Probability with Dice
Probability with dice is one of the most common topics in Class 10 CBSE Chapter 15 (Probability). A standard die has 6 faces numbered 1 to 6, and problems range from single die experiments to two-dice combinations.
When one die is thrown, there are 6 equally likely outcomes. When two dice are thrown simultaneously, there are 6 × 6 = 36 equally likely outcomes. Each outcome is an ordered pair (a, b) where a and b are the numbers on the first and second die.
Understanding the sample space and systematically counting favourable outcomes is the key to solving dice probability problems. These problems frequently appear in board exams (2–4 marks).
What is Probability with Dice - One Die, Two Dice, Sample Space & Solved Examples?
Standard Die: A cube with faces numbered 1, 2, 3, 4, 5, 6. Each face has an equal probability of landing face up = 1/6.
Sample Spaces:
- One die: S = {1, 2, 3, 4, 5, 6} → n(S) = 6
- Two dice: S = {(1,1), (1,2), ..., (6,6)} → n(S) = 36
- Three dice: n(S) = 6³ = 216 (not in CBSE Class 10 syllabus)
Properties of a standard die:
- Opposite faces sum to 7: (1,6), (2,5), (3,4)
- Even numbers: {2, 4, 6} — probability = 3/6 = 1/2
- Odd numbers: {1, 3, 5} — probability = 3/6 = 1/2
- Prime numbers: {2, 3, 5} — probability = 3/6 = 1/2
- Composite numbers: {4, 6} — probability = 2/6 = 1/3
- 1 is neither prime nor composite
Probability with Dice Formula
Probability Formula:
P(E) = Number of favourable outcomes / Total outcomes
Total outcomes:
- One die: 6
- Two dice: 36
Two-Dice Sum Table:
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
| P(sum) | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
Key fact: Sum of 7 is the most probable sum (probability = 6/36 = 1/6).
Derivation and Proof
Why there are 36 outcomes with two dice:
- Die 1 can show any of 6 values: {1, 2, 3, 4, 5, 6}.
- For EACH outcome of Die 1, Die 2 can show 6 values.
- By the multiplication principle: 6 × 6 = 36 total outcomes.
- Each outcome is an ordered pair (a, b) where a is Die 1 and b is Die 2.
- (2,3) and (3,2) are DIFFERENT outcomes.
Counting outcomes for sums:
- Sum = 2: only (1,1) → 1 way
- Sum = 3: (1,2), (2,1) → 2 ways
- Sum = 4: (1,3), (2,2), (3,1) → 3 ways
- The pattern increases by 1 until sum = 7 (6 ways), then decreases symmetrically.
- This is because for sum k (where 2 ≤ k ≤ 7): ways = k − 1. For sum k (where 8 ≤ k ≤ 12): ways = 13 − k.
Types and Properties
Common Types of Dice Problems in CBSE:
- Type 1: Single die — P(even), P(odd), P(prime), P(divisible by n), P(greater than k)
- Type 2: Sum of two dice — P(sum = k), P(sum > k), P(sum is even/odd)
- Type 3: Product of two dice — P(product = k), P(product is even)
- Type 4: Doublets — P(same number on both) = 6/36 = 1/6
- Type 5: At least one condition — P(at least one 6) = 1 − P(no 6) = 1 − 25/36 = 11/36
- Type 6: Difference of two dice — P(difference = k)
Solved Examples
Example 1: Single Die — Basic Probabilities
Problem: A die is thrown once. Find the probability of getting (a) a number greater than 4, (b) a prime number, (c) a number divisible by 3.
Solution:
Sample space: {1, 2, 3, 4, 5, 6}, n = 6
(a) P(> 4):
- Favourable: {5, 6} → m = 2
- P = 2/6 = 1/3
(b) P(prime):
- Primes: {2, 3, 5} → m = 3
- P = 3/6 = 1/2
(c) P(divisible by 3):
- Divisible by 3: {3, 6} → m = 2
- P = 2/6 = 1/3
Example 2: Two Dice — Probability of a Sum
Problem: Two dice are thrown. Find the probability that the sum of numbers is (a) 9, (b) at least 10, (c) less than 5.
Solution:
Total outcomes: 36
(a) P(sum = 9):
- Favourable: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- P = 4/36 = 1/9
(b) P(sum ≥ 10):
- Sum 10: (4,6), (5,5), (6,4) → 3
- Sum 11: (5,6), (6,5) → 2
- Sum 12: (6,6) → 1
- Total favourable = 6
- P = 6/36 = 1/6
(c) P(sum < 5):
- Sum 2: (1,1) → 1
- Sum 3: (1,2), (2,1) → 2
- Sum 4: (1,3), (2,2), (3,1) → 3
- Total favourable = 6
- P = 6/36 = 1/6
Example 3: Doublets — Same Number on Both Dice
Problem: Two dice are rolled. Find the probability of getting a doublet (same number on both).
Solution:
Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes
P(doublet) = 6/36 = 1/6
Answer: P(doublet) = 1/6
Example 4: Product of Numbers on Two Dice
Problem: Two dice are thrown. Find the probability that the product of the numbers is (a) 12, (b) even.
Solution:
(a) P(product = 12):
- Pairs with product 12: (2,6), (3,4), (4,3), (6,2) → 4 outcomes
- P = 4/36 = 1/9
(b) P(product is even):
- Product is even if at least one die shows an even number.
- P(product is odd) = P(both odd) = (3/6) × (3/6) = 9/36 = 1/4
- P(product is even) = 1 − 1/4 = 3/4
Verification: Count odd×odd pairs: (1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5) = 9. Even products = 36 − 9 = 27. P = 27/36 = 3/4 ✓
Example 5: At Least One Six
Problem: Two dice are thrown. Find the probability of getting at least one 6.
