Worksheets · Foundation and Higher

Probability language, frequency trees and expectation

8 exam-style questions, grades 1 to 6. Worked solutions and the marks are on the last page.

Question 1Non-calculator · 2 marks
(a) A counter is taken from a bag, its colour is recorded, and it is replaced. This is repeated 40 times. Red is recorded 17 times and blue 14 times. Every other result is yellow. How many times is yellow recorded? (2)
Question 2Calculator · 2 marks
(a) A fair eight-sided die has faces numbered 1 to 8. It is rolled 360 times. How many times would you expect a number greater than 5? (2)
Question 3Non-calculator · 2 marks
The probability that a train is on time is 0.35.
(a) Which word best describes the chance that the train is on time? (1)
1. Likely
2. Impossible
3. Unlikely
4. Even chance
(b) Work out the probability that the train is not on time. (1)
Question 4Non-calculator · 5 marks
A spinner lands on red 26 times and blue 22 times in 80 spins. All other results are yellow.
(a) Estimate the probability of yellow on the next spin. (3)
(b) Estimate the number of yellows in 500 further spins. (2)
Question 5Non-calculator · 4 marks
36 pupils are invited to a club. 27 attend, including 15 boys. Everyone attending is recorded as either a boy or a girl.
(a) Find the number of girls attending. (2)
(b) A pupil from the invited group is selected at random. Find the probability they attended. (2)
Question 6Non-calculator · 6 marks
A bag contains only red, green and blue counters. P(red) = 0.35 and P(green) = 0.4.
(a) Work out P(blue). (2)
(b) A counter is taken at random and replaced, 200 times. Estimate the number of times a green counter is taken. (2)
(c) There are 12 blue counters in the bag. How many counters are in the bag? (2)
Question 7Non-calculator · 6 marks
A drone landing system is tested 80 times. It lands in zone A 29 times and zone B 27 times. All other landings are in zone C.
(a) Estimate the probability of a zone C landing. (3)
(b) Estimate the number of zone C landings in 600 similar tests. (2)
(c) Why would 8000 tests under unchanged conditions usually give a more reliable probability estimate? (1)
Question 8Non-calculator · 4 marks
A biased dice is thrown. The probabilities of 1, 2, 3, 4 and 5 are each 0.15. At a fair, each throw costs 60p. A player who throws a 6 wins £2. Otherwise the player wins nothing.
(a) The dice is thrown 300 times. Work out an estimate of the organiser's profit. (4)

Worked solutions and marks

Question 1

(a) $9$

$40-17-14$
The frequencies of all outcomes add to the total number of trials.
Yellow frequency = 40 $-$ 17 $-$ 14 = 9.

P1 Establishing $40-17-14$ or an equivalent valid method.
A1 Correct answer: $9$

Question 2

(a) $135$

$3/8\times 360$
The favourable faces are 6, 7 and 8, so the probability is 3/8.
Expected frequency = number of trials $\times$ probability = 360 $\times$ 3/8 = 135.

P1 Establishing $3/8\times 360$ or an equivalent valid method.
A1 Correct answer: $135$

Question 3

(a) Unlikely

0.35 is between 0 and 0.5 on the probability scale.

B1 Correct answer: Unlikely.

(b) 0.65

On time and not on time cover every outcome, so their probabilities add up to 1: $1 - 0.35 = 0.65$ .

B1 0.65 or equivalent.

Question 4

(a) $2/5$

Find the unrecorded yellow frequency.
$80-26-22$
Divide yellow frequency by the number of trials.
$32/80$
Therefore $2/5$ .

M1 Find the unrecorded yellow frequency.
M1 Divide yellow frequency by the number of trials.
A1 Correct answer: $2/5$

(b) $200$

Multiply the estimated probability by the new trial count.
$500\times 0.4$
Therefore $200$ .

M1 Multiply the estimated probability by the new trial count.
A1 Correct answer: $200$

Question 5

(a) $12$

Subtract boys from the total attending.
$27-15$
Therefore $12$ .

M1 Subtract boys from the total attending.
A1 Correct answer: $12$

(b) $3/4$

Use the invited group, not only attendees, as the sample space.
$27/36$
Therefore $3/4$ .

M1 Use the invited group, not only attendees, as the sample space.
A1 Correct answer: $3/4$

Question 6

(a) $0.25$

The three probabilities add up to 1.
$1-0.35-0.4$
Therefore $0.25$ .

M1 The three probabilities add up to 1.
A1 Correct answer: $0.25$

(b) $80$

Multiply the probability by the number of trials.
$0.4\times 200$
Therefore $80$ .

M1 Multiply the probability by the number of trials.
A1 Correct answer: $80$

(c) $48$

Divide the number of blue counters by P(blue).
$12/0.25$
Therefore $48$ .

P1 Divide the number of blue counters by P(blue).
A1 Correct answer: $48$

Question 7

(a) $3/10$

Find the unlisted frequency for zone C.
$80-29-27$
Divide by the total number of tests.
$24/80$
Therefore $3/10$ .

M1 Find the unlisted frequency for zone C.
M1 Divide by the total number of tests.
A1 Correct answer: $3/10$

(b) $180$

Multiply the estimated probability by the new trial count.
$600\times 3/10$
Therefore $180$ .

M1 Multiply the estimated probability by the new trial count.
A1 Correct answer: $180$

(c) Random fluctuations usually have less effect on relative frequency in a larger sample.

Random fluctuations usually have less effect on relative frequency in a larger sample.

C1 Correct conclusion with supporting reasoning: Random fluctuations usually have less effect on relative frequency in a larger sample.

Question 8

(a) £30

The probabilities add up to 1.
$P(6) = 1 - 5 \times 0.15 = 0.25$
Expected number of 6s.
$0.25 \times 300 = 75$
Money in and money out.
$300 \times 0.60 = 180, \quad 75 \times 2 = 150$
$180 - 150 = 30$

P1 Finding $P(6) = 0.25$ .
P1 Expected number of wins, 75.
P1 Both totals: £180 taken and £150 paid out.
A1 Correct answer: £30.

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Probability language, frequency trees and expectation

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