Sample spaces & listing outcomes (product rule for counting) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator2 marks

Leo flips a coin and spins a fair three-sided spinner numbered 1, 2 and 3.

List all the possible outcomes. [2]

At least 3 correct outcomes, e.g. (H, 1), (H, 2), (T, 1) (ignore repeats for this mark)B1
All 6 outcomes listed with no repeats or extras: H1, H2, H3, T1, T2, T3 (accept any order and any clear notation, e.g. a table or 'Head, 1')B1
Answer: H1, H2, H3, T1, T2, T3
Examiner tip: Work systematically (all heads first, then all tails) so you do not miss or repeat an outcome.

2Grade 4–5FoundationNon-calculator4 marks

Spinner A is numbered 1, 2, 3, 4. Spinner B is numbered 1, 2, 3. Both spinners are fair.

Erin spins both spinners and multiplies the two numbers together to get her score. The grid shows some of the possible scores (rows: spinner A, columns: spinner B).

(a) Complete the grid. [1]

(b) Work out the probability that her score is an even number. [2]

(a) 6 (row 3, column 2), 4 (row 4, column 1) and 12 (row 4, column 3) all correctB1
(b) $\frac{a}{12}$ with $a<12$, or $\frac{8}{b}$ with $b>8$ (ft their completed grid)M1
(b) $\frac{8}{12}$ oe, e.g. $\frac{2}{3}$, 0.666… or 0.67 (ft their grid)A1
(c) $\frac{3}{12}$ oe, e.g. $\frac{1}{4}$ or 0.25 (ft their grid)B1
Answer: (a) row 3: 6 (under 2); row 4: 4 (under 1) and 12 (under 3) (b) $\frac{8}{12}=\frac{2}{3}$ (c) $\frac{3}{12}=\frac{1}{4}$
Examiner tip: Count from the completed grid — there are 12 outcomes, not 7 (4 + 3).

3Grade 4–5HigherNon-calculator2 marks

A café offers 4 starters, 6 main courses and 3 desserts.

A meal is made of one starter, one main course and one dessert.

Work out the number of different meals that can be chosen. [2]

$4 \times 6 \times 3$ (or e.g. $24 \times 3$ or $4 \times 18$)M1
72A1
Answer: 72
Examiner tip: Multiply the numbers of choices (product rule); adding them (4 + 6 + 3 = 13) scores 0.

4Grade 6–7HigherCalculator3 marks

A code is made of four digits chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

The first digit cannot be 0.
No digit can be used more than once.

Work out how many different codes are possible. [3]

9 choices for the first digit, e.g. $9 \times \ldots$ (or $10\times9\times8\times7 = 5040$ seen with intention to remove codes starting with 0)M1
$9 \times 9 \times 8 \times 7$ (or $5040 - 9\times8\times7$, i.e. $5040 - 504$)M1
4536A1
Answer: 4536
Examiner tip: After the first digit is chosen, 0 becomes available again — so the second digit also has 9 choices, not 8.

5Grade 4–5Non-calculator3 marks

Two fair six-sided dice are rolled and the scores are added together.

Work out the probability that the total is a prime number. [3]

Sample space of 36 outcomes drawn or listed (e.g. a grid of totals), or prime totals 2, 3, 5, 7, 11 identifiedM1
Counts the outcomes with a prime total, e.g. $1+2+4+6+2$ $(=15)$ (allow one counting error), or $\frac{a}{36}$ with $13 \le a \le 17$M1
$\frac{15}{36}$ oe, e.g. $\frac{5}{12}$ or 0.416… (accept 0.42)A1
Answer: $\frac{15}{36}=\frac{5}{12}$
Examiner tip: 1 is not prime, but 2 is — check both ends of your list.

6Grade 8–9HigherNon-calculator4 marks

How many even numbers between 100 and 999 have three different digits? [4]

Splits into cases: last digit 0, and last digit 2, 4, 6 or 8 (Alt: cases first digit odd and first digit even)P1
$9 \times 8 \times 1$ $(=72)$ for last digit 0 (Alt: $5 \times 8 \times 5$ $(=200)$ for first digit odd)P1
$8 \times 8 \times 4$ $(=256)$ for last digit 2, 4, 6 or 8 (Alt: $4 \times 8 \times 4$ $(=128)$ for first digit even)P1
328 caoA1
Answer: 328
Examiner tip: The first digit cannot be 0, so whether 0 has already been used as the last digit changes the count — that is why you need two cases.

7Grade 1–3FoundationNon-calculator5 marks

120 people visited a museum one day. 52 of them were children and the rest were adults.

18 of the children had visited the museum before. 41 of the adults had visited the museum before.

