MathsCBSE Class 10 · Probability

Probability Problems on Coins and Dice for Class 10

Updated 13 October 2026 · Fact-checked

Coin and dice probability means writing the full list of equally likely outcomes (2, 4 or 8 for coins, 6 or 36 for dice), counting the outcomes that suit the event, and dividing. The board asks MCQs, two- and three-part dice questions on sums, products and differences, coin-and-die combinations and case-based game questions.

Two dice give 36 ordered pairs; the six pairs with sum 7 lie on one diagonal, so P(sum 7) = 6/36 = 1/6.

Illustration drawn for QuestionBuddy.

Understand Probability Problems on Coins and Dice

Tossing a fair coin or throwing a fair die is the simplest random experiment in the chapter. A coin has two equally likely outcomes, H and T. A die has six equally likely faces, 1 to 6. Because every outcome has the same chance, you can use the classical definition directly: favourable outcomes divided by total outcomes.

The real work is writing the sample space correctly. When you toss two coins, each outcome of the first coin pairs with each outcome of the second, so you get $2 \times 2 = 4$ outcomes: HH, HT, TH, TT. Three coins give $2 \times 2 \times 2 = 8$ . Tossing one coin three times gives exactly the same 8 outcomes as tossing three coins once.

For two dice, each of the 6 faces of the first die pairs with each of the 6 faces of the second, giving $6 \times 6 = 36$ ordered pairs $(a, b)$ . Order matters because the dice are different objects: $(1, 4)$ and $(4, 1)$ are two separate outcomes. Only doublets such as $(5, 5)$ appear once. A coin and a die together give $2 \times 6 = 12$ outcomes, like H1, H2, ..., T6.

Why insist on ordered lists? Because they keep outcomes equally likely. If you write two coins as 'two heads, two tails, one of each', the last result can happen in two ways (HT and TH), so it is twice as likely. Using it as one outcome gives wrong answers. The same trap appears with dice: there are only 21 unordered pairs, but they are not equally likely.

Once the sample space is right, use ideas from the rest of the chapter. The complement rule $P (\overline{E}) = 1 - P (E)$ is quick for 'at least one' or 'not' events. Every answer must lie between $0$ and $1$ , and the probabilities of an event and its complement must add to $1$ , which is a free check.

The method holds only when the coin or die is fair (unbiased). If a question says a die has unusual markings, such as two faces showing 1, the six faces are still equally likely, so count faces, not numbers.

Probability Problems on Coins and Dice: formulas to remember

Outcomes for coins: $n coins (or 1 coin tossed n times) : 2^{n} outcomes; 1 \to 2, 2 \to 4, 3 \to 8$; Write the full list for 2 or 3 coins. Order matters: HT and TH are different.
Outcomes for dice: $1 die: 6, 2 dice (or 1 die thrown twice): 6 \times 6 = 36$; Outcomes are ordered pairs (a, b). Only doublets appear once.
Coin and die together: $2 \times 6 = 12 outcomes$; List as H1 to H6 and T1 to T6.
Number of pairs with a given sum: $pairs with sum s = 6 - ∣ s - 7∣, 2 \leq s \leq 12$; Sums 2 to 12 have 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 pairs. These add to 36.
Pairs with a given difference: $∣ a - b ∣ = d (d \geq 1) : 2 (6 - d) pairs; ∣ a - b ∣ = 0 : 6 pairs$; Difference 1: 10 pairs, 2: 8, 3: 6, 4: 4, 5: 2. Total with 6 doublets is 36.
Exactly k heads with 3 coins: $0 heads : 1, 1 : 3, 2 : 3, 3 : 1 (out of 8)$; Use for exactly, at least and at most questions on three coins.
Complement for at least one: $P (at least one) = 1 - P (none)$; For two dice, at least one six is 1 − 25/36 = 11/36.

How to solve Probability Problems on Coins and Dice questions

This method works for any coin or dice question, from a 1-mark MCQ to a 5-mark case.

1Identify the experiment: how many coins, how many dice, or a coin and a die together.
2Write the total number of equally likely outcomes (2, 4, 8, 6, 12 or 36) as a separate line; it earns a step mark.
3For coins write the full list; for two dice picture the 6 × 6 grid with rows for the first die and columns for the second.
4Read the event carefully: exactly, at least, at most, or, and, not, greater than.
5List the favourable outcomes row by row, counting (a, b) and (b, a) separately and doublets once; for 'or' count overlaps once.
6If listing is long, count the opposite event and use P(E) = 1 − P(not E).
7Write P(E) = favourable/total and reduce to lowest terms.
8Check: the answer lies between 0 and 1, and where you found both an event and its complement they add to 1.

