# People in an elevator .

#### robespierre722

##### New member
Hello , i know that this problem in kinda famous but there is a difference and i am not really sure how to finish it :
3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.
Each person is independent of the others to get off at any of the 5 floors above ground.
a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)
b)What is the probability that no person from the first elevator will want to go at the 5th floor.

a) So there are 5 possibilities for each person -> 5^6 possible cases
There are 6C2=15 possible pairs for that 3rd floor .
And now the funny part, i was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60
And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .

b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .

#### Jomo

##### Elite Member
b)What is the probability that no person from the first elevator will want to go at the 5th floor.

b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .
Why are you mentioning the 2nd elevator. From the beginning there could have been 6 elevators. However part a only talks about elevator 1 and 2. Does that change how you want to answer part a??
There is no elevator 2 in part b. 3 people get on elevator 1 on the ground floor of a six story building. What is the probability that no one gets off on the fifth floor?
1st of all I have a problem with this problem. Suppose I want to go to the 4th floor or even the 5th floor and then I remember that I left something in my car and I now want to go to the ground floor (has that ever happened to you?). I think that this needs to be included as an option, however we will not consider that possible. Each person has 5 possible floors to choose from with equal probability. So there are 6^5 ways for them to exit the elevator. How many of these ways do not include anyone going to the 5th floor? Figure that out and divide by 6^5 for your answer.

#### Jomo

##### Elite Member
Hello , i know that this problem in kinda famous but there is a difference and i am not really sure how to finish it :
3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.
Each person is independent of the others to get off at any of the 5 floors above ground.
a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)
We can consider that all 6 passengers are in the same elevator.
I would think of this problem as follows: You have a biased coin where p(heads) = 1/5 and p(tails) = 4/5. Do you see why this models your problem? Now toss the coin 6 times and find the p(getting exactly 2 heads)

#### JeffM

##### Elite Member
Hello , i know that this problem in kinda famous but there is a difference and i am not really sure how to finish it :
3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.
Each person is independent of the others to get off at any of the 5 floors above ground.
a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)
b)What is the probability that no person from the first elevator will want to go at the 5th floor.

a) So there are 5 possibilities for each person -> 5^6 possible cases
There are 6C2=15 possible pairs for that 3rd floor .
And now the funny part, i was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60
And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .

b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .
With a five-storey building, there are not five possibilities of exit for each person unless you are assuming that people get on and then get off on the ground floor before the elevator moves.

So my first question is whether the problem stipulates a five-storey or six-storey building?

Furthermore, we are not told whether people are equally likely to get off at each floor. It is important, as our summary of guidelines says, to give the original problem completely and exactly. Assuming, however, equal probability, the probability that any specific passenger will get off on floor three is (1/4). The probability that any specific passenger will get off on a floor other than 3 is (3/4). How many ways can we choose 2 from a total of 6?

By the way, what are you trying to say is the given answer? What you appear to be giving is way greater than 1.

#### robespierre722

##### New member
With a five-storey building, there are not five possibilities of exit for each person unless you are assuming that people get on and then get off on the ground floor before the elevator moves.

So my first question is whether the problem stipulates a five-storey or six-storey building?

Furthermore, we are not told whether people are equally likely to get off at each floor. It is important, as our summary of guidelines says, to give the original problem completely and exactly. Assuming, however, equal probability, the probability that any specific passenger will get off on floor three is (1/4). The probability that any specific passenger will get off on a floor other than 3 is (3/4). How many ways can we choose 2 from a total of 6?

By the way, what are you trying to say is the given answer? What you appear to be giving is way greater than 1.
It is a six-storey building as you said, sorry.
And they cannot exit on ground floor.
They have equal probability to exit any floor.
And i tried that option as well like: We can chose 2 from a total of 6 by : 6C2 and since we have 1/5 chance that they can exit the 3rd floor ->
-> 6C2*(1/5)^2*(4/5)^4=24.57% (wrong answer) .
I think that this answer was good in case it was not specified one floor (3rd floor) , like :"2 people can get off at the same floor " instead of " 2 people get off at the 3rd floor".
That 3rd floor beeing mentioned is my problem.

