Basic probability, expected outcomes, relative frequency – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator2 marks

A bag contains 5 red counters, 3 blue counters and 2 green counters. A counter is taken at random from the bag.

(a) Write down the probability that the counter is blue. [1]

(b) Work out the probability that the counter is not red. [1]

(a) $\frac{3}{10}$ oe (accept 0.3 or 30%)B1
(b) $\frac{5}{10}$ oe (accept $\frac{1}{2}$, 0.5 or 50%)B1
Answer: (a) $\frac{3}{10}$ (b) $\frac{1}{2}$
Examiner tip: Probabilities must be written as fractions, decimals or percentages — '3 out of 10' or '3:10' scores 0.

2Grade 1–3FoundationNon-calculator2 marks

A spinner can land on red, blue, green or yellow. The table shows the probability of some of the outcomes.

Colour	red	blue	green	yellow
Probability	0.25	0.3		0.15

Work out the probability that the spinner lands on green. [2]

$1-(0.25+0.3+0.15)$ or $1-0.7$M1
0.3 oeA1
Answer: 0.3
Examiner tip: The probabilities of all the outcomes add up to 1 — add the given ones first, then subtract from 1.

3Grade 4–5Non-calculator4 marks

Hana has a biased six-sided dice. The table shows the probability of some scores.

Score	1	2	3	4	5	6
Probability	0.1	0.15	0.2	0.25	$x$	$x$

(a) Work out the value of $x$. [2]

Hana is going to roll the dice 300 times.

(b) Work out an estimate for the number of times she will roll a 6. [2]

(a) $1-(0.1+0.15+0.2+0.25)$ $(=0.3)$ and $\div 2$, or $0.7+2x=1$M1
(a) 0.15 oeA1
(b) $300 \times$ their $x$M1
(b) 45 (ft their $x$ if $0A1
Answer: (a) $x=0.15$ (b) 45
Examiner tip: In (a) many students forget the remaining 0.3 is shared between TWO scores, so they give $x=0.3$.

4Grade 4–5Calculator4 marks

Ahmed drops a drawing pin 200 times. It lands point up 74 times.

(a) Work out the relative frequency of the pin landing point up. [1]

Bella drops the same pin 50 times. It lands point up 21 times.

(b) Whose results give the more reliable estimate of the probability of the pin landing point up? Give a reason for your answer. [1]

(c) Use all of Ahmed's and Bella's results to estimate how many times the pin would land point up if it were dropped 1000 times. [2]

(a) $\frac{74}{200}$ oe (accept 0.37)B1
(b) Ahmed, with reason referring to more trials / more drops (e.g. 200 > 50)C1
(c) $\frac{74+21}{200+50}$ $\left(=\frac{95}{250}=0.38\right)$ and $\times 1000$M1
(c) 380A1
Answer: (a) 0.37 (b) Ahmed — he did more trials (c) 380
Examiner tip: In (b) 'Ahmed because his is bigger' does not score — the reason must be about the number of trials.

5Grade 6–7HigherNon-calculator3 marks

A bag contains only red, blue and green counters.

The probability of taking a red counter at random is 0.35
The ratio of blue counters to green counters is $2:11$
There are 28 red counters in the bag.

Work out the number of green counters in the bag. [3]

$28 \div 0.35$ $(=80)$ or $1-0.35$ $(=0.65)$P1
$(80-28) \div 13 \times 11$ or $0.65 \div 13 \times 11$ $(=0.55)$, or $52 \div 13$ $(=4)$P1
44A1
Answer: 44
Examiner tip: Use the 28 red counters to find the total first — the ratio $2:11$ only applies to the non-red counters.

6Grade 6–7HigherCalculator4 marks

At a school fair, it costs 50p to play a game.

The probability of winning £5 is 0.04
The probability of winning £1 is 0.2
Otherwise, the player wins nothing.

200 people play the game once each.

Work out an estimate for the profit the school makes. [4]

$200 \times 0.04$ $(=8)$ or $200 \times 0.2$ $(=40)$, or expected winnings per game $0.04\times5$ $(=0.2)$ or $0.2\times1$ $(=0.2)$P1
$8 \times 5 + 40 \times 1$ $(=80)$ (expected prize money), or $0.04\times5+0.2\times1$ $(=0.4)$ per gameP1
complete method for profit, e.g. $200 \times 0.5 - 80$ or $200\times(0.5-0.4)$ (working in pence is fine: $10\,000-8000$)P1
£20 (accept 2000p)A1
Answer: £20
Examiner tip: Work in one unit throughout — mixing 50p with £5 is where most marks are lost.

7Grade 6–7HigherNon-calculator3 marks

A bag contains only red counters and blue counters.

