MathsICSE Class 10 · Probability

Dice and Coins Probability: Sample Spaces, Sums and Products (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Write the sample space first: one die has 6 outcomes, two coins have 4 (HH, HT, TH, TT) and two dice have 36 ordered pairs. Count the favourable outcomes in that list and divide. Treat (2, 5) and (5, 2) as different outcomes, just as HT and TH are different.

Understand Dice and Coins Probability

Coins and dice are the classic random experiments in the CISCE 2027 syllabus topic 'random experiments, sample space, events … simple problems on single events'. Throwing two dice together is one experiment with 36 outcomes, so each question is still about a single event.

Sample spaces.

One coin: ${H, T}$ , 2 outcomes.
Two coins (or one coin twice): ${H H, H T, T H, T T}$ , 4 outcomes.
Three coins: $2 \times 2 \times 2 = 8$ outcomes.
One die: ${1, 2, 3, 4, 5, 6}$ .
A die and a coin: $6 \times 2 = 12$ outcomes such as $(4, H)$ .
Two dice: 36 ordered pairs $(a, b)$ , where $a$ is the first die and $b$ the second.

Why order matters. HT and TH are different outcomes, so 'one head and one tail' has probability $\frac{2}{4} = \frac{1}{2}$ , not $\frac{1}{3}$ . In the same way, a sum of 7 comes from $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ : 6 outcomes.

The two-dice grid. Draw a $6 \times 6$ table (first die down, second die across) and write the sum or product in each cell. Then count cells.

Number of ways to get each sum:

Sum	2	3	4	5	6	7	8	9	10	11	12
Ways	1	2	3	4	5	6	5	4	3	2	1

The counts add to 36, rise to 6 at the sum 7 and fall again.

Useful shortcuts.

'At least one 6' $= 36 - 25 = 11$ outcomes (25 pairs have no 6).
'Both odd' $= 3 \times 3 = 9$ ; a product is odd only when both numbers are odd.
Doublets: 6. 'First greater than second': 15, by symmetry $\frac{36 - 6}{2}$ .
Impossible events still count: a sum of 1 or 13 has probability 0.

Dice and Coins Probability: formulas to remember

Outcomes: $2^{k} for k coins, 6 \times 6 = 36 for two dice$; Ordered outcomes.
Probability: $P (E) = \frac{n ( E )}{n ( S )}$; Count cells in the grid.
At least one: $P (at least one) = 1 - P (none)$; At least one 6: $1 - \frac{25}{36} = \frac{11}{36}$ .
Ways to make a sum s (two dice): $6 - ∣ s - 7∣$; For $2 \leq s \leq 12$ .

How to solve Dice and Coins Probability questions

Sample space, grid, count, divide.

1Identify the experiment and write $n (S)$ : 2, 4, 6, 8, 12 or 36.
2For two dice, sketch the $6 \times 6$ grid of sums (or products).
3List or shade the favourable outcomes as ordered pairs.
4Use 'none' or symmetry when it is quicker (at least one, first greater than second).
5Divide by $n (S)$ and simplify.
6Check: probabilities of all possible results add up to 1.

Dice and Coins Probability in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A dice game at a fair

Suppose a stall at a school fair lets you roll two dice, and you win a prize if the sum is 11 or 12.

What is your chance of winning?

Try it first, then show the solutionSolution

Sum 11: (5, 6), (6, 5); sum 12: (6, 6). 3 outcomes.
$P = \frac{3}{36} = \frac{1}{12}$ .

Answer: $\frac{1}{12}$

Memory hook: Big sums are rare: only 1 way to make 12.

Winning two tosses

Imagine a cricket captain calls 'heads' at the toss of two matches in a series.

What is the probability that both tosses land heads?

Try it first, then show the solutionSolution

Two tosses: ${H H, H T, T H, T T}$ .
Only HH is favourable: $\frac{1}{4}$ .

Answer: $\frac{1}{4}$

Memory hook: One coin twice = two coins once.

Tricks to remember Dice and Coins Probability

Short memory aids. Say each one aloud once, then use it on a question the same day.

Seven is the summit
Ways to make a sum climb 1, 2, 3, 4, 5, 6 up to 7, then fall back 5, 4, 3, 2, 1.
Use it when: Counting sums without the grid.
Order counts
HT ≠ TH and (2, 5) ≠ (5, 2): each die or coin has its own name.
Use it when: Avoiding $\frac{1}{3}$ for one head and one tail.
At least one? Count none
No 6 on either die: $5 \times 5 = 25$ , so at least one 6 is $36 - 25$ .
Use it when: 'At least' questions.

