(i) Complete the following table:
Event: 'Sum on 2 dice' 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 5/36 1/36
(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?

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
(i) From the table it can be observed that,
To get sum as 2, possible outcomes=(1,1)
To get sum as 3, possible outcomes=(2,1),(1,2)
To get sum as 4, possible outcomes=(3,1),(1,3),(2,2)
To get sum as 5, possible outcomes=(2,3),(3,2),(1,4),(4,1)
To get sum as 6, possible outcomes=(1,5),(5,1),(2,4),(4,2),(3,3)
To get sum as 7, possible outcomes=(1,6),(6,7),(3,4),(4,3),(2,5),(5,2)
To get sum as 8, possible outcome=(2,6),(6,2),(3,5),(5,3),(4,4)
To get sum as 9, possible outcomes=(3,6),(6,3),(4,5),(5,4)
To get sum as 10, possible outcome=(4,6),(6,4),(5,5)
To get sum as 11, possible outcome=(5,6),(6,5)
To get sum as 12, possible outcome=(6,6)
(ii) The probability of each of these sums will not be 1/11, as these sums are not equally likely.