Solution:
Method (complement):
- P(no 6 on either die) = (5/6) × (5/6) = 25/36
- P(at least one 6) = 1 − 25/36 = 11/36
Verification (direct count):
- Outcomes with 6 on Die 1: (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) → 6
- Outcomes with 6 on Die 2 (not already counted): (1,6),(2,6),(3,6),(4,6),(5,6) → 5
- Total = 11
- P = 11/36 ✓
Answer: P(at least one 6) = 11/36
Example 6: Sum is a Multiple of 3
Problem: Two dice are thrown. Find the probability that the sum is a multiple of 3.
Solution:
Possible sums that are multiples of 3: 3, 6, 9, 12
- Sum 3: (1,2),(2,1) → 2
- Sum 6: (1,5),(2,4),(3,3),(4,2),(5,1) → 5
- Sum 9: (3,6),(4,5),(5,4),(6,3) → 4
- Sum 12: (6,6) → 1
Total favourable = 12
P = 12/36 = 1/3
Answer: P(sum is multiple of 3) = 1/3
Example 7: Difference of Numbers
Problem: Two dice are thrown. Find the probability that the difference between the numbers is 2.
Solution:
Favourable outcomes (|a − b| = 2):
- (1,3), (2,4), (3,5), (4,6) → die 1 < die 2: 4 outcomes
- (3,1), (4,2), (5,3), (6,4) → die 1 > die 2: 4 outcomes
Total favourable = 8
P = 8/36 = 2/9
Answer: P(difference = 2) = 2/9
Example 8: Sum is Even
Problem: Two dice are thrown simultaneously. Find the probability that the sum is even.
Solution:
Sum is even when: both are even OR both are odd.
- Both even: 3 × 3 = 9 outcomes
- Both odd: 3 × 3 = 9 outcomes
- Total favourable = 18
P(even sum) = 18/36 = 1/2
Answer: P(sum is even) = 1/2
Example 9: Both Numbers Are Same and Even
Problem: Two dice are rolled. Find the probability that both show the same number and that number is even.
Solution:
Favourable outcomes: (2,2), (4,4), (6,6) → 3 outcomes
P = 3/36 = 1/12
Answer: P(even doublet) = 1/12
Example 10: Neither Number is 5
Problem: Two dice are thrown. Find the probability that neither die shows 5.
Solution:
- Each die has 5 non-5 outcomes: {1, 2, 3, 4, 6}
- Total favourable = 5 × 5 = 25
- P(neither is 5) = 25/36
Answer: P(neither shows 5) = 25/36
Real-World Applications
Real-life applications of dice probability:
- Board games: Games like Ludo, Monopoly, and Snakes & Ladders use dice probability to determine gameplay strategies.
- Statistics education: Dice are used to teach probability distributions, expected value, and the law of large numbers.
- Decision making: Understanding equally likely outcomes helps in fair decision making and random selection.
- Simulations: Dice rolls are used in Monte Carlo simulations for financial modeling and risk assessment.
- Gambling mathematics: Casino games like Craps are entirely based on two-dice probability distributions.
Key Points to Remember
- A standard die has 6 equally likely outcomes: {1, 2, 3, 4, 5, 6}.
- Two dice produce 36 outcomes (ordered pairs).
- Sum of 7 is the most likely sum with two dice (probability = 1/6).
- P(doublet) = 6/36 = 1/6.
- (2,3) and (3,2) are different outcomes — order matters.
- For "at least one" problems, use complement: P(at least one) = 1 − P(none).
- Sum is even when both dice show even or both show odd → P = 1/2.
- Product is even when at least one die is even → P = 3/4.
- Prime numbers on a die: {2, 3, 5}. Note: 1 is NOT prime.
- Composite numbers on a die: {4, 6}. Note: 1 is NOT composite either.
Practice Problems
- A die is thrown once. Find the probability of getting a composite number.
- Two dice are thrown. Find the probability of getting a sum of 5.
- Two dice are thrown. Find the probability that the product is odd.
- Two dice are thrown. Find the probability that at least one die shows 4.
- A die is thrown twice. Find the probability that the sum of the two numbers is at most 4.
- Two dice are thrown. Find the probability that the sum is a perfect square.
- Two dice are thrown. Find the probability of getting a doublet of even numbers.
Frequently Asked Questions
Q1. How many outcomes are there when two dice are thrown?
36 outcomes. Each die has 6 possibilities, and outcomes are ordered pairs (a,b). By the multiplication principle: 6 × 6 = 36.
Q2. Is (2,3) the same as (3,2) when throwing two dice?
No. (2,3) means die 1 shows 2 and die 2 shows 3. (3,2) means die 1 shows 3 and die 2 shows 2. They are different outcomes. This is why total outcomes = 36, not 21.
Q3. What is the most likely sum with two dice?
Sum of 7 is most likely with probability 6/36 = 1/6. The 6 ways are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Q4. What is the probability of getting a 7 on a single die?
0 (impossible). A standard die only has numbers 1 through 6. Getting 7 is an impossible event with probability 0.
Q5. How do you find P(at least one 6) with two dice?
Use the complement: P(at least one 6) = 1 − P(no 6) = 1 − (5/6)(5/6) = 1 − 25/36 = 11/36.
Q6. Is 1 a prime number on a die?
No. 1 is neither prime nor composite. The prime numbers on a standard die are 2, 3, and 5 only.
Q7. What is a doublet?
A doublet occurs when both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 doublets out of 36 outcomes, so P(doublet) = 1/6.
Q8. How do you find the probability that the sum is even?
The sum is even when both dice are even or both are odd. Both even: 3 × 3 = 9 ways. Both odd: 3 × 3 = 9 ways. Total = 18/36 = 1/2.
