Dostre said:
Homework Statement
Seven persons are staying at a hotel. The hotel has twenty-five floors, each with the same number of rooms. Each of the seven persons has been randomly assigned to one of the rooms in the hotel.
a. What is the probability that at least two people have rooms on the same floor? Show and explain how you calculate the answer.
b. Suppose you are one of the seven people. What is the probability that at least one of the other six persons has a room on the same floor as you? Assume there are at least 6 rooms on each floor of the hotel. Show and explain how you calculate the answer
c. Explain qualitatively why the answer to part b. is smaller than the answer to part a. In other words, give an intuitive argument or reason for the difference in probabilities.
Homework Equations
P(A)=1-P(A^c)
The Attempt at a Solution
Part a. Probability that at least two people have a room on the same floor is the complement of probability that no one has rooms on the same floor.
Event A=at least two people have rooms on the same floor
Event A^{c}=everybody lives on distinct floors.
P(A^c)=(25*24*23*22*21*20*19)/25^7=0.397
P(A)=1-P(A^c)=0.603
I think it is correct.
Part b. Using the same logic and taking into account that there are at least 6 rooms on each floor I get that there 24*6=184 rooms to choose, since I occupy the 25th floor.
Event A=at least one of the other six people has a room on the same floor with me
Event A^{c}=nobody lives on the same floor with me
Hence,
P(A^c)=(184*183*182*181*180*179)/184^6=0.92
P(A)=1-0.92=0.08
Is this correct?
Part c. I think it is because in part b order matters, but I am not sure. Help appreciated. Thanks.
For part (a) the answer depends on the number of rooms per floor. Say there are ##R## rooms on each floor. The first person to register gets some room on some floor, ##F_1##. That leaves ##24R## rooms on floors different from the first one, and ##25R-1## rooms altogether for the next person to occupy. So (given a random choice of rooms) the probability that the second person to register gets a room on a different floor (##F_2##) from the first is ##P(F_2 \neq F_1) = 24R/(25R-1)##. The probability that the third person's floor ##F_3## is different from the first two is ##23R/(25R-2)##, etc. So, in terms of the total number of rooms ##N = 25R## we have
P\{ \text{all different} \} = \frac{24 \cdot 23 \cdot 22 \cdots \cdot 19 R^6}{(25R-1)(25R-2) \cdots (25R-6)} = \frac{19381824}{48828125}\frac{1}{(1-\nu)(1-2\nu) \cdots (1-6\nu)}, \: \nu \equiv \frac{1}{N}
The first factor in the above is ##19381824/48828125 \doteq .3969397555## so gives your 0.397 answer; the second factor is a correction that is always > 1 but which will be pretty close to 1 for a large hotel. In any case, if you know ##R## you can compute it exactly. Of course, if ##R \to \infty## then the correction factor = 1 exactly, and your answer obtains. Your answer would also be correct if floors, rather than rooms, are selected at random, but that is not what the question implied.
You can do part (b) in terms of a hypergeometric distribution: from your point of view there are two types of rooms left for the other 6 to occupy: rooms of type I = other rooms on your floor, and rooms of type II = rooms on other floors. There are ##N_1 = R-1## rooms of type I and ##N_2 = 24R## rooms of type II. The complementary probability in (b) is the probability that in selecting 6 rooms without replacement (from the total ##25R-1## rooms) we obtain ##0## type I rooms. This is just the hypergeometric probability ##P(X = 0)##, where ##X## is a 'hypergeometric' random variable with parameters ##(N_1,N_2,n) = (R-1,24R,6)##.