Dice and Coins Probability: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: One die · 1 mark · MCQBoard level

A die is thrown once. The probability of getting a factor of 6 is:
(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{5}{6}$

Try it first, then show the solutionSolution

Factors of 6 on a die: 1, 2, 3, 6: $\frac{4}{6} = \frac{2}{3}$ .

Answer: (c)

Example 2: Two coins · 3 marks · Short answerBoard level

Two coins are tossed together. Write the sample space and find the probability of (a) both coins showing the same face, (b) a head on the first coin, (c) two tails.

Try it first, then show the solutionSolution

$S = {H H, H T, T H, T T}$ , $n (S) = 4$ .
(a) HH, TT: $\frac{2}{4} = \frac{1}{2}$ .
(b) HH, HT: $\frac{1}{2}$ .
(c) TT: $\frac{1}{4}$ .

Answer: (a) $\frac{1}{2}$ (b) $\frac{1}{2}$ (c) $\frac{1}{4}$

Example 3: Sums with two dice · 3 marks · Short answerHarder board level

Two dice are thrown together. Find the probability that (a) the sum is 7, (b) the sum is greater than 9, (c) both numbers are odd.

Try it first, then show the solutionSolution

$n (S) = 36$ .
(a) (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1): $\frac{6}{36} = \frac{1}{6}$ .
(b) Sums 10, 11, 12: $3 + 2 + 1 = 6$ : $\frac{1}{6}$ .
(c) Each die 1, 3 or 5: $3 \times 3 = 9$ : $\frac{9}{36} = \frac{1}{4}$ .

Answer: (a) $\frac{1}{6}$ (b) $\frac{1}{6}$ (c) $\frac{1}{4}$

Example 4: Products and multiples · 4 marks · Short answerHarder board level

Two dice are thrown together. Find the probability that (a) the product of the numbers is odd, (b) the product is a perfect square, (c) the sum is a multiple of 4.

Try it first, then show the solutionSolution

(a) Odd product needs both odd: 9 outcomes: $\frac{1}{4}$ .
(b) Squares: the 6 doublets (products 1, 4, 9, 16, 25, 36) and (1, 4), (4, 1) (product 4): 8 outcomes: $\frac{8}{36} = \frac{2}{9}$ .
(c) Sum 4: 3 ways; sum 8: 5 ways; sum 12: 1 way. $\frac{9}{36} = \frac{1}{4}$ .

Answer: (a) $\frac{1}{4}$ (b) $\frac{2}{9}$ (c) $\frac{1}{4}$

Example 5: A die and a coin · 3 marks · Short answerHOTS / competency

A die is thrown and a coin is tossed at the same time. Write the number of outcomes and find the probability of (a) a head with an even number, (b) a tail or a 6.

Try it first, then show the solutionSolution

$S = {(1, H), (1, T), \dots, (6, H), (6, T)}$ , $n (S) = 12$ .
(a) (2, H), (4, H), (6, H): $\frac{3}{12} = \frac{1}{4}$ .
(b) 6 outcomes with T, plus (6, H): $\frac{7}{12}$ .

Answer: (a) $\frac{1}{4}$ (b) $\frac{7}{12}$

Example 6: Case study: a board game · 5 marks · Case-basedHOTS / competency

In a board game, each player throws two dice together.
(a) A player needs at least one 6 to start. Find the probability.
(b) A player on square 32 needs a sum of exactly 8 to land on square 40. Find the probability.
(c) A bonus is given if the two numbers differ by at most 1. Find the probability.
(d) A penalty is given if neither number is more than 3. Find the probability.

Try it first, then show the solutionSolution

(a) No 6 on either die: $5 \times 5 = 25$ . At least one 6: $36 - 25 = 11$ : $\frac{11}{36}$ .
(b) (2, 6), (3, 5), (4, 4), (5, 3), (6, 2): $\frac{5}{36}$ .
(c) Difference 0: 6 doublets; difference 1: (1, 2), (2, 3), (3, 4), (4, 5), (5, 6) and their reverses, 10. Total 16: $\frac{16}{36} = \frac{4}{9}$ .
(d) Both from 1, 2, 3: $3 \times 3 = 9$ : $\frac{1}{4}$ .

Answer: (a) $\frac{11}{36}$ (b) $\frac{5}{36}$ (c) $\frac{4}{9}$ (d) $\frac{1}{4}$

Where students lose marks in Dice and Coins Probability

The slips examiners see most often, and what to write instead.

