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

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The discussion focuses on calculating the probability that at least two out of six friends choose the same park from six available options. The approach involves defining Event A as at least two friends going to the same park and Event B as all friends choosing different parks. The probability is calculated using the formula P(A) = 1 - P(B), where P(B) is determined by the number of ways to assign friends to different parks, resulting in the equation 1 - (6!/6^6). Additionally, there is a side conversation about the variance of a random variable, with participants expressing confusion about the possibility of negative variance, which is clarified as not being possible. The thread concludes with a focus on probability calculations and a brief mention of mathematical modeling.
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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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