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Thread: Elevator and passengers

  1. #1
    Senior Member Vinod's Avatar
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    Elevator and passengers

    Hello,
    An elevator starts with 5 passengers and stops at 8 floors. I want to find the probability that no two or more passengers leave at the same floor assuming that all arrangements of discharging the passengers have same probability.
    Solution will be worked out shortly. Meanwhile if any forumites know the answer, he may reply with detailed answer.
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    Senior Member Vinod's Avatar
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    Re: Elevator and passengers

    Hello,
    My answer is $\frac{6720}{9192}$
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    Re: Elevator and passengers

    Quote Originally Posted by Vinod View Post
    An elevator starts with 5 passengers and stops at 8 floors. I want to find the probability that no two or more passengers leave at the same floor assuming that all arrangements of discharging the passengers have same probability.
    Solution will be worked out shortly. Meanwhile if any forumites know the answer, he may reply with detailed answer.
    Note that we treat the people as identical for the purposes of counting in this question.
    In other words, we do not care who gets off at any floor but the number that get off.
    We want only one at any floor.

    There are $\dbinom{5+8-1}{5}=792$ ways for five people to exit the elevator at eight floors.
    There are There are $\dbinom{8}{5}=56$ ways to pick five floors from eight possible.
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    Senior Member Vinod's Avatar
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    Re: Elevator and passengers

    Quote Originally Posted by Plato View Post
    Note that we treat the people as identical for the purposes of counting in this question.
    In other words, we do not care who gets off at any floor but the number that get off.
    We want only one at any floor.

    There are $\dbinom{5+8-1}{5}=792$ ways for five people to exit the elevator at eight floors.
    There are There are $\dbinom{8}{5}=56$ ways to pick five floors from eight possible.
    Hello Plato,
    The various arrangements of discharging the passengers to any one floor may be denoted by symbols like (3,1,1) to be interpreted as the event that three passengers leave together at a certain floor, one passenger at another floor and finally the last one passenger at still another floor. So there are seven possible arrangements ranging from (5) to (1,1,1,1,1). Now calculate the total probability by adding probability for each arrangement. E.g. In how many ways all the five passengers gets off at any one floor. The answer is in $\binom{8}{1}$ ways. That is the way i calculated the answer $\frac{6720}{9192}$
    Last edited by Vinod; Jul 29th 2018 at 07:32 AM.
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    Re: Elevator and passengers

    Quote Originally Posted by Vinod View Post
    Hello,
    An elevator starts with 5 passengers and stops at 8 floors. I want to find the probability that no two or more passengers leave at the same floor assuming that all arrangements of discharging the passengers have same probability.
    That statement means that only one passenger gets off on any given floor

    Quote Originally Posted by Vinod View Post
    The various arrangements of discharging the passengers to any one floor may be denoted by symbols like (3,1,1) to be interpreted as the event that three passengers leave together at a certain floor, one passenger at another floor and finally the last one passenger at still another floor. So there are seven possible arrangements ranging from (5) to (1,1,1,1,1). Now calculate the total probability by adding probability for each arrangement. E.g. In how many ways all the five passengers gets off at any one floor. The answer is in $\binom{8}{1}$ ways. That is the way i calculated the answer $\frac{6720}{9192}$
    So from the above I conclude that you have no idea what this is about.
    This is a standard multi-selection question.
    The number of ways to place $N$ identical items into $k$ different cells is $\dbinom{N+k-1}{N}$.
    Here we are counting the number ways the five passengers get off on five different floors or only one on each floor.
    Now there are $\binom{8}{5}=56$ ways to select the combination of five floors.
    Now there are $\binom{5+8-1}{5}=792$ ways for passengers to exit the elevator without any conditions on any of eight different floors. In other words there are $792$ ways that all five ways to get off at any one of eight floors to each passenger get off at a different floor. Thus $792$ counts the numbers of ways the seven partitions of five:$(~\;5~,\;4+1~,\;3+1+1~,\;2+1+1+1~,\;3+2~, \;2+2+1~,\;1+1+1+1+1~)$ can be placed into the eight cells.
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    Re: Elevator and passengers

