# Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately

**Solution:**

Given, two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 are thrown and the sum of the numbers on them is noted.

We have to find the probability of getting each sum from 2 to 9 separately.

When the two dice are thrown. The possible outcomes are

(1,1) (1,1) (1,2) (1,2) (1,3) (1,3)

(2,1) (2,1) (2,2) (2,2) (2,3) (2,3)

(3,1) (3,1) (3,2) (3,2) (3,3) (3,3)

(4,1) (4,1) (4,2) (4,2) (4,3) (4,3)

(5,1) (5,1) (5,2) (5,2) (5,3) (5,3)

(6,1) (6,1) (6,2) (6,2) (6,3) (6,3)

Total number of possible outcomes = 36

Probability = number of favourable outcomes / number of possible outcomes

a) probability of getting a sum of 2

Favourable outcome = {(1,1) (1,1)}

Number of favourbale outcome = 2

Number of possible outcome = 36

Probability = 2/36

= 1/18

Therefore, the probability of getting a sum of 2 is 1/18.

b) probability of getting a sum of 3

Favourable outcomes = {(1,2) (1,2) (2,1) (2,1)}

Number of favourable outcomes = 4

Number of possible outcomes = 36

Probability = 4/36

= 1/9

Therefore, the probability of getting a sum of 3 is 1/9.

c) probability of getting a sum of 4

Favourable outcomes = {(1,3) (1,3) (2,2) (2,2), (3, 1), (3, 1)}

Number of favourable outcomes = 6

Number of possible outcomes = 36

Probability = 6/36

= 1/6

Therefore, the probability of getting a sum of 4 is 1/9.

d) probability of getting a sum of 5

Favourable outcomes = {(2,3) (2,3) (3,2) (3,2) (4,2) (4,2)}

Number of favourable outcomes = 6

Number of possible outcomes = 36

Probability = 6/36

= 1/6

Therefore, the probability of getting a sum of 5 is 1/6.

e) probability of getting a sum of 6

Favourable outcomes = {(5,1) (5,1) (3,3) (3,3) (4,2) (4,2)}

Number of favourable outcomes = 6

Number of possible outcomes = 36

Probability = 6/36

= 1/6

Therefore, the probability of getting a sum of 6 is 1/6.

f) probability of getting a sum of 7

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

Number of favourable outcomes = 6

Number of possible outcomes = 36

Probability = 6/36

= 1/6

Therefore, the probability of getting a sum of 7 is 1/6.

g) probability of getting a sum of 8

Favourable outcomes = {(5,3) (5,3) (6,2) (6,2)}

Number of favourable outcomes = 4

Number of possible outcomes = 36

Probability = 4/36

= 1/9

Therefore, the probability of getting a sum of 8 is 1/9.

h) probability of getting a sum of 9

Favourable outcomes = {(6,3)(6,3) }

Number of favourable outcomes = 2

Number of possible outcomes = 36

Probability = 2/36

= 1/18

Therefore, the probability of getting a sum of 9 is 1/18.

**✦ Try This: **Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 0, 4, 2, 5, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 8.

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 14

**NCERT Exemplar Class 10 Maths Exercise 13.3 Problem 23**

## Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately

**Summary:**

Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. The probability of getting each sum from 2 to 9 separately are 1/18, 1/9, 1/6, 1/6, 1/6, 1/6, 1/9, 1/18

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