Two numbers a and b are chosen at random from the set {1 ,2, 3, ......, 3n}. The probability that $a^3 +b^3$ is divisible by 3, is

$\frac{1}{3}$

The number of ways of choosing two numbers from the given set is ${^{3n}C}_2$.

Let us divide given 3n numbers into the groups $G_1, G_2$ and $G_3$ as follows:

$G_1 : 3, 6, 9, ...., 3n $

$G_2 : 1, 4, 7, 10, ....., 3n -1$

$G_3 : 2, 5, 8, 11, ...., 3n -2$

We have,$ a^3 + b^3 = (a + b)^3 -3ab (a + b)$

Therefore, $a^3 + b^3 $ will be divisible by 3, if a + b us divisible by 

(i) Both the number belong to the first group.

(ii) One of the two numbers belongs to second group is one of them belongs to the third group.

∴ Favourable number of elementary events = ${^nC}_2 + {^nC}_1 + {^nC}_1$

Hence, required probability $=\frac{^nC_2  + {^nC}_1×{^nC}_1}{^{3n}C_2}=\frac{1}{3}$