Dice Problems in Probability for Competitive Exams

Dice Problems in Probability with Solutions

Direction (7-19): Two dice are thrown simultaneously. Find the probability of: 

Q.7. Getting sum of at least 11

ans 1/12

(A) 1/11

(B) 1/10

(C) 1/9

(D) 1/12

getting sum of atleast 11:

Let E₇ = event of getting sum of atleast 11. The events of the sum of atleast 11 will be E₇ = [(5, 6), (6, 5), (6, 6)] = 3

Therefore, probability of getting ‘sum of atleast 11’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
       = 3/36
       = 1/12

Q.8. Getting six as a product

(A) 1/9

(B) 1/8

(C) 1/6

(D) 1/10

getting six as a product:

Let E₁ = event of getting six as a product. The number whose product is six will be E₁ = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4

Therefore, probability of getting ‘six as a product’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 4/36
       = 1/9

Q.9. Getting a multiple of 3 as the sum

(A) 1/2 

(B) 1/8

(C) 1/3

(D) 2/9

getting a multiple of 3 as the sum:

Let E₈ = event of getting a multiple of 3 as the sum. The events of a multiple of 3 as the sum will be E₈ = [(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3) (6, 6)] = 12

Therefore, probability of getting ‘a multiple of 3 as the sum’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 12/36
       = 1/3
 

Q.10. getting sum ≤ 3

(A) 2/25

(B) 1/12

(C) 1/16

(D) 2/16

getting sum ≤ 3:

Let E₂ = event of getting sum ≤ 3. The number whose sum ≤ 3 will be E₂ = [(1, 1), (1, 2), (2, 1)] = 3

Therefore, probability of getting ‘sum ≤ 3’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 3/36
       = 1/12

Q.11. Getting a total of at least 10

(A) 1/6

(B) 5/8

(C) 7/9

(D) 1/10

getting a total of atleast 10:

Let E₉ = event of getting a total of atleast 10. The events of a total of atleast 10 will be E₉ = [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6

Therefore, probability of getting ‘a total of atleast 10’

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

Q.12. Getting sum ≤ 10

(A) 10/12

(B) 12/10

(C) 11/12

(D) 15/17

getting sum ≤ 10:

Let E₃ = event of getting sum ≤ 10. The number whose sum ≤ 10 will be E₃=

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),

(6, 1), (6, 2), (6, 3), (6, 4)] = 33

Therefore, probability of getting ‘sum ≤ 10’

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 33/36
       = 11/12

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