Two numbers $x$ and $y$ are chosen at random without replacement from the first $30$ natural numbers. The probability that

$x^2-y^2$ is divisible by$3$ is

Total number of ways of choosing two numbers out of $30$ is $n\left(S\right)={ }^{30}{C}_2$

E=  Event of selecting two numbers x, y such that $x^2-y^2$  is divisible  by 3

We rewrite the first $30$ natural numbers in three rows as follows:

$\begin{array}{l}\text{ Row I: }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1,4,7,10,13,16,19,22,25,28\\ \text{ Row II: }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2,5,8,11,14,17,20,23,26,29\\ \text{ Row III: }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3,6,9,12,15,18,21,24,27,30\end{array}$

For $x^2-y^2$ to be divisible by $3$, either both $x$and $y$ must be chosen from the same row or exactly one of$x, y$ from Row $I$and the other from Row$II.$ Thus, the number of favourable ways

$n\left(E\right)=3\left({ }^{10}{C}_2\right)+10\times 10=235$

$\therefore$ probability of the required event

$=\frac{235}{{ }^{30}{C}_2}=\frac{235\times 2}{30\times 29}=\frac{47}{87}$