Dice probability questions and answers present methods and worked examples for finding the likelihood of different outcomes when one or more dice are rolled through simple and effective steps. This guide reviews the basic idea that probability measures how likely an event is to happen and demonstrates its use through solved problems involving outcomes, favourable events, sample space and equal chances on a die. From straightforward calculations to application-based questions, each solution focuses on clear steps, counting outcomes and helpful shortcuts. Worked examples with brief explanations help strengthen understanding and exam preparation.

A standard die has six faces numbered 1 to 6 and every face is equally likely to land face-up on a fair roll.
Single die: Total outcomes = 6. P(specific number) = 1/6.
Two dice: Total outcomes = 36. Always count ordered pairs, not sums.
Doublet: Both dice show the same number; 6 such outcomes exist.
Sum of 7 is the most probable sum (6/36); sums of 2 and 12 are the least probable (1/36 each).
Sample space when two dice are rolled simultaneously:
Q1: A die is thrown once. What is the probability of getting a number greater than 4?
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
Solution:
Correct option: B) 1/3
Numbers greater than 4 in {1, 2, 3, 4, 5, 6} are {5, 6} ⇒ 2 favourable outcomes
Probability = 2/6 = 1/3
Q2: Aditi rolls a die once. What is the probability that she gets a number that is a multiple of 3?
Solution:
Multiples of 3 in {1, 2, 3, 4, 5, 6} = {3, 6} ⇒ 2 favourable outcomes
Probability = 2/6 = 1/3
Q3: Rohan recorded the outcomes of 150 throws of a die and found that 6 appeared 28 times. Based on this data, what is the experimental probability of getting a 6?
Solution:
Experimental probability = Number of times 6 appeared ÷ Total throws
= 28/150
= 14/75 (in simplest form)
Q4: A die is thrown. What is the probability of getting a number that is neither a prime nor 1?
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
Solution:
Correct option: B) 1/3
Primes on a die: {2, 3, 5}. Excluding primes and 1 leaves {4, 6} ⇒ 2 outcomes
Probability = 2/6 = ⅓
Q5: Two dice are thrown simultaneously. What is the probability of getting a sum of 7?
(A) 1/9
(B) 1/6
(C) 5/36
(D) 1/12
Solution:
Correct option: B) 1/6
Favourable outcomes for sum 7: (1,6), (2,5), (3,4),(4,3),(5,2),(6,1) = 6 outcomes
Probability = 6/36 = 1/6
Q6: Assertion (A): The probability of getting a sum of 2 when two dice are thrown is the same as the probability of getting a sum of 7.
Reason (R): A sum of 2 can only occur as (1,1), while a sum of 7 can occur in six different ways.
(A) Both A and R are true, and R is the correct explanation of why A is false.
(B) Both A and R are true, and R explains A.
(C) A is true, R is false.
(D) A is false, R is true, but R does not explain A.
Solution:
Correct option: A
Assertion is false: P(sum = 2) = 1/36, while P(sum = 7) = 6/36; these are not equal. Reason is true and correctly explains why they differ.
Q7: Two dice are thrown together. What is the probability of getting a doublet?
(A) 1/36
(B) 1/6
(C) 1/12
(D) 1/3
Solution:
Correct option: B) 1/6
Doublets: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) = 6 outcomes
Probability = 6/36 = ⅙
Q8: In a school fair game in Bengaluru, players roll two dice. They win a prize if the sum of the two numbers is a prime number.
(a) List the possible prime sums.
(b) Find the probability that a player wins.
Solution:
(a) Possible sums range from 2 to 12. Prime sums among these: 2, 3, 5, 7, 11.
(b) Favourable outcomes:
Sum 2: (1,1) → 1; Sum 3: (1,2),(2,1) → 2; Sum 5: (1,4),(4,1),(2,3),(3,2) → 4; Sum 7: 6 outcomes; Sum 11: (5,6),(6,5) → 2
Total favourable = 1 + 2 + 4 + 6 + 2 = 15
Probability = 15/36 = 5/12
Q9: Two dice are thrown simultaneously. Find the probability that the difference of the numbers shown is 3.
Solution:
Pairs with a difference of 3 (either order): (1,4),(4,1),(2,5),(5,2),(3,6),(6,3) = 6 outcomes
Probability = 6/36 = 1/6
Q10: Two dice are thrown together. Find the probability that the sum of the numbers is neither a multiple of 3 nor a multiple of 4.
Solution:
Multiples of 3 as sums: 3,6,9,12 → outcomes: 2+5+4+1 = 12
Multiples of 4 as sums: 4,8,12 → outcomes: 3+5+1 = 9
Overlap (multiples of both 3 and 4, i.e. sum = 12): 1 outcome
Total multiples of 3 or 4 = 12 + 9 − 1 = 20
Favourable (neither) = 36 − 20 = 16
Probability = 16/36 = 4/9
Q11: Which is more likely when two dice are thrown: getting a sum of 9, or getting a sum of 10? Justify with the actual probabilities.
Solution:
Sum 9: (3,6),(6,3),(4,5),(5,4) = 4 outcomes → P = 4/36 = 1/9
Sum 10: (4,6),(6,4),(5,5) = 3 outcomes → P = 3/36 = 1/12
Since 1/9 > 1/12, a sum of 9 is more likely than a sum of 10.
Q12: A die is thrown twice, and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 3 has appeared at least once?
(A) 1/5
(B) 2/5
(C) 1/3
(D) 0
Solution:
Correct option: B) 2/5
Outcomes giving a sum of 8 : (2,6),(6,2),(3,5),(5,3),(4,4) → 5 outcomes
Of these, the outcomes containing at least one 3 are: (3,5),(5,3) → 2 outcomes
Conditional probability = 2/5
A fair die is rolled once. Find the probability of getting:
(a) an even number
(b) a prime number
(c) a number greater than 4.
Two dice are thrown together. Find the probability that the sum of the numbers is 8.
Two dice are rolled simultaneously. Find the probability of getting at least one six.
Two dice are thrown together. Find the probability that both dice show odd numbers.
Two dice are rolled. Find the probability that the sum is a prime number.
Two dice are thrown together. Find the probability that the difference between the numbers is 2.
Two dice are rolled simultaneously. Find the probability that the product of the numbers is a multiple of 6.
The probability of rolling any specific number (1 through 6) on a fair single die is 1/6, since there is exactly one favourable outcome out of six equally likely outcomes.
When two dice are thrown simultaneously, there are 6 × 6 = 36 total possible outcomes.
The sum of 7 is the most likely outcome, with 6 favourable combinations out of 36 total outcomes, a probability of 6/36 = 1/6, higher than any other possible sum.
A doublet is an outcome where both dice show the same number, such as (1,1), (2,2), (3,3), (4,4), (5,5) or (6,6). There are 6 doublets out of 36 total outcomes, giving a probability of 1/6.
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