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Math Help - Probability problem

  1. #1
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    Probability problem

    Hello,

    I would appreciate some help on how to get started on this problem.

    Consider a 45-story building. Person A lives on the 20th floor. Let's say 5 people (including person A) entered an elevator and the elevator is going up and everyone pressed a different floor (button), what is the probabiltiy that person A was the 1st one to exit the elevator? The 2nd? The 3rd? The 4th? The 5th (i.e., the last)?


    I know that if the building had only 20 floors then the probability of person A exiting the elevator last would be 1.
    Also if there are only 2 people, then is the probability of person A leaving 1st 20/45 ?
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  2. #2
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    Quote Originally Posted by satzer View Post
    Hello,

    I would appreciate some help on how to get started on this problem.

    Consider a 45-story building. Person A lives on the 20th floor. Let's say 5 people (including person A) entered an elevator and the elevator is going up and everyone pressed a different floor (button), what is the probabiltiy that person A was the 1st one to exit the elevator? The 2nd? The 3rd? The 4th? The 5th (i.e., the last)?


    I know that if the building had only 20 floors then the probability of person A exiting the elevator last would be 1.
    Also if there are only 2 people, then is the probability of person A leaving 1st 20/45 ?
    Excluding person A and the 20th floor, there are 44 floors and 4 people. We will assume every subset of 4 floors taken from the set {1, 2, ..., 19, 21, 22, ..., 45} is equally likely, so there are \binom{44}{4} equally likely possibilities.

    Suppose i people get off on floors 1, 2, ..., 19 and j get off on floors 21, 22, ..., 45, where i + j = 4. Then the number of ways this can done is
    \binom{19}{i} \binom{25}{j},

    and the probability of this occurrence is

    \frac{\binom{19}{i} \binom{25}{j}}{\binom{44}{4}}.
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  3. #3
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    Thank you very much. This is a great answer!
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