Three houses are available in a locality and three different persons each independently apply for one house, choosing without consulting the others. What is the probability that all three applicants apply for exactly the same house?

Introduction / Context:
This problem illustrates probability with independent choices among a small number of options. Each of three persons chooses a house from three available houses, and we are interested in the chance that all three people end up choosing the same house. Such questions relate directly to the idea of outcomes in a discrete sample space where all choices are equally likely and independent.

Given Data / Assumptions:

• There are exactly 3 distinct houses available.
• There are 3 distinct persons applying for houses.
• Each person chooses one house only.
• Each person chooses independently and each house is equally likely to be chosen by any person.
• We want the probability that all three choose the same house.

Concept / Approach:
The key concept is to model each person's choice as an independent event with three equally likely outcomes. The total number of possible ordered triples of choices is 3^3. We then count the favorable outcomes in which all three chosen houses are identical. Finally, we divide the favorable count by the total count to obtain the required probability.

Step-by-Step Solution:
Step 1: Represent each person's choice as one of three houses, say H1, H2 or H3.Step 2: Since there are 3 independent choices and each has 3 outcomes, the total number of possible ordered outcomes is 3 * 3 * 3 = 3^3 = 27.Step 3: For all three applicants to choose the same house, they may all choose H1, all choose H2, or all choose H3.Step 4: Thus, the favorable ordered triples are (H1, H1, H1), (H2, H2, H2) and (H3, H3, H3).Step 5: The number of favorable outcomes is therefore 3.Step 6: Probability = favorable outcomes / total outcomes = 3 / 27 = 1 / 9.
Verification / Alternative check:
We can verify this answer by checking the complement event and ensuring probabilities sum correctly. The complement event is that the three applicants are not all choosing the same house. Its probability is 1 minus our answer, which is 1 minus 1/9 = 8/9. There is no conflict with any other probabilities because all 27 outcomes are accounted for and the sample space is uniform.

Why Other Options Are Wrong:
2/9 would imply 6 favorable outcomes, but only three same house outcomes exist. 8/9 is the probability that they are not all in the same house, not the event in question. 7/9 also does not correspond to any natural count in this context. 1/3 would require 9 favorable outcomes, which is not supported by any reasonable enumeration of possibilities.

Common Pitfalls:
A common mistake is to treat the persons as indistinguishable, which leads to undercounting the total outcomes. Another error is to think that the first person fixes the house, and then to incorrectly count subsequent probabilities without considering all three houses symmetrically. It is safer to count all ordered triples and then identify favorable ones systematically.

Final Answer:
The required probability that all three applicants apply for the same house is 1/9.