Probability Problems on Coins and Dice 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.

Two dice: 36 ordered pairs

Imagine you are playing a board game at home with your cousins. On your turn you throw two fair dice together. Your piece needs a total of exactly 10 to reach home, and a doublet (both dice showing the same number) gives you an extra turn.

On one throw of the two dice, find the probability that (i) you reach home, (ii) you get an extra turn, (iii) you either reach home or get an extra turn.

Try it first, then show the solutionSolution

Total outcomes = $6 \times 6 = 36$ .
(i) Sum 10: $(4, 6), (5, 5), (6, 4)$ , which is 3 outcomes.
$P (home) = \frac{3}{36} = \frac{1}{12}$ .
(ii) Doublets: $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ , which is 6 outcomes.
$P (extra turn) = \frac{6}{36} = \frac{1}{6}$ .
(iii) $(5, 5)$ is in both lists, so count it once: $3 + 6 - 1 = 8$ outcomes.
$P = \frac{8}{36} = \frac{2}{9}$ .

Answer: (i) $\frac{1}{12}$ (ii) $\frac{1}{6}$ (iii) $\frac{2}{9}$

Memory hook: Home or extra turn: list both, strike the shared (5, 5) once.

Three coins: 8 outcomes

Suppose your class team plays a three-match cricket series against another section. Before each match a fair coin is tossed, and your captain calls heads every time. Each toss is a fresh, fair toss.

Find the probability that your captain (i) wins at least two of the three tosses, (ii) wins no toss at all, (iii) wins the toss in the first match but loses at least one of the other two.

Try it first, then show the solutionSolution

Outcomes (first, second, third toss): HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, total 8.
(i) At least two heads: HHH, HHT, HTH, THH, which is 4 outcomes.
$P = \frac{4}{8} = \frac{1}{2}$ .
(ii) No head: TTT only, so $P = \frac{1}{8}$ .
(iii) First toss H and at least one T later: HHT, HTH, HTT, which is 3 outcomes.
$P = \frac{3}{8}$ .

Answer: (i) $\frac{1}{2}$ (ii) $\frac{1}{8}$ (iii) $\frac{3}{8}$

Memory hook: Three tosses, three matches: 2, 4, 8 outcomes, and order of matches matters.

Tricks to remember Probability Problems on Coins and Dice

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

Doubling coins
Every extra coin doubles the list: 2, 4, 8. For three coins remember 1-3-3-1 for 0, 1, 2, 3 heads.
Use it when: Exactly, at least and at most head questions on three coins.
The sum mountain
Sums 2 to 12 climb 1, 2, 3, 4, 5, 6 to the peak at 7, then fall 5, 4, 3, 2, 1.
Use it when: Any 'sum is ...' question on two dice, without drawing the grid.
Swap and count twice
For two dice, if a ≠ b then (a, b) has a twin (b, a). Doublets have no twin.
Use it when: Avoid undercounting differences, products and sums.
None is easier
See 'at least one'? Count 'none' and subtract from 1.
Use it when: At least one head, at least one six, at least one even number.

Probability Problems on Coins and Dice: solved examples

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

Example 1: Prime or greater than 4 on one die · 1 mark · MCQBoard level

A fair die is thrown once. The probability of getting a prime number or a number greater than 4 is

(a) $\frac{1}{2}$

(b) $\frac{2}{3}$

(d) $\frac{1}{3}$

Try it first, then show the solutionSolution

Total outcomes = 6.
Primes: 2, 3, 5. Greater than 4: 5, 6.
Favourable (5 counted once): 2, 3, 5, 6, which is 4 outcomes.
$P = \frac{4}{6} = \frac{2}{3}$ .

Answer: (b) $\frac{2}{3}$

Example 2: Assertion-reason: sum 7 on two dice · 1 mark · Assertion-reasonBoard level

Assertion (A): When two fair dice are thrown together, the probability that the sum of the numbers is 7 is $\frac{1}{6}$ .

Reason (R): The possible sums 2, 3, ..., 12 are 11 equally likely outcomes.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(d) A is false but R is true.

Try it first, then show the solutionSolution

Total outcomes = 36 ordered pairs.
Sum 7: $(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)$ , 6 pairs, so $P = \frac{6}{36} = \frac{1}{6}$ . A is true.
The sums are not equally likely: sum 2 has 1 pair but sum 7 has 6 pairs. R is false.

Answer: (c) A is true but R is false.

