# Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer

**Solution:**

We use the basic concepts of probability to solve the problem.

(i) Number of possible outcomes to get the sum as 2 = (1,1) = 2

Number of possible outcomes to get the sum as 3 = (2,1), (1, 2) = 2

Number of possible outcomes to get the sum as 4 = (2, 2), (1, 3), (3,1) = 3

Number of possible outcomes to get the sum as 5 = (3, 2), (2, 3), (4,1), (1, 4) = 4

Number of possible outcomes to get the sum as 6 = (5,1), (1, 5), (3, 3), (4, 2), (2, 4) = 5

Number of possible outcomes to get the sum as 7 = (4, 3), (3, 4), (6,1), (1, 6), (5, 2), (2, 5) = 6

Number of possible outcomes to get the sum as 8 = (4, 4), (6, 2), (2, 6), (5, 3), (3, 5) = 5

Number of possible outcomes to get the sum as 9 = (5, 4), (4, 5), (6, 3), (3, 6) = 4

Number of possible outcomes to get the sum as 10 = (5, 5), (6, 4), (4, 6) = 3

Number of possible outcomes to get the sum as 11 = (6, 5), (5, 6) = 11

Number of possible outcomes to get the sum as 12 = (6, 6) = 1

Thus, the table is:

Event ‘Sum of 2 dice’ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |

(ii) Probability of each of them is not 1/11 as these are not equally likely.

Check out more about terms of probability.

**Video Solution:**

## Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.

### NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 Question 22 - Chapter 15 Exercise 15.1 Question 22:

**Summary:**

A student argues that ‘there are 11 number of outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. (i) P(3) = 2/36, P(4) = 3/36, P(5) = 4/36, P(6) = 5/36, P(7) = 6/36, P(9) = 4/36, P(10) = 3/36, and P(11) = 2/36; (ii) Probability of each of them is not 1/11 as these are not equally likely.