A three digit number is formed at random using the digits 1, 2, 3, 4 and 5 without repetition. What is the probability that the number formed is divisible by 5?

Introduction / Context:
This question involves forming three digit numbers from a given set of digits without repetition. The focus is on divisibility by 5, which depends entirely on the last digit of the number. By understanding how many numbers can be formed in total and how many of those end with the required digit, we can find the probability efficiently.

Given Data / Assumptions:

• Available digits are 1, 2, 3, 4 and 5.
• Each digit can be used at most once in the number.
• A three digit number is formed, so the first digit cannot be zero, but there is no zero in the set anyway.
• We are interested in numbers that are divisible by 5.

Concept / Approach:
A number is divisible by 5 if and only if its units digit is 0 or 5. Since 0 is not in the list of digits, any valid number divisible by 5 must end with 5. That means the last digit is fixed as 5. The first and second digits are then arranged using the remaining four digits. We use permutations because the order of digits in a number matters. Finally, the probability is favourable permutations divided by total permutations.

Step-by-Step Solution:
Total digits available = 5. Total three digit numbers without repetition = 5P3 = 5 * 4 * 3 = 60. For divisibility by 5, the last digit must be 5. Fix the units digit as 5. Remaining digits for the first two places are 1, 2, 3 and 4, which is 4 digits. Number of ways to arrange the first two digits from these 4 digits = 4P2 = 4 * 3 = 12. So favourable numbers divisible by 5 = 12. Required probability = favourable / total = 12 / 60 = 1 / 5.

Verification / Alternative check:
We can double check by noticing that in all three digit numbers formed without repetition from five digits, each of the five digits is equally likely to appear in the last place. Therefore, the probability that the last digit is 5 is simply 1 out of 5. Since divisibility by 5 depends only on the last digit here, this reasoning again gives probability 1 / 5, confirming our detailed calculation.

Why Other Options Are Wrong:
1/4 would be correct only if there were four possible digits that could make the number divisible by 5, which is not true. 1/2 and 2/5 are too large and would imply that a very high proportion of all possible numbers are divisible by 5, which contradicts the exact count of 12 favourable numbers out of 60.

Common Pitfalls:
A common mistake is to forget that digits are used without repetition and to treat the number of possibilities as 5^3 instead of 5P3. Another error is to think that any occurrence of the digit 5 in the number makes it divisible by 5, whereas only the last digit matters. Focusing on the units place rule for divisibility by 5 is the key to solving this problem correctly.

Final Answer:
The probability that the three digit number formed is divisible by 5 is 1/5.