(a) Complete the frequency tree. [3]

(b) One of the 120 people is chosen at random. Work out the probability that this person had visited the museum before. [2]

(a) 68 (adults)B1
(a) 34 (children not visited)B1
(a) 27 (adults not visited) (ft their 68 − 41)B1
(b) $18+41$ $(=59)$ seen, or $\frac{a}{120}$ with $a<120$M1
(b) $\frac{59}{120}$ oe (accept 0.491… or 0.49)A1
Answer: (a) 68, 34, 27 (b) $\frac{59}{120}$
Examiner tip: Each pair of branches must add up to the number in the box they come from — check this before moving on.

8Grade 1–3FoundationNon-calculator3 marks

An ice-cream van sells four flavours: mint, vanilla, strawberry and chocolate.

Priya buys a cone with two scoops. The two scoops must be different flavours. The order of the scoops does not matter.

(a) List all the possible combinations of two flavours Priya could choose. [2]

(b) Priya chooses one of these combinations at random. Write down the probability that her combination includes chocolate. [1]

(a) At least 4 correct different combinations listed (allow initials, e.g. MV, MS, MC, VS, VC, SC)B1
(a) All 6 combinations listed with no repeats and no extras (e.g. MV and VM count as the same combination; a same-flavour pair such as MM is an extra)B1
(b) $\frac{3}{6}$ oe, e.g. $\frac{1}{2}$, 0.5, 50% (ft their list from (a) if at least 4 different combinations)B1
Answer: (a) MV, MS, MC, VS, VC, SC (b) $\frac{3}{6}=\frac{1}{2}$
Examiner tip: Work systematically: pair mint with each of the others, then vanilla with the ones left, and so on — students who list in a random order usually miss one or repeat one (e.g. MV and VM).

9Grade 4–5Non-calculator5 marks

$\xi$ = {whole numbers from 1 to 15}

A = {multiples of 3}

B = {factors of 30}

(a) Complete the Venn diagram to show all the numbers in $\xi$. [3]

(b) A number is chosen at random from $\xi$. Work out the probability that the number is in A or in B or in both. [2]

(a) 3, 6, 15 (only) in the overlap of A and BB1
(a) 9, 12 in A only and 1, 2, 5, 10 in B onlyB1
(a) 4, 7, 8, 11, 13, 14 outside both circles (and every number 1–15 placed exactly once)B1
(b) Counts the numbers inside the circles, e.g. $3 + 2 + 4 = 9$ or $5 + 7 - 3$, written as a numerator over 15, or $1 - \frac{6}{15}$ (ft their Venn diagram)M1
(b) $\frac{9}{15}$ oe, e.g. $\frac{3}{5}$, 0.6, 60%A1
Answer: (a) Overlap: 3, 6, 15; A only: 9, 12; B only: 1, 2, 5, 10; outside: 4, 7, 8, 11, 13, 14 (b) $\frac{9}{15}=\frac{3}{5}$
Examiner tip: Write each number in exactly once: 3, 6 and 15 belong in the overlap, so don't also write them in the 'A only' or 'B only' regions — and remember the numbers outside both circles.

10Grade 4–5Calculator4 marks

Ravi rolls two fair four-sided dice. Each dice is numbered 1, 2, 3 and 4.

His score is the difference between the two numbers (the larger number subtract the smaller number). If the two numbers are the same, his score is 0.

(a) Work out the probability that Ravi's score is 1. [2]

(b) Ravi rolls the two dice 80 times. Work out an estimate for the number of times his score is 0. [2]

(a) A correct sample space (e.g. a 4 by 4 grid of differences or a list of 16 outcomes) or 6 outcomes with a difference of 1 identified out of a total of 16M1
(a) $\frac{6}{16}$ oe, e.g. $\frac{3}{8}$, 0.375A1
(b) $\frac{4}{16} \times 80$ oe, e.g. $\frac{1}{4}\times 80$ (ft their sample space for P(score 0))M1
(b) 20 (do not accept 20 out of 80 or $\frac{20}{80}$ as the final answer)A1
Answer: (a) $\frac{6}{16}=\frac{3}{8}$ (b) 20
Examiner tip: There are 16 equally likely outcomes, not 4 — the scores 0, 1, 2 and 3 are not equally likely, so draw the full 4 by 4 grid; and in (b) the answer is a number of times (20), not a probability.

11Grade 4–5HigherNon-calculator3 marks

A sandwich shop lets customers make their own sandwich.

Each sandwich has one type of bread, one filling and one sauce.

There are 5 types of bread and 4 sauces. The shop says there are 160 different sandwiches that can be made.

(a) Work out how many fillings the shop has. [2]

(b) The shop now also allows customers to choose 'no sauce'. Nothing else changes.

Work out how many different sandwiches can now be made. [1]

(a) $160 \div (5\times4)$ oe, e.g. $160\div 20$ or $160 \div 5 \div 4$, or $5\times4\times n = 160$M1
(a) 8A1
(b) 200 (ft $5 \times$ their (a) $\times 5$)B1
Answer: (a) 8 (b) 200
Examiner tip: The product rule works backwards too: divide the total by the other choices — and 'no sauce' is an extra option, so there are now 5 sauce choices, not 4.

