# Two dice probability questions

In this chapter, we will solve questions related to throwing of two dice.

We know that in a dice there are six numbers ( 1 to 6 ). When we throw two dice simultaneously, there are total of 6 x 6 = 36 possible outcomes.

The sample space for experiment is given as;
S = { (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) }

The outcome of getting same number on both the dice is called doublets.
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

I hope you understood the basic concept. Let’s solve some problems;

## Questions on two dice probability

Question (01)
Two dice are thrown simultaneously. Find the probability of getting (1, 1) and (1, 5).

Solution
Total possible outcome = 36

We want to get (1, 1) or (1, 5) as final solution.
Number of favorable outcome = 2

Probability = Number of favorable outcome / total possible outcome

Probability = 2 / 36 = 1/18

Hence, 1/18 is the required probability

Question 02
Two dice are thrown simultaneously. Find the probability of getting doublets.

Solution
When two dice are thrown, the doublets are;
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Number of favorable outcome = 6

Total possible outcomes = 36

Probability = 6 / 36 = 1 / 6

Hence, 1/6 is the required probability.

Question 03
In an experiment of throwing two dice simultaneously, find the probability for;
(a) getting sum of 4
(b) getting sum of 9

Solution
(a) Let A be the event to get sum of 4

The event A is represented as;
A = { (1, 3), (2, 2), (3, 1) }

Number of favorable outcome = 3
Total possible outcome = 36

Probability = 3 / 36 = 1/12

(b) Let B be the event of getting sum 9

B = { (3, 6), (4, 5), (5, 4), (6, 3) }

Number of favorable outcome = 4
Total number of possible outcome = 36

Probability = 4 / 36 = 1/9

Hence, 1/9 is the required probability.

Question 04
In an experiment of throwing two dice simultaneously, Find the probability of getting even number on first dice.

Solution
Let A be the event of getting even number on first dice.

A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }

Number of favorable outcome = 18

Total possible outcome = 36

Probability = 18 / 36 = 1/2

Hence, 1/2 is the probability of getting even number on first dice.

Question 05
Two dice are thrown simultaneously. Find the following probability;

(a) getting sum divisible by 7
(b) getting sum at least 11
(c) getting even multiple on first and odd multiple on second dice

Solution
(a) getting sum divisible by 7

Let A be the even of getting sum divisible by 7.

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) }

Total possible outcome = 6

Number of favorable outcome = 36

Probability = 6 / 36 = 1/6

(b) getting sum at least 11

Let B be the event of getting sum atleast 11.
B = {(5, 6), (6, 5), (6 , 6) }

Number of favorable outcome = 3
Total possible outcome = 36

Probability = 3/36 = 1/12

(c) getting even multiple on first and odd multiple on second dice

C = {(2, 1), (2, 3), (2, 5),
(4, 1), (4, 3), (4, 5),
(6, 1), (6, 3), (6, 5),}

Number of favorable outcome = 9

Total possible outcome = 36

Probability = 9/36 = 1/4

Hence, 1/4 is the required probability.