(i) When digits are repeated : In a 4-digit number greater than 5000, thousandth place can be filled up by either 5 or 7 So thousandth place can be filled in 2 ways. Since the digits can be repeated, therefore the remaining three places can be filled in $5^3$ = 125 ways. ∴ Total number of numbers formed = 2 × 125 = 250. But in 250 numbers, 5000 is also included. So, total number of numbers greater than 5000 = 250 – 1 = 249 A number is divisible by 5 if the digit at unit place is either 0 or 5. For a 4-digit number greater than 5000 and divisible by 5, the unit and thousandth place can be filled in 4 ways. The hundredth and tenth place are to be filled in $5^2$ = 25 ways. ∴ Number of numbers formed = 4 × 25 = 100. Also, in 100 numbers, 5000 is included. So, total number of numbers greater than 5000 and divisible by 5 = 100 – 1 = 99 Thus required probability = $\frac{99}{249}$ = $\frac{33}{83}$ (ii) When digits are not repeated : In a 4-digit number greater than 5000, thousandth place can be filled up by either 5 or 7. So, the hundredth, tenth and unit place may be filled in 4 × 3 × 2 = 24 ways. ∴ Total number of exhaustive cases = 2 × 24 = 48. A number greater than 5000 and is divisible by 5 when unit place is either 0 or 5 and thousandth place is either 5 or 7. The unit and thousandth place can be filled in 3 ways. The tenth and hundredth place can be filled in 3 × 2 = 6 ways. Number of favourable cases = 3 × 6 = 18. Thus, required probability = $\frac{18}{48}$ = $\frac{3}{8}$