A town has 6 parks. One day, 6 friends, who are unaware of each

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Discussion Overview

The discussion revolves around a probability problem involving 6 friends randomly choosing from 6 parks and the likelihood that at least two of them select the same park. The scope includes mathematical reasoning and probability theory.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant proposes defining Event A as at least 2 friends going to the same park and Event B as all friends going to different parks, suggesting that P(A) = 1 - P(B).
  • Another participant agrees with the approach and emphasizes that counting the ways for Event B is simpler.
  • A participant calculates the total ways the friends can choose parks as 6^6 and the ways for Event B as 6!, leading to the expression for P(A) as 1 - (6!/6^6).
  • There is a question raised about the possibility of variance (Var(X)) being negative, with some participants expressing confusion about this concept.
  • One participant states their preference to avoid questions about variance, asserting that they do not believe it can be negative.

Areas of Agreement / Disagreement

There is general agreement on the approach to solving the probability problem, but uncertainty remains regarding the nature of variance, with differing opinions on whether it can be negative.

Contextual Notes

Participants have not fully resolved the question regarding variance, and there may be assumptions or definitions that are not explicitly stated in the discussion.

weiji
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A town has 6 parks. One day, 6 friends, who are unaware of each other's decision, choose a park randomly and go there at the same time. What is the probability that at least 2 of them go to the same park?
Any idea for this kind of question? I have an idea, but I don't know how to put in a mathematical way.
Let Event A = at least 2 of them go to the same park ; Event B = all of them go to the different park. ( They are not going to meet each other )
So, P(A) = 1 - P(B).. Then i don't know how to continue. =(
 
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Hello again! :smile:
weiji said:
Let Event A = at least 2 of them go to the same park ; Event B = all of them go to the different park. ( They are not going to meet each other )
So, P(A) = 1 - P(B).

Yes, that's exactly the right way to do it … this is a case where "not A" is a lot easier to count than A! :smile:

ok, now you need to count the number of ways they can go to different parks.

How many ways are there of putting someone in park #1?

Then how many ways are there of putting someone (someone else, of course) in park #2?

And so on … :wink:​
 


Yes, I got it. Thanks. The total ways they can go to parks are 6^(6), because 6 persons, each person has 6 options. Then event B, it would be 6! ways. Therefore the solution is 1 - (6!/6^6).

Thanks again, tiny-tim. I am picking up a probability course for this semester, and I am really thankful for your help.

By the way, is it possible for Var(X) to be negative? If it is negative, we will never get the standard deviation. But someone told me, it can be negative. This is really making me confuse.
 
Hi weiji! :smile:

(try using the X2 tag just above the Reply box :wink:)
weiji said:
Yes, I got it. Thanks. The total ways they can go to parks are 6^(6), because 6 persons, each person has 6 options. Then event B, it would be 6! ways. Therefore the solution is 1 - (6!/6^6).

Yup! :biggrin:
By the way, is it possible for Var(X) to be negative? If it is negative, we will never get the standard deviation. But someone told me, it can be negative. This is really making me confuse.

hmm … I usually avoid questions about Var(X) :redface: …

no, I don't think it can be negative.
 


Yea, me too. By the way, do you know anything about mathematical modeling?
 
Nope! :smile:
 

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