I         II        III       IV      V 

 

1        2        3        4       5

6        7        8        9      10

........................................

........................................

5n-4   5n-3   5n-2   5n-1    5n

 

x³ + y³ is divisible by 5 if x + y is divisible by 5.

 

Hence x and y can be choosen from I and IV , II and III or both from V.

No. of ways of choosing x and y from I and IV = (n)(n) = n²

No. of ways of choosing x and y from II and III = (n)(n) = n²

No. of ways of choosing x and y from V = ⁿC₂ = n(n-1)/2

 

Total no. of required ways = n² + n² + n(n-1)/2 = n(5n-1)/2

 

Total no. of ways of choosing x and y from 5n numbers = ⁵ⁿC₂ = 5n(5n-1)/2

 

Required probability = {n(5n-1)/2}/{5n(5n-1)/2} = 1/5