Probability - Cominations and Integer Valued Vectors

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Homework Help Overview

The problem involves calculating the probability that three people checking into hotels in a town with five hotels each choose a different hotel. The context is rooted in combinatorial probability.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use combinations to find the probability but questions the correctness of their approach after comparing it to a different method. Some participants suggest alternative methods involving permutations and probabilities based on sequential choices.

Discussion Status

Participants are exploring various methods to approach the problem, including combinatorial reasoning and sequential probability calculations. There is acknowledgment of different perspectives on how to solve the problem, but no consensus has been reached yet.

Contextual Notes

The original poster expresses confusion about their initial method and seeks validation or alternative approaches. There is an indication of a desire to understand the problem better without simply receiving a solution.

AsianMan
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This problem comes from Sheldon Ross's book "A First Course in Probability (6th ed)."

There are 5 hotels in a certain town. If 3 people check into hotels in a day, what is the probability that they each check into a different hotel?


Attempt at a solution:

There are 5C3 = 10 different combinations of hotels where each individual person picks a different hotel.

I also decided that there were 7C4 = 35 possible ways for 3 individuals to choose from the 5 hotels, if more than 1 can stay in the same hotel. I got this answer because there are (n+r-1)C(r-1) distinct nonnegative integer-valued vectors (x1,x2,...,xr) satisfying x1 + x2 + ... + xr = n, where n = 3 and r = 5.

Therefore, I got 10/35 as my answer, but the answer is actually .48 (rounded?)

Interestingly, I got very close this answer mistakenly at first by dividing 5C3 by 7C2.
 
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Your'e making it too complicated

\frac {P(5,3)}{5^{3}}
 
The first person arrives and checks into any hotel. The second person arrives and checks into a hotel. What is the probability that person checks into a different hotel? The third person arrives. What is the probability this person checks into yet a different hotel? The probability that they check into three different hotels is the product of those two probabilities.. This is exactly the same as Random Variable gives- although, Random Variable, it would be better not to just "give" answers. Especially in the "coursework and homework sections".
 
Wow, I'm embarrassed.

Thanks guys!

BTW, is there any way to do it the way that I was doing it?
 

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