Example 3: Two dice: sum, difference and an 'or' event · 3 marks · Short answerHarder board level

Two different fair dice are thrown together. Find the probability that (i) the sum of the numbers is a multiple of 3, (ii) the numbers differ by exactly 2, (iii) the sum is 8 or the product is 12.

Try it first, then show the solutionSolution

Total outcomes = $6 \times 6 = 36$ .
(i) Sums 3, 6, 9, 12 have 2, 5, 4, 1 pairs: total 12, so $P = \frac{12}{36} = \frac{1}{3}$ . (1 mark)
(ii) $(1, 3), (2, 4), (3, 5), (4, 6)$ and their reverses: 8 pairs, so $P = \frac{8}{36} = \frac{2}{9}$ . (1 mark)
(iii) Sum 8: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ . Product 12: $(2, 6), (3, 4), (4, 3), (6, 2)$ .
$(2, 6)$ and $(6, 2)$ are in both, so favourable = $5 + 2 = 7$ .
$P = \frac{7}{36}$ . (1 mark)

Answer: (i) $\frac{1}{3}$ (ii) $\frac{2}{9}$ (iii) $\frac{7}{36}$

Example 4: A coin and a die together · 3 marks · Short answerBoard level

A fair coin is tossed and a fair die is thrown at the same time. Find the probability of getting (i) a head and an even number, (ii) a tail or a number greater than 4, (iii) neither a head nor a prime number.

Try it first, then show the solutionSolution

Sample space: H1, ..., H6, T1, ..., T6, total 12. (½ mark)
(i) H2, H4, H6: 3 outcomes, $P = \frac{3}{12} = \frac{1}{4}$ . (½ mark)
(ii) All 6 tails, plus H5, H6: $6 + 2 = 8$ . (½ mark)
$P = \frac{8}{12} = \frac{2}{3}$ . (½ mark)
(iii) Neither head nor prime means tail with 1, 4 or 6: T1, T4, T6. (½ mark)
$P = \frac{3}{12} = \frac{1}{4}$ . (½ mark)

Answer: (i) $\frac{1}{4}$ (ii) $\frac{2}{3}$ (iii) $\frac{1}{4}$

Example 5: Three coins: exactly, at least one of each, and a pattern · 3 marks · Short answerHarder board level

Three fair coins are tossed together (think of them as first, second and third coin). Find the probability that (i) exactly two heads appear, (ii) at least one head and at least one tail appear, (iii) the first and the third coins show the same face.

Try it first, then show the solutionSolution

Sample space: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, total 8. (½ mark)
(i) Exactly two heads: HHT, HTH, THH, so $P = \frac{3}{8}$ . (½ mark)
(ii) The opposite event is 'all same face': HHH, TTT, which is 2 outcomes. (½ mark)
$P = 1 - \frac{2}{8} = \frac{6}{8} = \frac{3}{4}$ . (½ mark)
(iii) First and third coins match: HHH, HTH, THT, TTT, which is 4 outcomes. (½ mark)
$P = \frac{4}{8} = \frac{1}{2}$ . (½ mark)

Answer: (i) $\frac{3}{8}$ (ii) $\frac{3}{4}$ (iii) $\frac{1}{2}$

Example 6: Points from a die thrown twice · 5 marks · Long answerHarder board level

A fair die is thrown twice. Let $a$ be the first number and $b$ the second, and let $P$ be the point $(a, b)$ in the coordinate plane. Find the probability that (i) the distance of $P$ from the origin is less than 5 units, (ii) $P$ lies on the line $y = x + 1$ , (iii) the product $ab$ is even.

Try it first, then show the solutionSolution

Total outcomes = 36. (1 mark)
(i) Distance from origin is $a^{2} + b^{2}$ ; less than 5 means $a^{2} + b^{2} < 25$ .
$a = 1$ : $b^{2} < 24$ , $b = 1, 2, 3, 4$ (4). $a = 2$ : $b^{2} < 21$ , $b = 1$ to $4$ (4).
$a = 3$ : $b^{2} < 16$ , $b = 1, 2, 3$ (3). $a = 4$ : $b^{2} < 9$ , $b = 1, 2$ (2). $a = 5, 6$ : none.
Favourable = 13, $P = \frac{13}{36}$ . (2 marks)
(ii) $b = a + 1$ : $(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)$ , so $P = \frac{5}{36}$ . (1 mark)
(iii) Product is odd only when both are odd: $3 \times 3 = 9$ pairs.
$P (even) = 1 - \frac{9}{36} = \frac{27}{36} = \frac{3}{4}$ . (1 mark)

Answer: (i) $\frac{13}{36}$ (ii) $\frac{5}{36}$ (iii) $\frac{3}{4}$

Example 7: Case study: a dice game at a fair · 4 marks · Case-basedHOTS / competency

At a school fair, a player throws two fair dice together. Under the current rule, the player wins if the sum of the numbers is 9 or more.