The probability of taking a red counter at random is $\frac{2}{5}$

5 more red counters are put into the bag. The probability of taking a red counter at random is now $\frac{1}{2}$

How many counters were in the bag at the start? [3]

Red at start $=\frac{2}{5}n$ (or $2k$ red out of $5k$)M1
$\frac{\frac{2}{5}n+5}{n+5}=\frac{1}{2}$ oe, e.g. $2(2k+5)=5k+5$, or using the unchanged blue counters $\frac{3}{5}n=\frac{1}{2}(n+5)$M1
25A1
Answer: 25
Examiner tip: Remember to add the 5 extra counters to the total as well as to the red counters.

8Grade 1–3FoundationNon-calculator3 marks

A fair spinner has 8 equal sections numbered 1, 2, 3, 4, 5, 6, 7 and 8. The spinner is spun once.

(a) Choose the word from the list that best describes the probability that the spinner lands on an even number.

impossible unlikely evens likely certain [1]

The arrows P, Q, R and S are marked on the probability scale below.

(b) Which arrow shows the probability that the spinner lands on a number less than 7? [1]

(a) evens (accept even chance, 50-50, fifty-fifty, $\frac{1}{2}$)B1
(b) SB1
(c) Q (do not accept P: landing on 3 is possible, so the probability is not 0)B1
Answer: (a) evens (b) S (c) Q
Examiner tip: In (c), $\frac{1}{8}$ is small but not zero, so the arrow at 0 (impossible) is wrong.

9Grade 4–5FoundationNon-calculator5 marks

120 people watched a film at a cinema.

52 of the people were adults.
The rest of the people were children.
18 of the adults bought popcorn.
28 of the children did not buy popcorn.

(a) Use this information to complete the frequency tree. [3]

One of the people who bought popcorn is chosen at random.

(b) Work out the probability that this person is an adult. [2]

(a) 52 (adults) and 68 (children) in the first pair of boxesB1
(a) 18 (bought popcorn) and 34 (no popcorn) on the adults branchB1
(a) 40 (bought popcorn) and 28 (no popcorn) on the children branch (ft their 68 − 28 for the 40)B1
(b) denominator 58 (= 18 + 40, ft their tree), e.g. $\frac{\ldots}{58}$M1
(b) $\frac{18}{58}$ oe, e.g. $\frac{9}{29}$ (accept 0.31 or better; do not accept 18 : 58 or '18 out of 58' for this mark)A1
Answer: (a) adults 52 (popcorn 18, no 34); children 68 (popcorn 40, no 28) (b) $\frac{18}{58}=\frac{9}{29}$
Examiner tip: In (b) the person comes only from the popcorn buyers, so divide by 58, not by 120 or 52.

10Grade 4–5Calculator5 marks

Kai thinks his six-sided dice might be biased. He rolls it 200 times.

After every 20 rolls he works out the relative frequency of rolling a 6 so far. The graph shows his results.

(a) How many 6s did Kai roll in his first 20 rolls? [1]

(b) Write down the best estimate Kai can give for the probability of rolling a 6 with his dice. Give a reason for your answer. [1]

Kai is going to roll the dice another 500 times.

(d) Work out an estimate for the number of 6s he will roll. [2]

(a) 7B1
(b) 0.3 and reason, e.g. 'it is from the most rolls / all 200 rolls' or 'the graph has levelled off at 0.3' (accept answers in range 0.29 to 0.31 with reason)C1
(c) Yes, with reason comparing with a fair dice, e.g. 'a fair dice would give about $\frac{1}{6}$ ≈ 0.17, but 0.3 is much higher after 200 rolls' or '60 sixes is far more than the 33 or 34 expected from a fair dice' (do not accept 'yes' with no comparison with $\frac{1}{6}$, 0.17 or the expected number of 6s)C1
(d) $0.3 \times 500$ (ft their (b))M1
(d) 150 (ft their (b) × 500)A1
Answer: (a) 7 (b) 0.3, from the most rolls (c) Yes: 0.3 is far above $\frac{1}{6}\approx0.17$ (d) 150
Examiner tip: A reason about 'more rolls' needs to say which value it refers to; the estimate is the last point on the graph, not the first or an average of the points.

11Grade 1–3FoundationNon-calculator4 marks

A fair spinner has 5 equal sections numbered 1, 2, 3, 4 and 5.

(a) Write down the probability that the spinner lands on a number greater than 3. [1]

Leo is going to spin the spinner 150 times.

(b) Work out how many times Leo should expect the spinner to land on 4. [2]

Leo spins the spinner 150 times. It lands on 4 a total of 38 times.

Leo says, 'This proves the spinner is not fair.'

