Two numbers a and b are chosen at random from the set {1, 2, 3, ....5n}. The probability that $a^4-b^4$  is divisible by 5, is

$\frac{17n-5}{5(5n-1)}$

The number of ways of choosing a and b from the given set of 5n integers is ${^{5n}C}_2$.
Let us divide the given set of 5n integers in 5 groups as follows:

$G_1:1, 6, 11,..., 5n-4$

$G_2 : 2,7,12,..., 5n-3 $

$G_3: 3, 8, 13,..., 5n-2$

$G_4:4, 9, 14, ..., 5n-1 $

$G_5:5, 10, 15, ..., 5n$

We have, $a^4-b^4 = (a - b) (a+b) (a^2 + b^2)$

So, we observe that $a^4-b^4$ will be divisible by 5, if both a and i belong to the last group or if they belong to any of the remaining four groups. Thus, the number of favourable elementary events is ${^nC}_2 +{^{4n}C}_2.$

Hence, required probability $=\frac{^nC_2 + {^{4n}C}_2}{^{5n}C_2}=\frac{17n-5}{5(5n-1)}$