Solution:

On the off chance that 2 dices are tossed, the potential occasions are:

(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)

Along these lines, the complete quantities of occasions: 6×6 = 36

(I) It is offered that to get the aggregate as 2, the likelihood is 1/36 as the solitary potential results = (1,1)

For getting the aggregate as 3, the potential occasions (or results) = E (total 3) = (1,2) and (2,1)

In this way, P(sum 3) = 2/36

Essentially,

E (aggregate 4) = (1,3), (3,1), and (2,2)

In this way, P (total 4) = 3/36

E (aggregate 5) = (1,4), (4,1), (2,3), and (3,2)

In this way, P (total 5) = 4/36

E (total 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

Thus, P (total 6) = 5/36

E (total 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

Thus, P (total 7) = 6/36

E (total 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

Thus, P (total 8) = 5/36

E (total 9) = (3,6), (6,3), (4,5), and (5,4)

Thus, P (total 9) = 4/36

E (total 10) = (4,6), (6,4), and (5,5)

Thus, P (total 10) = 3/36

E (total 11) = (5,6), and (6,5)

Thus, P (total 11) = 2/36

E (total 12) = (6,6)

Thus, P (total 12) = 1/36

Thus, the table will be as:

Occasion:

Total on 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) The contention isn’t right as it is as of now legitimized in (I) that the quantity of all potential results is 36 and not 11.