If we throw a die once, then possible outcomes (s), are
S = { 1, 2, 3, 4, 5, 6 }
⇒    n(E) = 6
(i) Let E be the favourable outcomes of getting an even number, then
E = { 2, 4, 6 }
⇒ n(S) = 3

Therefore, P(E) = 

(ii) Let F be the favourable outcomes of getting a number greater than 3, then
F = { 4, 5, 6 }
⇒    n( F) = 3

Therefore,  P(E) = 

(iii) Let G be the favourable outcomes of getting a number between 3 and 6.
G = { 4, 5 }
⇒    n(G) = 2

Therefore,  P(G) =