If {x, y} is a subset of the first 30 natural numbers, then the probability, that $x^3+y^3$is divisible by $3,$

$x^3+y^3$ is divisible by $3$
$\Rightarrow$x+y is  divisible by $3 \Rightarrow$x,y are multiples of $3$ or one
leaves remainder $1$ and the other $2$ when divided by $3. 3, 6, 9, , 30$
are multiples of $3, 1, 4, 7, , 28$ leave remainder$1, 2, 5, 8,......., 29$ leave remainder $2$.
$\therefore$Required probability = $=\frac{\left({10}_{C_1}\right)\left({10}_{C_1}\right)+\left({10}_{C_2}\right)}{\left({30}_{C_2}\right)}=\frac{145}{15\times 29}=\frac{1}{3}$