given: a pair of dice have been thrown, so the number of elementary events in sample space is 6²=36

formula:

therefore n(S)=36

(i) let E be the event that the sum 8 appears

E= {(2,6) (3,5) (4,4) (5,3) (6,2)}

n(E)=5

(ii) let E be the event of getting a doublet

E= {(1,1) (2,2) (3,3) (4,4) (5,5) (6,6)}

n(E)=6

(iii) let E be the event of getting a doublet of prime numbers

E= {((2,2) (3,3) (5,5)}

n(E)=3

(iv) let E be the event of getting a doublet of odd numbers

E= {(1,1) (3,3) (5,5)}

n(E)=3

(v) let E be the event of getting sum greater than 9

E= {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}

n(E)=6

(vi) let E be the event of getting even on first die

E= {(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)}

n(E)=18

(vii) let E be the event of getting even on one and multiple of three on other

E= {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}

n(E)=11

(viii) let E be the event of getting neither 9 or 11 as the sum

E= {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n(E)=6

(ix) let E be the event of getting sum less than 6

E= {(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1)}

n(E)=10

(x) let E be the event of getting sum less than 7

E= {(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)}

n(E)=15

(xi) let E be the event of getting more than 7

E= {(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)}

n(E)=15

(xii) let E be the event of getting neither a doublet nor a total of 10

E’ be the event that either a doublet or a sum of ten appears

E’= {(1,1) (2,2) (3,3) (4,6) (5,5) (6,4) (6,6) (4,4)}

n(E’) =8

Therefore P(E)=1-P(E’)

(xiii) let E be the event of getting odd number on first and 6 on second

E= {(1,6) (5,6) (3,6)}

n(E)=3

(xiv) let E be the event of getting greater than 4 on each die

E= {(5,5) (5,6) (6,5) (6,6)}

n(E)=4

(xv) let E be the event of getting total of 9 or 11

E= {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n(E)=6

(xvi) let E be the event of getting total greater than 8

E= {(3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)}

n(E)=10