12Grade 6–7HigherCalculator3 marks

A company gives each customer a code.

Each code is 2 letters followed by some digits.

The letters are chosen from the 26 letters A to Z.
The digits are chosen from 0 to 9.
Letters and digits can be repeated.

The company needs at least 1 000 000 different codes. Every code must have the same number of digits.

Work out the smallest number of digits each code needs. You must show your working. [3]

$26 \times 26$ or 676 seen as the number of letter pairsM1
$676 \times 10^n$ worked out for at least one value of $n$, e.g. $676\,000$ for 3 digits or $6\,760\,000$ for 4 digits, or $1\,000\,000 \div 676$ (= 1479.2…)M1
4 digits, supported by $676\,000 \lt 1\,000\,000$ (3 digits not enough) and $6\,760\,000 \geq 1\,000\,000$, or by $10^3 \lt 1479.2… \leq 10^4$A1
Answer: 4 digits ($26^2 \times 10^3 = 676\,000$ is too few; $26^2 \times 10^4 = 6\,760\,000$ is enough)
Examiner tip: Use the product rule ($26\times26\times10\times10\ldots$), not addition — and to justify the smallest number you must show that one fewer digit gives too few codes.

13Grade 6–7HigherNon-calculator4 marks

Five friends, Amy, Ben, Cara, Dev and Eli, sit in a row of five seats.

(a) Amy and Ben must sit in the two end seats.

Work out the number of different ways the five friends can sit. [2]

(b) Another day, the five friends sit in the five seats in a random order.

Work out the probability that Amy and Ben sit in the two end seats. [2]

(a) $2 \times 3 \times 2 \times 1$ oe, e.g. $2 \times 6$ (2 ways for the ends, 6 ways for the middle three)M1
(a) 12A1
(b) $5\times4\times3\times2\times1$ or 120 seen as the total number of arrangements, or $\frac{2}{5}\times\frac{1}{4}$ oe (P(Amy and Ben at the ends) worked seat by seat)M1
(b) $\frac{12}{120}$ oe, e.g. $\frac{1}{10}$, 0.1 (ft their (a) over 120)A1
Answer: (a) 12 (b) $\frac{12}{120}=\frac{1}{10}$
Examiner tip: Amy and Ben can swap ends, so there are 2 ways for the ends — students who give 6 have forgotten to multiply by 2.

14Grade 8–9HigherNon-calculator4 marks

In a football league, every team plays every other team twice: once at home and once away.

Last season there were 132 matches in the league.

(a) Work out how many teams were in the league. [3]

(b) Next season, 2 more teams will join the league. The rule stays the same.

How many more matches will there be next season than last season? [1]

(a) Recognises that each team has $n - 1$ home matches (one against each other team), so the total is $n(n-1)$; may be seen as a correct total for any trial value, e.g. 10 teams gives $10 \times 9 = 90$M1
(a) Forms $n(n-1) = 132$ (or $n^2 - n - 132 = 0$), or finds two consecutive whole numbers with product 132, e.g. $12 \times 11$M1
(a) 12 (from correct working; ignore a negative solution of -11 if rejected)A1
(b) 50 (from $14 \times 13 - 132 = 182 - 132$; ft their $(n+2)(n+1) - 132$)B1
Answer: (a) 12 teams (b) 50 more matches
Examiner tip: Count matches with the product rule: each of the $n$ teams hosts each of the other $n-1$ teams once, giving $n(n-1)$ — dividing by 2 is wrong here because every pair meets twice.

15Grade 4–5FoundationNon-calculator4 marks

Maya flips three fair coins.

(a) List all the possible outcomes. [2]

(b) Write down the probability that she gets exactly two heads. [1]

“You can get 0, 1, 2 or 3 heads. That is four possibilities, so the probability of getting three heads is $\frac{1}{4}$.”

Explain why Maya is wrong. [1]

(a) At least 5 correct different outcomes listed (e.g. HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)B1
(a) All 8 outcomes listed with no repeatsB1
(b) $\frac{3}{8}$ oe (ft their list from (a) if at least 5 different outcomes)B1
(c) Explains that the four numbers of heads are not equally likely, e.g. '1 head can happen in 3 ways but 3 heads in only 1 way' or 'there are 8 equally likely outcomes so P(3 heads) = $\frac{1}{8}$'. Do not accept 'the probability is $\frac{1}{8}$' without reference to the 8 outcomes or to unequal likelihoodC1
Answer: (a) HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (b) $\frac{3}{8}$ (c) The numbers of heads 0, 1, 2, 3 are not equally likely; there are 8 equally likely outcomes and only 1 of them (HHH) gives three heads, so P(3 heads) = $\frac{1}{8}$.
Examiner tip: Treat each coin separately (HHT, HTH and THH are different outcomes) — counting 'two heads and a tail' as one outcome is why students get 4 outcomes instead of 8.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro