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 6³ = (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 6³ = (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(E₁) =
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(E₂) =
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(E₃) =
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(E₄) =
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(E₅) =
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(E₆) =
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.