Three students are picked at random from a school with 1000 students. What is the probability that these three have the same birthday (same day and month)? Assume 365 equally likely birthdays and independence.

Introduction / Context:
This is a variant of the birthday problem. We want the probability that all three selected students share the same day and month of birth. The school size (1000) only ensures that sampling without replacement effects are negligible compared to assuming independent birthdays across students.

Given Data / Assumptions:

• 365 possible birthdays (ignoring leap day), equally likely.
• Three students chosen; independence of birthdays.

Concept / Approach:
Fix the first student’s birthday arbitrarily. For the other two to match, each must equal the first student’s day and month. Therefore multiply the two independent matching probabilities: (1/365)·(1/365) = 1/365^2.

Step-by-Step Solution:

Verification / Alternative check:
If we insisted on unordered counting across 365 categories, this matches the multinomial perspective: the only favorable case has all three in the same category, giving the same probability.

Why Other Options Are Wrong:
3/365 or 3/1000 are unrelated heuristics; 1/365 is the two-student match probability, not three.

Common Pitfalls:
Accidentally computing the probability that at least two match; or using 366 (leap year) when the question specifies a non-leap model.

Final Answer:
1/(365)^2