Mistake: Treating HT and TH as one outcome
The coins are different objects, so the order of results matters.
Wrong working
$S = {H H, H T, T T}$ , so $P (one of each) = \frac{1}{3}$
Right working
$S = {H H, H T, T H, T T}$ , so $\frac{2}{4} = \frac{1}{2}$
Do this: Name the coins or dice 'first' and 'second'.
Usually costs: The whole part.
Mistake: Not writing the sample space
Without $S$ , the total is often wrong.
Wrong working
$P (even) = \frac{2}{3}$ for one die
Right working
$S = {1, \dots, 6}$ ; $P = \frac{3}{6} = \frac{1}{2}$
Do this: Write $S$ (or the $6 \times 6$ grid) first.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q1(xiii), where candidates confused the whole sample space with the even numbers.
Mistake: Missing reversed pairs
(2, 6) and (6, 2) are both in the 36 outcomes.
Wrong working
Sum 8: (2, 6), (3, 5), (4, 4): 3 ways
Right working
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2): 5 ways
Do this: For each pair with different numbers, write its reverse at once.
Usually costs: 1 mark per part.
Mistake: Unsimplified fractions
Answers are expected in lowest terms.
Wrong working
$\frac{9}{36}$
Right working
$\frac{1}{4}$
Do this: Cancel at the end.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q8(ii) and 2025 Q9(ii).

Exam tips

Asked in 2024 Q1(xiii) (1 mark): the probability of an even number on a die, set as assertion and reason; the denominator must be the whole sample space, 6.
For two dice, a quick $6 \times 6$ grid of sums takes a minute and prevents missing pairs.
Write favourable outcomes as ordered pairs; the list earns the method mark.
Check that complementary answers add to 1, for example $\frac{11}{36} + \frac{25}{36} = 1$ .

Practice questions on Dice and Coins Probability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
When two coins are tossed together, the number of possible outcomes is:
(a) 2
(b) 3
(c) 4
(d) 6
Hint
HT and TH are different.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
A die is thrown once. Find the probability of getting (a) a number between 2 and 6, (b) a multiple of 7.
Hint
'Between' excludes 2 and 6.
Answer
(a) 3, 4, 5: $\frac{1}{2}$ (b) none: 0
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When two dice are thrown together, the probability that the sum is 1 is 0.
Reason (R): The smallest possible sum with two dice is 2.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
(1, 1) gives 2.
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
Two dice are thrown together. Find the probability that (a) the first die shows a larger number than the second, (b) the sum is 5 or 10.
Hint
(a) Remove the doublets and halve.
Answer
(a) $\frac{36 - 6}{2} = 15$ : $\frac{5}{12}$ (b) $4 + 3 = 7$ : $\frac{7}{36}$
Q5 · 2 marks · Short answerHarder board level
Two dice are thrown together. Find the probability that the product of the numbers is at least 20.
Hint
Start from (4, 5).
Answer
(4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6): $\frac{8}{36} = \frac{2}{9}$
Q6 · 2 marks · Short answerHarder board level
A die is thrown once. Find the probability that the number $x$ shown satisfies $x^{2} - 5 x + 6 = 0$ .
Hint
Factorise.
Answer
$(x - 2) (x - 3) = 0$ , so $x = 2$ or 3: $\frac{2}{6} = \frac{1}{3}$
Q7 · 3 marks · Short answerHOTS / competency
Three coins are tossed together. Write the sample space and find the probability of (a) at least two heads, (b) exactly one tail.
Hint
8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Answer
(a) HHH, HHT, HTH, THH: $\frac{4}{8} = \frac{1}{2}$ (b) HHT, HTH, THH: $\frac{3}{8}$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Outcomes when two dice are thrown?Show answer
36
2.Ways to get a sum of 7?Show answer
6
3.Sample space for two coins?Show answer
{HH, HT, TH, TT}
4.P(at least one 6) with two dice?Show answer
$\frac{11}{36}$
5.Number of doublets?Show answer
6

Dice and Coins Probability: frequently asked questions

How many outcomes are there when two dice are thrown?

There are 6 × 6 = 36 ordered outcomes, from (1, 1) to (6, 6). (2, 5) and (5, 2) count as different outcomes.

What is the probability of getting one head and one tail with two coins?

The sample space is {HH, HT, TH, TT}; HT and TH are both favourable, so the probability is 2/4 = 1/2.

What is the most likely sum when two dice are thrown?

A sum of 7, which can be made in 6 ways out of 36, giving a probability of 1/6.