random variation may give a different result.	2	For a fair coin, $P(\text{tail})=\dfrac{1}{2}$ . The expected frequency is $90\times\dfrac{1}{2}=45$ , but individual results are random.

Tier 2 · Standard

A fair spinner has six equal sectors labelled A, A, A, B, C and D. It is spun

420

times. Work out an estimate of how many times it lands on A and an estimate of how many times it lands on B. Hence estimate the difference between these numbers.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	A: $210$ times; B: $70$ times; difference: $140$ .	3	The expected frequency of A is $420\times\dfrac{3}{6}=210$ . The expected frequency of B is $420\times\dfrac{1}{6}=70$ . The expected difference is $210-70=140$ .

Tier 3 · Hard

A fair spinner has five equal sectors, two red and three blue. In a contest, Ruby scores one point for red on each of her three spins. Ben scores one point for blue on each of his two spins. Work out each player's expected score and decide whether the contest is fair in terms of expected score.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	Ruby's expected score is $\dfrac{6}{5}$ and Ben's expected score is $\dfrac{6}{5}$ . The contest is fair in terms of expected score because their expected scores are equal.	4	Ruby scores on a spin with probability $2/5$ , so her expected score is $3\times\dfrac{2}{5}=\dfrac{6}{5}$ . Ben scores with probability $3/5$ , so his expected score is $2\times\dfrac{3}{5}=\dfrac{6}{5}$ . Equal expected scores make the contest fair in terms of expected score.

Tier 2 · Standard

A biased six-sided die has probability

0.08

of landing on

6

. The expected number of sixes in a planned experiment is

36

. Work out the number of rolls planned and the expected number of rolls that do not land on

6

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$450$ rolls are planned; the expected number of non-sixes is $414$ .	3	If $n$ rolls are planned, $0.08n=36$ , so $n=36\div0.08=450$ . The probability of a non-six is $0.92$ , giving an expected frequency of $450\times0.92=414$ ; equivalently, subtract the $36$ expected sixes from $450$ .

Tier 3 · Hard

A processing job is chosen at random from ten equally likely job cards. Five cards show a time of

2

minutes, three cards show

4

minutes and two cards show

x

minutes. The expected processing time is

3.4

minutes. Work out

x

. A new method reduces each

x

-minute job by

1.5

minutes. Work out the expected total processing time for

800

jobs after this change.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$x=6$ minutes; expected total processing time $=2480$ minutes.	4	Use the expected time: $\dfrac{5(2)+3(4)+2x}{10}=3.4$ . Hence $22+2x=34$ and $x=6$ . The reduced long-job time is $4.5$ minutes, so the new expected time is $\dfrac{10+12+2(4.5)}{10}=3.1$ minutes. For $800$ jobs, the expected total is $800\times3.1=2480$ minutes.

Tier 3 · Hard

A company routes each new request using a wheel with sectors of

72^\circ

108^\circ

and

180^\circ

for teams A, B and C respectively. The company expects

2500

requests. One worker can handle

125

requests. Teams A, B and C are assigned

4

6

and

9

workers respectively. Work out the expected number of requests for each team and decide whether every team has enough workers for its expected workload.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Expected requests: A $500$ , B $750$ , C $1250$ . No. Teams A and B have exactly enough capacity, but team C can handle only $1125$ requests, an expected shortfall of $125$ .	4	Use each sector angle as a fraction of $360^\circ$ . The expected numbers are $2500(72/360)=500$ , $2500(108/360)=750$ and $2500(180/360)=1250$ . The assigned capacities are $4(125)=500$ , $6(125)=750$ and $9(125)=1125$ , so only team C is short.

Q10

Tier 3 · Hard

A box holds

15

equally likely job cards labelled A, B or C. The number of A cards is twice the number of B cards. Over

840

selections with replacement, the expected number of A selections is

224

more than the expected number of B selections. Work out how many cards have each label and the expected number of C selections.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	A cards: $8$ ; B cards: $4$ ; C cards: $3$ . Expected C selections: $168$ .	4	Let the number of B cards be $b$ , so there are $2b$ A cards. The difference in their expected frequencies is $840(2b/15-b/15)=56b$ . Hence $56b=224$ , giving $b=4$ . There are $8$ A cards and $15-8-4=3$ C cards, so the expected C frequency is $840(3/15)=168$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-11	3H	Q19	3	Allowed	Higher	QP MS
2021-11	3H	Q6	4	Allowed	Higher	QP MS
2019-06	2F	Q16	5	Allowed	Foundation	QP MS
2024-11	1H	Q2	4	Non-calculator	Higher	QP MS
2023-11	3H	Q2	2	Allowed	Higher	QP MS
2022-11	1F	Q23	2	Non-calculator	Foundation	QP MS
2024-06	2F	Q11	1	Allowed	Foundation	QP MS
2023-11	3F	Q21	2	Allowed	Foundation	QP MS
2023-11	2F	Q11	2	Allowed	Foundation	QP MS

Other points in P Probability

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