A five digit number is formed using the digits 0, 1, 2, 3 and 4 without repetition. The first digit cannot be 0. What is the probability that the number formed is divisible by 5?

Introduction / Context:
This question is a blend of basic number theory and counting methods. We form five digit numbers using the digits 0, 1, 2, 3 and 4 without repetition, under the restriction that the first digit of the number cannot be 0. We are asked to find the probability that such a number is divisible by 5. A number is divisible by 5 if and only if its last digit is 0 or 5; in this problem, the digit 5 is not available, so the last digit must be 0.

Given Data / Assumptions:

Available digits: 0, 1, 2, 3 and 4.

Each digit can be used at most once (no repetition).

The first digit of the number cannot be 0.

The number formed must be five digits long.

We want the probability that the number is divisible by 5.

Concept / Approach:
We first count the total number of valid five digit numbers we can form under the no repetition and non zero leading digit constraint. Then we count the number of such numbers that are divisible by 5. Since 5 is not among the allowed digits, divisibility by 5 is possible only when the last digit is 0. The required probability is the ratio of the favourable count to the total count.

Step-by-Step Solution (Total Numbers):
Step 1: Total permutations of the 5 distinct digits without any restriction = 5! = 120.Step 2: Count invalid permutations where the first digit is 0. If 0 is fixed at the first position, the remaining 4 digits can be arranged in 4! = 24 ways.Step 3: Therefore, valid five digit numbers (with first digit not 0) = 120 - 24 = 96.

Step-by-Step Solution (Favourable Numbers):
Step 1: For divisibility by 5, last digit must be 0, because 5 is not one of the digits.Step 2: Fix 0 in the last position. The remaining four positions (from left to right) must be filled by the digits 1, 2, 3 and 4.Step 3: Since these four digits are all non zero, any arrangement of them in the first four positions yields a valid five digit number.Step 4: Number of ways to arrange 1, 2, 3 and 4 in the first four positions = 4! = 24.Step 5: Therefore, the number of favourable five digit numbers divisible by 5 = 24.Step 6: Probability = favourable / total = 24 / 96 = 1 / 4.

Verification / Alternative check:
We can double check by noticing that all valid five digit numbers (with these digits and a non zero first digit) are equally likely if we choose a random arrangement. Since the last digit can equally likely be any of the five digits subject to the non zero first digit constraint, and only one choice for the last digit (0) gives divisibility by 5, the fraction of such numbers is effectively 1 out of 4 possible non leading positions, aligning with the more explicit counting result of 1 / 4.

Why Other Options Are Wrong:
The probability 1/5 would be correct only if any digit could appear in the last place with equal likelihood and digits 0 and 5 were available. However, 5 is missing here. The values 2/5, 3/5 and 1/2 are too large and imply that a much higher fraction of numbers are divisible by 5 than is actually the case. Only 1/4 matches the exact counting of valid and favourable permutations.

Common Pitfalls:
Students sometimes forget to remove the invalid numbers with leading 0 when counting the total. Others mistakenly assume that both 0 and 5 are available as last digits or that the five digit number can start with 0. Carefully handling the first digit restriction and the divisibility rule ensures a correct solution.

Final Answer:
The probability that the five digit number formed is divisible by 5 is 1/4.