(i) Find the probability that the sum is exactly 7. (1 mark)

(ii) Find the probability that a player wins under the current rule. (1 mark)

(iii) The organisers want the chance of winning to be at least $\frac{2}{5}$ . They consider changing the rule to 'sum of 8 or more'. Find the probability of losing under the new rule and state whether it meets their aim. (2 marks)

Try it first, then show the solutionSolution

Total outcomes = 36.
(i) Sum 7 has 6 pairs, so $P = \frac{6}{36} = \frac{1}{6}$ . (1 mark)
(ii) Sums 9, 10, 11, 12 have $4 + 3 + 2 + 1 = 10$ pairs, so $P (win) = \frac{10}{36} = \frac{5}{18}$ . (1 mark)
(iii) Sum 8 or more: $5 + 4 + 3 + 2 + 1 = 15$ pairs, so $P (win) = \frac{15}{36} = \frac{5}{12}$ . (½ mark)
$P (lose) = 1 - \frac{5}{12} = \frac{7}{12}$ . (½ mark)
Compare $\frac{5}{12}$ with $\frac{2}{5}$ : $5 \times 5 = 25 > 2 \times 12 = 24$ , so $\frac{5}{12} > \frac{2}{5}$ . (½ mark)
So the new rule meets the aim. (½ mark)

Answer: (i) $\frac{1}{6}$ (ii) $\frac{5}{18}$ (iii) $P (lose) = \frac{7}{12}$ ; winning chance $\frac{5}{12} > \frac{2}{5}$ , so yes.

Where students lose marks in Probability Problems on Coins and Dice

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

Mistake: Treating two dice outcomes as unordered
Students list pairs like {2, 5} once, giving 21 outcomes that are not equally likely.
Wrong working
Sum 7: {1,6}, {2,5}, {3,4}, so $P = \frac{3}{21} = \frac{1}{7}$ .
Right working
Sum 7: $(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)$ , so $P = \frac{6}{36} = \frac{1}{6}$ .
Do this: Always use 36 ordered pairs and count each swap separately.
Usually costs: Whole question, since the total is wrong.
Mistake: Counting a doublet twice
When applying 'swap and count twice', students also double the doublets.
Wrong working
Sum 8: $(2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (4, 4)$ , 6 pairs, $P = \frac{1}{6}$ .
Right working
$(4, 4)$ is one outcome, so 5 pairs and $P = \frac{5}{36}$ .
Do this: Doublets appear once in the grid. Check sums with the 1-2-3-4-5-6-5-4-3-2-1 pattern.
Usually costs: 1 mark per part.
Mistake: Misreading 'at least' and 'at most'
Students read 'at least two heads' as 'exactly two heads'.
Wrong working
Three coins, at least two heads: HHT, HTH, THH, $P = \frac{3}{8}$ .
Right working
Include HHH: 4 outcomes, $P = \frac{4}{8} = \frac{1}{2}$ .
Do this: Translate first: at least two means 2 or 3; at most one means 0 or 1.
Usually costs: 1 mark.
Mistake: Adding probabilities of overlapping events
For 'or' events, outcomes in both events are counted twice.
Wrong working
Sum 8 or product 12: $\frac{5}{36} + \frac{4}{36} = \frac{9}{36}$ .
Right working
$(2, 6)$ and $(6, 2)$ are in both, so $5 + 4 - 2 = 7$ and $P = \frac{7}{36}$ .
Do this: List the favourable outcomes of both events and strike out repeats.
Usually costs: 1 mark.

Exam tips

Write the total number of outcomes (for example 'Total outcomes = 36') on its own line; it carries a step mark.
List the favourable outcomes explicitly in 2- and 3-mark answers, so the examiner can give method marks even if you slip in counting.
Reduce every fraction to lowest terms; MCQ options usually show the simplified value only.
For case-based parts that ask you to compare or decide, give the probability and then a one-line conclusion in words.
When a die or coin is thrown again, treat it exactly like two dice or two coins thrown together.

