# Probability for Rolling Three Dice

Probability

for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots

in each (three) dies.

When three dice are thrown simultaneously/randomly, thus number of event can be 6^{3} = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces.

Worked-out problems involving probability for rolling three dice:

**1.** Three dice are thrown together. **Find the probability of:**

(i) getting a total of 5

(ii) getting a total of atmost 5

(iii) getting a total of at least 5.

(iv) getting a total of 6.

(v) getting a total of atmost 6.

(vi) getting a total of at least 6.

**Solution:**

Three different dice are thrown at the same

time.

Therefore, total number of possible outcomes will be 6^{3} = (6 × 6 × 6) = 216.

**(i)
getting a total of 5:**

Number of events of getting a total of 5 =

6

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

2, 1), (2, 1, 2) and (1, 2, 2)

Therefore, probability of getting a total

of 5

Number of favorable outcomes

P(E_{1}) =

Total number of possible outcome

= 6/216

= 1/36

**(ii) getting a total of
atmost 5:**

Number of events of getting a total of atmost

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 atmost 5

Number of favorable outcomes

P(E_{2}) =

Total number of possible outcome

= 10/216

= 5/108

**(iii)
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

Number of favorable outcomes

P(E_{3}) =

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

**(iv)
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

Number of favorable outcomes

P(E_{4}) =

Total number of possible outcome

= 10/216

= 5/108

**(v)
getting a total of atmost 6:**

Number of events of getting a total of atmost

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 atmost 6

Number of favorable outcomes

P(E_{5}) =

Total number of possible outcome

= 20/216

= 5/54

**(vi)
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

Number of favorable outcomes

P(E_{6}) =

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

These examples

will help us to solve different types of problems based on probability for

rolling three dice.