    1 2 3 4 5 6 7 8
    1 1 1 1 1 0 0 0 : 1
    1 1 1 1 0 1 0 0 : 2
    1 1 1 1 0 0 1 0 : 3
    .....
    0 0 1 0 1 1 1 1 : n-1
    0 0 0 1 1 1 1 1 : n = ?
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    Senior Member Vinod's Avatar
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    Re: Elevator and passengers

    Quote Originally Posted by DenisB View Post
    1 2 3 4 5 6 7 8
    1 1 1 1 1 0 0 0 : 1
    1 1 1 1 0 1 0 0 : 2
    1 1 1 1 0 0 1 0 : 3
    .....
    0 0 1 0 1 1 1 1 : n-1
    0 0 0 1 1 1 1 1 : n = ?
    Hello,
    First passenger can select any one floor in 8 ways. Second passenger can select any one floor in 7 ways because no 2 or more passenger leave at the same floor.Third passenger can select any one floor in 6 ways.Fourth passenger can select any one floor in 5 ways. And last passenger can select any one floor in 4 ways. Hence the total number of ways of no 2 or more passengers leave at the same floor are 8*7*6*5*4=6720 ways.
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    Re: Elevator and passengers

    Quote Originally Posted by Vinod View Post
    Hello,
    First passenger can select any one floor in 8 ways. Second passenger can select any one floor in 7 ways because no 2 or more passenger leave at the same floor.Third passenger can select any one floor in 6 ways.Fourth passenger can select any one floor in 5 ways. And last passenger can select any one floor in 4 ways. Hence the total number of ways of no 2 or more passengers leave at the same floor are 8*7*6*5*4=6720 ways.
    What you've neglected is that we don't care which passengers get off where. Just about the distribution of floors.

    So we have to divide your number, 6720, but the total number of ways the 5 passengers themselves can be permuted. This is $5! = 120$

    and lo and behold $\dfrac{6720}{120} = 56$ which is what Plato told you in post #3
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    Senior Member Vinod's Avatar
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    Re: Elevator and passengers

    Quote Originally Posted by romsek View Post
    What you've neglected is that we don't care which passengers get off where. Just about the distribution of floors.

    So we have to divide your number, 6720, but the total number of ways the 5 passengers themselves can be permuted. This is $5! = 120$

    and lo and behold $\dfrac{6720}{120} = 56$ which is what Plato told you in post #3
    Hello,
    William Feller, Prof. of Mathematics, Princeton University, Author of the book"An Introduction To Probability Theory and its Applications Volume 1 gave the probability =$8^{-5}(8*7*6*5*4).$. He calculated the total sample space is $8^5$.I made mistake in computing the total sample space.
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    Re: Elevator and passengers

    Quote Originally Posted by Vinod View Post
    Hello,
    William Feller, Prof. of Mathematics, Princeton University, Author of the book"An Introduction To Probability Theory and its Applications Volume 1 gave the probability =$8^{-5}(8*7*6*5*4).$. He calculated the total sample space is $8^5$.I made mistake in computing the total sample space.
    so you had an answer and a method you liked, you ignored everything posted, why even ask for help here?
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    Re: Elevator and passengers

    fer da general case:

    p = passengers, f = floors

    [f! / (f - p)!] / f^p where f => p
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    Re: Elevator and passengers

    Quote Originally Posted by Vinod View Post
    Hello,
    William Feller, Prof. of Mathematics, Princeton University, Author of the book"An Introduction To Probability Theory and its Applications Volume 1 gave the probability =$8^{-5}(8*7*6*5*4).$. He calculated the total sample space is $8^5$.I made mistake in computing the total sample space.
    That method does not produce the probability every arrangement of passenger discharges have equal probability as the problem dictates. This implies that instead each passenger has equal probability of choosing a particular floor, which is a vastly different problem. Plato gave you the solution to the problem you originally asked. The solution you have offered is to a completely different problem.
    Thanks from Plato
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