**In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) 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) Let E be event of getting the sum as 8**

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

We know that, Probability P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 5/36

**(ii) E ⟶ event of getting a doublet**

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

Total no. of possible outcomes = 36

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

**(iii) E ⟶ event of getting a doublet of prime no’s**

No. of favorable outcomes = 3 {(2, 2) , (3, 3) , (5, 5)}

Total no. of possible outcomes = 36

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

**(iv) E ⟶ event of getting a doublet of odd no’s**

No. of favorable outcomes = 3 {(1, 1) , (3, 3) , (5, 5)}

Total no. of possible outcomes = 36

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

**(v) E ⟶ event of getting a sum greater than 9**

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

Total no. of possible outcomes = 36

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

**(vi) E ⟶ event of getting an even no. on first**

No. of favorable outcomes = 18 {(2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 5) , (2, 6) , (4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 5) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6)}

Total no. of possible outcomes = 36

P(E) = 18/36 = 1/2

**(vii) E ⟶ event of getting an even no. on one and a multiple of 3 on other**

No. of favorable outcomes = 11 {(2, 3) , (2, 6) , (4, 3) , (4, 6) , (6, 3) , (6, 6) , (3, 2) , (3, 4) , (3, 4) , (3, 6) , (6, 2) , (6, 4)}

Total no. of possible outcomes = 36

P(E) = 11/36

**(viii) Bar E ⟶ event of getting neither 9 nor 11 as the sum of numbers on faces**

E ⟶ getting either 9 or 11 as the sum of no’s on faces

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

Total no. of possible outcomes = 36

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

P(Bar E) = 1 - P = 1 - 1/6 = 5/6

(ix) **E ⟶ event of getting a sum less than 6**

No. of favorable outcomes = 10 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (2, 1) , (2, 2) , (2, 3) , (3, 1) , (3, 2) , (4, 1)}

Total no. of possible outcomes = 36

P(E) = 10/36 = 5/18

**(x) E ⟶ event of getting a sum less than 7**

No. of favorable outcomes = 15 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (3, 1) , (3, 2) , (3, 3) , (4, 1) , (4, 2) , (5, 1)}

Total no. of possible outcomes = 36

P(E) = 15/36 = 5/12

**(xi) E ⟶ event of getting a sum more than 7**

No. of favorable outcomes = 15 {(2, 6) , (3, 5) , (3, 6) , (4, 4) , (4, 5) , (4, 6) , (5, 3) , (5, 4) , (5, 5) , (5, 6) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6)}

Total no. of possible outcomes = 36

P(E) = 15/36 = 5/12

**(xii) E ⟶ event of getting a 1 at least once**

No. of favorable outcomes = 11 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (3, 1) , (4, 1) , (5, 1) , (6, 1)}

Total no. of possible outcomes = 36

P(E) = 11/36

**(xiii) E ⟶ event of getting a no other than 5 on any dice**

No. of favourable outcomes = 25 {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 6) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 6) , (3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 6) , (4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 6)}

Total no. of possible outcomes = 36

P(E) = 25/36