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

  • Thread starter weiji
  • Start date
  • Tags
    Friends
In summary, the conversation discusses the probability of at least 2 out of 6 friends choosing the same park on a given day. The solution is found by counting the number of ways they can go to different parks, which is 6^(6), and subtracting it from the total number of ways they can go to any park, which is 6!. The resulting fraction, 1 - (6!/6^6), gives the probability of at least 2 of them going to the same park. Additionally, the conversation briefly touches on the topic of Var(X) and its potential for being negative.
  • #1
weiji
7
0
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. =(
 
Physics news on Phys.org
  • #2
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:
 
  • #3


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
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.
 
  • #5


Yea, me too. By the way, do you know anything about mathematical modeling?
 
  • #6
Nope! :rofl:
 

1. How many possible ways can the 6 friends divide themselves into the 6 parks?

There are 720 possible ways for the 6 friends to divide themselves into the 6 parks. This can be calculated by using the formula for permutations, which is n! (n factorial), where n is the number of objects. In this case, it would be 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.

2. Is there a way for all 6 friends to end up in the same park?

Yes, it is possible for all 6 friends to end up in the same park. This can be achieved if all 6 friends choose the same park to visit.

3. What is the probability of all 6 friends choosing different parks?

The probability of all 6 friends choosing different parks is 1/720 or approximately 0.0014. This can be calculated by taking the number of ways for all 6 friends to choose different parks (720) and dividing it by the total number of possible ways for them to divide themselves among the 6 parks (720).

4. If 2 of the friends decide to visit the same park, how many possible ways can the other 4 friends divide themselves among the remaining 5 parks?

There are 120 possible ways for the other 4 friends to divide themselves among the remaining 5 parks. This can be calculated by using the formula for combinations, which is nCr (n choose r), where n is the total number of objects and r is the number of objects being chosen. In this case, it would be 5C4 = 5! / (4! x 1!) = 5.

5. Can the 6 friends visit all 6 parks in a specific order?

Yes, the 6 friends can visit all 6 parks in a specific order. This can be achieved if the 6 friends agree on a specific order in which they will visit the parks, such as park 1, park 2, park 3, park 4, park 5, and park 6.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
11
Views
355
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
1K
  • Sci-Fi Writing and World Building
Replies
18
Views
2K
  • Quantum Interpretations and Foundations
Replies
22
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
14
Views
859
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
530
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
2K
Back
Top