Practice questions on Probability Problems on Coins and Dice

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

Q1 · 2 marks · Very short answerBoard level
Three fair coins are tossed together. Find the probability of getting (i) at most two heads, (ii) exactly one tail.
Hint
List all 8 outcomes. At most two heads is everything except HHH.
Answer
(i) 7 outcomes, so $\frac{7}{8}$ . (ii) HHT, HTH, THH, so $\frac{3}{8}$ .
Q2 · 3 marks · Short answerHarder board level
Two different fair dice are thrown together. Find the probability that (i) the product of the numbers is a prime number, (ii) the sum is a perfect square, (iii) the numbers differ by at least 3.
Hint
A product is prime only when one die shows 1 and the other a prime. Perfect square sums possible are 4 and 9.
Answer
(i) $(1, 2), (2, 1), (1, 3), (3, 1), (1, 5), (5, 1)$ : $\frac{6}{36} = \frac{1}{6}$ . (ii) Sum 4: 3 pairs, sum 9: 4 pairs, $\frac{7}{36}$ . (iii) Differences 3, 4, 5 give $6 + 4 + 2 = 12$ , so $\frac{1}{3}$ .
Q3 · 3 marks · Short answerHarder board level
A special fair die has its six faces marked 0, 1, 1, 2, 2, 3. It is thrown twice and the two numbers are added to get a total score. Find the probability that the total score is (i) 4, (ii) 2, (iii) 5 or more.
Hint
There are still 36 equally likely face pairs. Count faces: one 0, two 1s, two 2s, one 3.
Answer
(i) $(1, 3)$ : $2 \times 1 = 2$ , $(3, 1)$ : 2, $(2, 2)$ : $2 \times 2 = 4$ ; total 8, $\frac{2}{9}$ . (ii) $(0, 2)$ : 2, $(2, 0)$ : 2, $(1, 1)$ : 4; total 8, $\frac{2}{9}$ . (iii) $(2, 3)$ : 2, $(3, 2)$ : 2, $(3, 3)$ : 1; total 5, $\frac{5}{36}$ .
Q4 · 3 marks · Short answerHarder board level
A fair die is thrown twice, giving $a$ on the first throw and $b$ on the second. Find the probability that (i) $a > 2 b$ , (ii) one number is exactly double the other, (iii) $a + b$ is even.
Hint
Fix $b$ and list possible $a$ . For (iii), both numbers must be odd or both even.
Answer
(i) $b = 1$ : $a = 3, 4, 5, 6$ ; $b = 2$ : $a = 5, 6$ ; total 6, $\frac{1}{6}$ . (ii) $(2, 1), (4, 2), (6, 3)$ and reverses: 6, $\frac{1}{6}$ . (iii) $9 + 9 = 18$ , $\frac{1}{2}$ .
Q5 · 3 marks · Short answerHOTS / competency
Ritu says: 'When three coins are tossed, the number of heads can be 0, 1, 2 or 3. So the probability of exactly two heads is $\frac{1}{4}$ .' Find her error, give the correct probability of exactly two heads, and find the probability of at least one head.
Hint
Are the four results 0, 1, 2, 3 heads equally likely? Write the full sample space.
Answer
Error: 0, 1, 2, 3 heads are not equally likely (they occur 1, 3, 3, 1 times out of 8). Exactly two heads: HHT, HTH, THH, so $\frac{3}{8}$ . At least one head $= 1 - P (TTT) = 1 - \frac{1}{8} = \frac{7}{8}$ .

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.How many outcomes when three coins are tossed?Show answer
8
2.How many outcomes when two dice are thrown?Show answer
36
3.Which sum of two dice is most likely, and with how many pairs?Show answer
Sum 7, with 6 pairs.
4.Probability of a doublet with two dice?Show answer
$\frac{6}{36} = \frac{1}{6}$
5.Probability of at least one head when two coins are tossed?Show answer
$1 - \frac{1}{4} = \frac{3}{4}$
6.How many outcomes when a coin is tossed and a die is thrown?Show answer
12

Probability Problems on Coins and Dice: frequently asked questions

Is throwing one die twice the same as throwing two dice together?

Yes. Both give 36 equally likely ordered pairs. The first throw plays the role of the first die and the second throw the role of the second die.

Why are (2, 5) and (5, 2) counted separately?

The two dice are different objects, so 'first die 2, second die 5' and 'first die 5, second die 2' are different results. Counting them separately keeps all 36 outcomes equally likely, which the classical definition needs.

Do I need to draw the full 36-outcome table in the exam?

Not always. Writing the total as 36 and clearly listing the favourable pairs is enough. Drawing the grid on rough work helps you avoid missing pairs.

Are questions on tossing four or more coins asked?

The syllabus asks for simple problems, so coin questions normally use one, two or three coins. Listing more outcomes than that is rarely needed at this level.

What if the die has unusual numbers on its faces?

The six faces are still equally likely if the die is fair. Count faces, not distinct numbers, so a number appearing on two faces is twice as likely.