#### robespierre722

##### New member
We can consider that all 6 passengers are in the same elevator.
I would think of this problem as follows: You have a biased coin where p(heads) = 1/5 and p(tails) = 4/5. Do you see why this models your problem? Now toss the coin 6 times and find the p(getting exactly 2 heads)
I tried that option as well like: We can chose 2 from a total of 6 by : 6C2 and since we have 1/5 chance that they can exit the 3rd floor ->
-> 6C2*(1/5)^2*(4/5)^4=24.57% (wrong answer)
I think that this answer was good in case it was not specified one floor (3rd floor) , like :"2 people can get off at the same floor " instead of " 2 people get off at the 3rd floor".
That 3rd floor beeing mentioned is my problem.
And yes , you can consider of course 1 elevator and 6 persons for a)

I wrote the only way i can get the given answer ( 0.3072) is like this :
a) So there are 5 possibilities for each person -> 5^6 possible cases
There are 6C2=15 possible pairs for that 3rd floor .

(Wrong but it seems right
I was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60 ways they can exit
And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . )

(The right answer but i don't understand why since i only have the (0.3072)
If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .
so the only way to get the given answer is by this formula :
((4P1+4P2+4P3+4P4)*5*6C2)/(5^6)=0.3072 , i don't understand the logic behind those (4P1+4P2+4P3+4P4) )

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#### Dr.Peterson

##### Elite Member
"The third floor being mentioned" is not your problem. Anything else would be a harder problem, not an easier one.

I think your answer of 6C2*(1/5)^2*(4/5)^4=24.57% is correct, as far as I can see. All your other ideas seem to be your attempts to make the supposedly correct answer, which is 5/4 of that, come out as your answer. Is it possible, instead, that it is just wrong? Might someone have just raised 4/5 to the wrong power, or something like that?

If you haven't stated the problem exactly as given to you, please do so, so we can make sure we are dealing with the right question. Also tell us how you know that 0.3072 is supposed to be correct.

#### Jomo

##### Elite Member
I tried that option as well like: We can chose 2 from a total of 6 by : 6C2 and since we have 1/5 chance that they can exit the 3rd floor ->
-> 6C2*(1/5)^2*(4/5)^4=24.57% (wrong answer)
I disagree with that being the wrong answer. Can you show why it is wrong or why another answer is correct. I stand by my answer.

#### robespierre722

##### New member
"The third floor being mentioned" is not your problem. Anything else would be a harder problem, not an easier one.

I think your answer of 6C2*(1/5)^2*(4/5)^4=24.57% is correct, as far as I can see. All your other ideas seem to be your attempts to make the supposedly correct answer, which is 5/4 of that, come out as your answer. Is it possible, instead, that it is just wrong? Might someone have just raised 4/5 to the wrong power, or something like that?

If you haven't stated the problem exactly as given to you, please do so, so we can make sure we are dealing with the right question. Also tell us how you know that 0.3072 is supposed to be correct.
I just solved some probability problems for my exam and i only have the final answer which is 0.3072 in this case . For all the problems i solved until now the final answer was good so i presume that 0.3072 is good as well. I will try to translate word by word :

An elevator stops at P+1, P+2, P+3, P+4 and P+5 (5 levels) . The elevator is at ground floor P with 3 people A, B , C .
A second elevator is in the same building at floor P+5, with 3 persons D, E, F. This elevator stops at P+4, P+3, P+2, P+1, and P.
a)What is the probability that exactly 2 people will get off at P+3 ? (regardless of the elevator)
answer: 0.3072
b)What is the probability that no person from the first elevator will want to go at P+5 .
answer: 0.008
My attempts to solve those two are above, but they don't match the final answer.