# Probability for Rolling Two Dice – Examples | How to find Probability of Rolling Two Dice?

Probability is a possibility of outcome for a nonoccurrence event. That means, when we are not sure about the outcome or result of an event, at that moment we can apply the probability method to the event. So that, we will know the chances of outcome of an event. For example, if we are trying to flip a coin and we don’t know the result or outcome of a coin that means, either it may be heads or tails. In such a case, we can use the probability method. If you want to know What is the Probability for Rolling Two Dices

Get to know the Probability When Two Dice are Rolled, Solved Examples on How to Calculate the Two Dice Rolling Probability, etc.  Also, find the Possible Outcomes Whe Two Dice are Rolled by checking out the Probability Table.

## Two Dice Rolling Probability

In order to determine the probability of a dice roll we need to know two things namely

• Size of the Sample Space or Set of Possible Outcomes
• How often an Event Occurs

If you throw a single die the sample space is equal to values on the die i.e. (1, 2, 3, 4, 5, 6). Since th die is fair each number in the set occurs only once. To obtain the probability of rolling any number on the die we divide the event frequency by the size of sample space.

In the same way, When Two Dice are Rolled calculating the Probability becomes difficult. Here, Rolling One Die is independent of the other. One roll has no effect on the other and while dealing with the independent events we use the multiplication rule.

## Possibilities of Outcomes When Two Dice are Rolled

As said above, when we throw two dice, there is the possibility to get 36 outcomes. Have a look at the possibilities below.
Probability of an event = Number of favorable outcomes/ Total number of outcomes
Two dice are thrown at a time. Here, one die is x and another one is y.
X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}

Probability Table for Rolling Two Dice

The Possible Outcomes When Two Dice are Rolled is given below. Total Possible Outcomes is equal to the Product of sample space of the first die(6) and the sample space of the second die(6) that is 36.

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

Possibility of outcomes are {(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), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Total number of outcomes are 36.

### Two Dice Probability Examples

1. Two dice are rolled. Find the Probability of the sum of scores is an even number?

Solution:
Two dices are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
Add the scores of the two dices. That is
{(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)}.
So, the number of possibilities of the sum of even numbers is 18.
The probability of an event = number of favorable outcomes/ total number of outcomes.

Probability of sum of an even numbers = 18 / 36 = 1 / 2.

2. Two dice are rolled. Find the Probability of the sum of scores is an odd number?

Solution:
Two dices are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of two dices are 36.
Add the scores of the two dices. That is
{(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}.
So, the number of possibilities of the sum of odd numbers is 18.
The probability of an event = number of favorable outcomes/ total number of outcomes.

Probability of sum of an odd numbers = 18 / 36 = 1 / 2.

3. Two dice are rolled. Find the probability of the sum of 2, 4, and 12?

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
(i) Number of favorable outcomes of the sum of 2 is (1,1).
So, only 1 outcome.
The total number of outcomes = 36.
So, the probability of an event = number of favorable outcomes/ total number of outcomes.
Probability of sum of 2 = 1/36.
(ii) Number of favorable outcomes of sum of 4 are {(1,3), (2,2), (3,1)}.
So, the number of favorable outcomes is 3.
The total number of outcomes = 36.
So, the probability of an event = number of favorable outcomes/ total number of outcomes.
Probability of sum of 4 = 3/36 = 1/12.
(iii) Number of favorable outcomes of the sum of 12 are {(6,6)}.
So, a number of favorable outcomes is 1.
The total number of outcomes = 36.
So, the probability of an event = number of favorable outcomes/ total number of outcomes.
Probability of sum of 12 = 1/36.

4. Two dice are rolled. P is the event that the sum of the numbers shown on the two dice is 5, and Q is the event that at least one of the dice shows up a 3?
Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
P is sum of 5. That is, P = {(1,4), (2,3), (3,2), (4,1)}.
Q is at least one of the dice shows up 3. That is, Q = {(1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)}
(i) Mutually Exclusive
P ∩ Q = {(2, 3), (3, 2)} ≠ ∅.
Hence, P and Q are not mutually exclusive.
(ii) Exhaustive
P∪ Q ≠ S.

