Question

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 $\frac{1}{11}$. Do you agree with this argument?

Justify your answer.

Answer 1

It can be observed that, there are 11 outcomes. so

To get the sum as 2, possible outcomes = (1, 1) the probability is 1/11 but the probability of getting 3 is

To get the sum as 3, possible outcomes = (2, 1) and (1,2) 2/11 as it is not 1/11 so im not agree

To get the sum as 4, possible outcomes = (3, 1),(1, 3),(2, 2)

To get the sum as 5, possible outcomes = (4, 1), (1,4), (2, 3), (3, 2)

To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2),

(3,3)

To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2),

(3,4), (4,3)

To get the sum as 8, possible outcomes = (6,2),(2,6),(3,5),(5,3),

(4,4)

To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4)

To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)

To get the sum as 11, possible outcomes = (5, 6), (6, 5)

To get the sum as 12, possible outcomes = (6, 6)

Event:

Sum of two dice

2

3

4

5

6

7

8

9

10

11

12

Probability

Probability of each of these sums will not be as these sums are not equally likely.