When repetition of digits is not allowed The thousands place can be filled with either of the two digits $5$ or $7$ .

The remaining $3$ places can be filled with any of the remaining $4$ digits.

$\therefore$ Total number of $4\, -$ digit numbers greater than $5000=2 \times 4 \times 3 \times 2=48$

When the digit at the thousands place is $5$ , the units place can be filled only with $0$ and the tens and hundreds places can be filled with any two of the remaining $3$ digits.

$\therefore$ Here, number of $4 \,-$ digit numbers starting with $5$ and divisible by $5$

$=3 \times 2=6$

When the digit at the thousands place is $7$ , the units place can be filled in two ways ( $0$ or $5$ ) and the tens and hundreds places can be filled with any two of the remaining $3$ digits.

$\therefore$ Here, number of $4\,-$ digit numbers starting with $7$ and divisible by $5$.

$=1 \times 2 \times 3 \times 2=12$

$\therefore$ Total number of $4\,-$ digit numbers greater than $5000$ that are divisible by $5=6+12=18$

Thus, the probability of forming a number divisible by $5$ when the repetition of digits is not allowed is $\frac{18}{48}=\frac{3}{8}$.