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

1. Jan 24, 2010

### weiji

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. =(

2. Jan 24, 2010

### tiny-tim

Hello again!
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!

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 …

3. Jan 28, 2010

### weiji

Re: Combinatorial

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.

4. Jan 28, 2010

### tiny-tim

Hi weiji!

(try using the X2 tag just above the Reply box )
Yup!
hmm … I usually avoid questions about Var(X)

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

5. Jan 28, 2010

### weiji

Re: Combinatorial

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

6. Jan 28, 2010

Nope! :rofl: