Form a 4-digit number without repetition using digits {1, 3, 5, 7, 9}. What is the probability that the number is divisible by 5?

Introduction / Context:
A number is divisible by 5 iff its last digit is 0 or 5. Here, the available digits are {1,3,5,7,9} and each can be used at most once, so the last digit must be 5.

Given Data / Assumptions:

• Digits: 1, 3, 5, 7, 9; no repetition.
• All 4-digit permutations using these digits are equally likely.

Concept / Approach:
Count favorable permutations with 5 in the units place; divide by total permutations of length 4 from 5 distinct digits.

Step-by-Step Solution:
Total 4-digit numbers = 5P4 = 5 × 4 × 3 × 2 = 120.Favorable: fix last digit = 5 → arrange remaining 4 digits in the thousands–hundreds–tens positions = 4! = 24.Probability = 24 / 120 = 1/5.

Verification / Alternative check:
Symmetry: each of the 5 digits is equally likely to occupy the last place under uniform random permutation; hence probability the last digit is 5 equals 1/5.

Why Other Options Are Wrong:
Fractions larger than 1/5 ignore the single admissible last digit; 2/3 and 1/3 are not justified by the counting.

Common Pitfalls:
Accidentally allowing repetition or considering numbers ending with 0 (digit 0 not available).

Final Answer:
1/5