Let T be the set of all students: T = {B1...B20, G1...G10} ⟹ n(T) = 30
Let G be the set of all girls: G = {G1...G10} ⟹ n(G) = 10
Let B be the set of all boys: B = {B1...B20} ⟹ n(B) = 20
Let BE be the set of all blue eyed students: BE = {B1...B10, G1...G5} ⟹ n(BE) = 15
.....as (10 boys + 5 girls)
We want to find P(G ∪ BE)
P(G ∪ BE) = P(G) + P(BE) - P(G∩BE) .......................... (Since sets G and BE are overlapping)
P(G ∪ BE) = n(G)/n(T) + n(BE)/n(T) - n(G∩BE)/n(T)
P(G ∪ BE) = 10/30 + 15/30 - 5/ 30
P(G ∪ BE) = 20/30 = 2/3
I hope these more formal steps help, Rod has already excellently explained in his answer.