prob 2.

there are n numbers each of the form 3n,3n+1,3n+2...

x+y is divisible by 3 in two cases
first when x and y are both of the form 3n and the other when one of them is of the form 3n+1 and the other of the form 3n+2....
for the first case.....
x and y can be chosen in ⁿC₂ ways

for the second case:
x and y both can be chosen in n ways.....

so total favourable events=ⁿC₂+n²
sample space=³ⁿC₂

therefore probability=ⁿC₂+n²³ⁿC₂