(a) $\frac{2}{5}$ oe (accept 0.4, 40%; do not accept '2 out of 5' or 2 : 5)B1
(b) $\frac{1}{5} \times 150$ oe, e.g. $150 \div 5$M1
(b) 30A1
(c) No, with reason, e.g. 'results vary each time, so you would not get exactly 30' or '38 is not far from 30' or 'he would need many more spins to be sure'C1
Answer: (a) $\frac{2}{5}$ (b) 30 (c) No — results vary randomly; 38 is close to 30 and does not prove bias
Examiner tip: An expected value is what happens on average: getting 38 instead of 30 is normal variation, not proof of bias.

12Grade 6–7HigherNon-calculator4 marks

A spinner can land on red, blue, green or yellow only.

The probability it lands on red is twice the probability it lands on blue.
The probability it lands on green is 0.1 more than the probability it lands on blue.
The probability it lands on yellow is 0.26

Kofi is going to spin the spinner 250 times.

Work out an estimate for how many more times the spinner will land on red than on green. [4]

forms an equation using the probabilities summing to 1, e.g. $2x + x + (x + 0.1) + 0.26 = 1$M1
$x = 0.16$ (P(blue)), or P(red) $= 0.32$ and P(green) $= 0.26$A1
$(0.32 - 0.26) \times 250$, or $0.32 \times 250$ $(=80)$ and $0.26 \times 250$ $(=65)$ (ft their $x$ if $0 < x < 0.25$)M1
15A1
Answer: 15
Examiner tip: Use the fact that the four probabilities must add up to 1 to find $x$ first, then remember the question asks for the difference, not the number of reds.

13Grade 6–7HigherNon-calculator3 marks

Mia flips a coin 40 times. The relative frequency of the coin landing heads is 0.45

Mia then flips the coin $n$ more times. The coin lands heads on every one of these $n$ flips.

The relative frequency of heads for all of Mia's flips is now 0.6

Work out the value of $n$. [3]

$0.45 \times 40$ $(=18)$ heads in the first 40 flipsM1
forms an equation, e.g. $\frac{18 + n}{40 + n} = 0.6$ oe, or a complete trial-and-improvement method showing $\frac{18+n}{40+n}$ evaluated for at least two values of $n$M1
$n = 15$ (a correct answer with no incorrect working scores 3 marks)A1
Answer: $n = 15$
Examiner tip: Both the number of heads and the total number of flips go up by $n$ — adding $n$ only to the top is the most common slip.

14Grade 4–5Calculator5 marks

A factory makes light bulbs. A sample of 250 bulbs is tested and 6 of them are faulty.

(a) The factory makes 12 000 bulbs in one day. Work out an estimate for the number of faulty bulbs made that day. [2]

(b) The manager says, 'Exactly this number of bulbs made today will be faulty.' Explain why the manager is wrong. [1]

The factory claims that at least 99% of its bulbs work.

(a) $\frac{6}{250} \times 12\,000$ oe, e.g. $12\,000 \div 250 \times 6$M1
(a) 288A1
(b) it is only an estimate based on a sample / the actual number could be different (accept 'the sample may not be representative')C1
(c) $\frac{244}{250}$ or $\frac{6}{250}$ converted to a percentage or decimal, e.g. 97.6% work or 2.4% faulty, or 99% of 250 $= 247.5$M1
(c) No, with correct comparison, e.g. 97.6% < 99% or 2.4% > 1% or 244 < 247.5C1
Answer: (a) 288 (b) It is only an estimate from a sample (c) No: only 97.6% of the sample worked, which is less than 99%
Examiner tip: In (c) a bare 'No' scores nothing — you must show the percentage that worked (or were faulty) and compare it with 99% (or 1%).

15Grade 6–7HigherCalculator5 marks

A bag contains 12 green counters and some counters that are not green.

Jake takes a counter at random from the bag, records its colour and puts it back. He does this 200 times.

He takes a green counter 48 times.

(a) Work out an estimate for the total number of counters in the bag. [3]

Jake now puts 10 more green counters into the bag.

(b) Use your answer to part (a) to estimate the probability that a counter taken at random from the bag is now green. [2]

(a) $\frac{48}{200}$ oe $(=0.24)$ as an estimate for P(green)M1
(a) $12 \div 0.24$ oe, or $\frac{12}{T} = \frac{48}{200}$ oeM1
(a) 50A1
(b) $\frac{12 + 10}{\text{their } 50 + 10}$M1
(b) $\frac{22}{60}$ oe, e.g. $\frac{11}{30}$ (accept 0.366… or 0.37)A1
Answer: (a) 50 (b) $\frac{22}{60} = \frac{11}{30}$
Examiner tip: In (b) the total goes up by 10 as well as the number of greens: $\frac{22}{50}$ is a common wrong answer.

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