Dice Problems in Probability for Competitive Exams

Here are sharing dice problems related to probability with solutions, in this blog. These questions and answers are important for competitive exams. Here are the explained dice problems in probability in detail which will help you to improve your performance.

You just need to understand the topic with these problems and their solutions. So, keep the practice of these questions and strong your preparation. You should visit on Dice Probability Formulas for Competitive Exams to understand using dice formulas to solve dice problems.

Dice Problems in Probability with Solutions

Consider the following points while solving problems:

p(E) = Probability of Event.

n(E) = Total number of favorable outcomes.

n(S) = Total number of Possible outcomes.

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
Direction (1 to 6):  Three dice are thrown together. Find the probability of:

Q.1. Getting a total of 6.

(A) 20/108

(B) 30/100

(C) 5/108

(D) 6/120

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of 6:

Number of events of getting a total of 6 = 10

i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

Therefore, probability of getting a total of 6

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 10/216
       = 5/108

Q.2. Getting a total of 5.

(A) 1/36

(B) 2/18

(C) 1/8

(D) 1/30

Q.3. Getting a total of at most 6.

(A) 6/54

(B) 53/54

(C) 5/54

(D) 54/53

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of at most 6:

Number of events of getting a total of at most 6 = 20

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

Therefore, probability of getting a total of at most 6

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 20/216
       = 5/54

Q.4. Getting a total of at most 5

(A) 4/100

(B) 5/108

(C) 6/36

(D) 18/54

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of at most 5:

Number of events of getting a total of at most 5 = 10

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2).

Therefore, probability of getting a total of at most 5

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 10/216
       = 5/108

Q.5. getting a total of at least 6.

(A) 8/90

(B) 7/65

(C) 6/36

(D) 5/108

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of at least 6:

Number of events of getting a total of less than 6 (event of getting a total of 3, 4 or 5) = 10

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1) (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1).

Therefore, probability of getting a total of less than 6

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
 
       = 10/216
       = 5/108
 Therefore, probability of getting a total of at least 6 = 1 - P(getting a total of less than 6)

= 1 - 5/108

= (108 - 5)/108

= 103/108

Q.6. Getting a total of at least 5.

(A) 36/54

(B) 35/45

(C) 53/54

(D) 54/55

Therefore, total number of possible outcomes will be 6³ = (6 × 6 × 6) = 216.

getting a total of at least 5:

Number of events of getting a total of less than 5 = 4

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).

Therefore, probability of getting a total of less than 5

$$P(E) ={Number \ of \ favorable \ outcomes\over Total \ number \ of \ possible \ outcome }$$
       = 4/216
       = 1/54
 Therefore, probability of getting a total of at least 5 = 1 - P(getting a total of less than 5)

= 1 - 1/54

= (54 - 1)/54

= 53/54

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