Therefore, P and Q are not exhaustive events.

5. Two dice are thrown simultaneously. Find the probability of (i) doublet (ii) product of 6 (iii) Divisible by 4 (iv) total of at least 10 (v) sum of 8?

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of two dices is 36.
(i) Doublets
Possibilities of doublets are {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} = 6.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 6 / 36 = 1 / 6.
(ii) Product of 6.
Possibilities of product of 6 is {(1,6), (2,3), (3,2), (6,1)} = 4.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 4 / 36 = 1 / 9.
(iii) Divisible by 4.
Possibilities of Divisible by 4 is {(1,4), (2,2), (2,6), (3,1), (3,5), (4,1), (4,4), (5,3), (6,2), (6,6)} = 10.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 10 / 36 = 5 / 18.
(iv) Total of at least 10.
Possibilities of Total of at least 10 are {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} = 6.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 6 / 36 = 1 / 6.
(v) Sum of 8.
Possibilities of sum of 8 are {(2,6), (3,5), (4,4), (5,3), (6,2)} = 5.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 5 / 36.

6. Two dice are thrown. Find the probability of
(i) multiple of 4.
(ii) multiple of 5.
(iii) prime number as the sum.
(iv) product as 2.
(v) sum as < = 5.
(vi) getting a multiple of 4 on one die and multiple of 2 on another die.

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
(i) Multiple of 4.
Possibilities of multiple of 4 are {(1,3), (2,2), (2,6), (3,1), (3,5), (4,4), (5,3), (6, 2), (6,6)} = 9.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Multiple of 4 = 9 / 36 = 1 / 4.
(ii) Multiple of 5.
Possibilities of multiple of 5 are {(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)} = 7.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Multiple of 5 = 7 / 36.
(iii) Prime number as sum
Prime numbers are 1,2,3,5,7,11,13…..
Possibilities of prime number as sum = {(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.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Prime number as sum = 15 / 36 = 5 / 12.
(iv) Product as 2
Possibilities of product as 2 are {(1,2), (2,1)} = 2.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Product as 2 = 2 / 36 = 1 / 18.
(v) sum as < = 5.
Possibilities of sum as < = 5 are {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} = 10.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of sum as <=5 = 10 / 36 = 5 / 18.
(vi) getting a multiple of 4 on one die and multiple of 2 on another die.
Possibilities of getting a multiple of 4 on one die and multiple of 2 on another die are {(4, 2), (4, 4), (4,6), (2,4), (6,4)} = 5.
Probability of an event = number of favorable outcomes / total number of outcomes.

Probability of getting a multiple of 4 on one die and multiple of 2 on another die = 5 / 36.

7. Two dice are thrown. Find out the (i) the odds in favor of getting the sum 4, and (ii) the odds against getting the sum 3.

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.

(i) The odds in favor of getting the sum 4.
Possibilities of getting the sum 4 = {(1,3), (2,2), (3,1)} = 3.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of getting the sum 4 = 3 / 36 = 1 / 12.
Odds in favor of getting the sum 4 = probability of getting the sum 4 / (1 – probability of getting the sum 4).
Odds in favor of getting the sum 4 = (1 / 12) / (1 – (1 / 12).
= (1 / 12) / (11 / 12).
=1 / 11.

Finally, the odds I favor of getting the sum of 4 is equal to 1 / 11.

(ii) the odds against getting the sum 3.
Possibilities of getting the sum 3 = {(1,2), (2,1)} = 2.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of getting the sum 3 = 2 / 36 = 1 / 18.
Odds in favor of getting the sum 3 = probability of getting the sum 3 / (1 – the probability of getting the sum 3).
Odds in favor of getting the sum 3 = (1 / 18) / (1 – (1 / 18).
= (1 / 18) / (17 / 18).
=1 / 17.

Finally, the odds I favor of getting the sum 3 is equal to 1 / 17.

8. Two dice are thrown. Find the probability that the numbers on the two dices are different?

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
Possibilities of the numbers on the two dices are different = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5)} = 30.
Probability of an event = number of favorable outcomes / total number of outcomes.

Probability of the numbers on the two dices are different = 30 / 36 = 5 / 6.