Question

If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated? (ii) the repetition of digits is not allowed?

Answer 1

(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.
Total number of 4-digit numbers greater than 5000 =$2\times 5\times 5\times 5-1$
$=250-1=249$
[In this case, 5000 cannot be counted; so 1 is subtracted]
A number is divisible by 5 if the digit at its units place is either 0 or 5.
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.
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.
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.
Here, number of 4-digit numbers starting with 7 and divisible by 5
$=1\times 2\times 3\times 2=12$
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 $=\frac{18}{48}=\frac{3}{8}$