If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7,

Solution:

(i) When the digits are repeated

Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5.

The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed.

$\therefore$ Total number of 4-digit numbers greater than $5000=2 \times 5 \times 5 \times 5-1$

= 250 − 1 = 249

[In this case, 5000 can not be counted; so 1 is subtracted]

A number is divisible by 5 if the digit at its units place is either 0 or 5.

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by $5=2 \times 5 \times 5 \times 2-1=100-1=99$

Thus, the probability of forming a number divisible by 5 when the digits are repeated is $=\frac{99}{249}=\frac{33}{83}$.

(ii) 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 × 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 × 2 × 3 × 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 $\frac{18}{48}=\frac{3}{8}$.