Dice Problems in Probability for Competitive Exams

Dice Problems in Probability with Solutions

Q.13. getting an even number as the sum

(A) 1/8

(B) 1/2

(C) 1/9

(D) 1/10

getting an even number as the sum:

Let E₁₀ = event of getting an even number as the sum. The events of an even number as the sum will be E₁₀ = [(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 18

Therefore, probability of getting ‘an even number as the sum

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 18/36
       = 1/2
 

Q.14. Getting a doublet

(A) 2/3

(B) 4/8

(C) 1/6

(D) 4/8

getting a doublet: Let E₄ = event of getting a doublet. The number which doublet will be E₄ = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6

Therefore, probability of getting ‘a doublet’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 6/36
       = 1/6

Q.15. Getting a prime number as the sum

(A) 2/12

(B) 3/12

(C) 5/12

(D) 12/5

getting a prime number as the sum:

Let E₁₁ = event of getting a prime number as the sum. The events of a prime number as the sum will be E₁₁ = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15

Therefore, probability of getting ‘a prime number as the sum’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 15/36
       = 5/12

Q.16. Getting a sum of 8

(A) 6/36

(B) 8/56

(C) 7/40

(D) 5/36

getting a sum of 8:

Let E₅ = event of getting a sum of 8. The number which is a sum of 8 will be E₅ = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5

Therefore, probability of getting ‘a sum of 8’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 5/36

Q.17. Getting a doublet of even numbers

(A) 1/18

(B) 1/16

(C) 1/10

(D) 1/12

getting a prime number as the sum:

Let E₁₁ = event of getting a prime number as the sum. The events of a prime number as the sum will be E₁₁ = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15

Therefore, probability of getting ‘a prime number as the sum’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 15/36
       = 5/12
 

Q.18. Getting sum divisible by 5

(A) 8/36

(B) 6/36

(C) 5/36

(D) 7/36

getting sum divisible by 5:

Let E₆ = event of getting sum divisible by 5. The number whose sum divisible by 5 will be E₆ = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7

Therefore, probability of getting ‘sum divisible by 5’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 7/36
 

Q.19. Getting a multiple of 2 on one die and a multiple of 3 on the other die

(A) 11/36

(B) 12/36

(C) 10/54

(D) 12/54

getting a multiple of 2 on one die and a multiple of 3 on the other die:

Let E₁₃ = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E₁₃ = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11

Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 11/36 

Feel free and ask me in the comment section related dice problems in probability if you face any problem.