Question

Given 10 rooms and 5 people, ask each person to choose one room to stay. What is the possibility that no people select same room?

Answer 1

i think its 1/10 * 1/9* 1/8 * 1/7* 1/6
though probability is not my strong point
a second opinion is essential!

Answer 2

Thanks. So the possibility is very little based on your equation. I was thinking it is P(10,5)/10^5=0.3

Answer 3

The number of ways to assign rooms, allowing multiple people to choose the same room is 10^5
the number of ways to assign rooms so only 1 person per room is 10*9*8*7*6
this ratio is your answer.

Answer 4

Same here. While then what is the possibility that there are two people select one room? C(10,1)*C(9,3)/10^5?

Answer 5

I think it's 10*(9*8*7) for the numerator. i.e. P(9,3) not C(9,3)

Answer 6

2nd opinion welcome!

Answer 7

Agree. P(9,3) instead of C(9,3)

Answer 8

identical to "birthday problem" with 10 days and 5 people.

Answer 9

pretty pretty sure it is
\[\frac{10P5}{10^{10}}\] but i would not swear to it

Answer 10

Ooh, confusing. I'll try...

Answer 11

well that is wrong. try
\[\frac{10\times 9\times 8\times 7\times 6}{10^5}\]

Answer 12

you can read the "birthday problem" here
http://en.wikipedia.org/wiki/Birthday_problem#Understanding_the_problem then change 365 to 10 and 23 to 5

Answer 13

I think it is kind of birthday problem. While if ask this:
Given these 10 rooms and 5 people, what is the (expected) number of rooms have two or more people in it?

Answer 14

first person picks a room. he has 10 choice. then second person. now he has 9 choices etc

Answer 15

to the number of ways they can all pick different rooms is
\[10\times 9\times 8\times 7\times 6\times 5\]

Answer 16

and the number of way total that 5 people can pick from 10 rooms is
\[10^5\]

Answer 17

so pretty sure that my second answer (not the first) is right

Answer 18

For the chance of no same room, It is P(10,5)/10^5=0.3, isn't it?

Answer 19

oh and since phi said exactly the same thing i am really pretty sure

Answer 20

@phi not sure why you stopped the numerator at 6 though. fifth person needs to pick a different room as well

Answer 21

oops now i see why! sorry

Answer 22

I'm sure about the first problem. But I'm also sure I have a bug in the 2nd question, the prob of two people picking the same room.

Answer 23

i only see one question. hmmm

Answer 24

Second question is :
what is the possibility that there are two people select one room? C(10,1)*P(9,3)/10^5?

Answer 25

Third one is :):
Given these 10 rooms and 5 people, what is the (expected) number of rooms have two or more people in it?

Answer 26

On the 3rd question, prob of multiple occupancy is 1- prob(unique rooms) = 1-0.3=0.7

Answer 27

0.7 is the prob of there is people share one room.
the question is , how many rooms we can expect to see that are shared by more than one people

Answer 28

Back to the 2nd q. I think we need a factor of 5*4 to account for the different pairs of people, so the numerator is 10*9*8*7*5*4

Answer 29

phi: Agree. For second question it is 10*9*8*7*5*4/10^5

Answer 30

Do you know?

Answer 31

Now I am little bit sure about the second question
Just no idea for the third one

Answer 32

question #2 is 1 - question 1

Answer 33

unless i messed that up too!

Answer 34

I interpreted it as exactly 2 which is different from not 1

Answer 35

oh yes. exactly one room has two people in it is not what i said for sure

Answer 36

Q3: how about this?
1 * (.7 * 0.3^4) + 2*(.7^2 * 0.3^3) + 3*(0.7^3 * 0.3^2)+ 4*(0.7^4 * 0.3)+5*(0.7^5)

Answer 37

= 1.25

Answer 38

let me be quiet that has nothing to do with it sorry

Answer 39

expected value is a weighted sum: number of rooms * probability of event occurring

Answer 40

I am trying to understand your equation phi

Answer 41

It was meant as an idea.... I do not believe it is correct.

Answer 42

Bad news on 2nd q. 10*9*8*7*5*4/10^5 = 1.008. So this is definitely wrong.

Answer 43

Now I am thinking the second question is
C(5,2)*C(10,1)*P(9,3)/10^5=0.72, right?

Answer 44

Sorry, it is 0.504 based on previous equation

Answer 45

Q3: expected # of shared rooms = 0.8146
1 1 1 1 1 0.3024
2 1 1 1 0.5040
2 2 1 0.1080
3 1 1 0.0720
3 2 0.0090
4 1 0.0045
5 0.0001
--------
1.0000
expected # shared = 1*0.5040 + 2*0.108 + 1*0.072 + 2*0.009 + 0.0045 + 0.0001

Answer 46

Thank you Phi!