S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),...........,(2,6)(3,1),(3,2)..............(3,6)(4,1),(4,2).............(4,6),(5,1),(5,2).............(5,6)(6,1),(6,2)...........(6,6)}
A={(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)}
B={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
C={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}
(i) Clearly, A∩B=ϕ
∴ A and B are mutually exclusive.
Thus the given statement is true.
(ii)A∩B=ϕ and A∪B=S
∴ A and B are mutually exclusive and exhaustive.
Thus the given statement is true.
(iii)B′={(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)}=A
Thus the given statement is true.
(iv) A∩C={(2,1),(2,2),(2,3),(4,1)}≠ϕ
∴ A and C are not mutually exclusive.
Thus the given statement is false..
(v) A∩B′=A∩A=A
∴A∩B′≠ϕ
∴ A and B' are not mutually exclusive.
Thus the given statement is false.
(vi) A′∪B′∪C=S
However B′∩C={(2,1),(2,2),(2,3),(4,1)}≠ϕ
Therefore events A′,B′ and C are not mutually exclusive and exhaustive.
Thus the given statement is false.