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.