Total no. of possible outcomes when 2 dice are thrown = 6×6 = 36 which 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) }

**(i) E ⟶ event of getting sum that turn up is 8**

No. of possible outcomes = 36

No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 5/36

**(ii) Let E ⟶ event of obtaining a total of 6**

No. of favourable outcomes = 5

{(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}

P(E) = 5/36

(iii) Let E ⟶ event of obtaining a total of 10.

No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}

P(E) = 3/36 = 1/12

**(iv) Let E ⟶ event of obtaining the same no. on both dice**

No. of favourable outcomes = 6 {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}

P(E) = 3/36=1/12

**(v) E ⟶ event of obtaining a total more than 9**

No. of favourable outcomes = 6 {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}

P(E) = 6/36 = 1/6

**(vi) The maximum sum is 12 (6 on 1**^{st} + 6 on 2^{nd}**)**

So, getting a sum of no’s appearing on the top of the two dice as 13 is an impossible event.

∴ Probability is 0

**(vii) Since, the sum of the no’s appearing on top of 2 dice is always less than or equal to 12, it is a sure event.**

Probability of sure event is 1.

So, the required probability is 1.