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Math Help - Help with combinations and Permutations Question

  1. #1
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    Help with combinations and Permutations Question

    There are 13 guests at a hotel and 3 rooms, one with 2 beds, one with 3 beds, and one with 4 beds. If nine guests are chosen at random and placed randomly into one of the rooms, assuming you are one of the guests, what is the probability that you will be chosen and will be in the room with 2 beds?

    If someone please explain the logic behind solving this question, I'd be real grateful because I have no idea where to start.
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  2. #2
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    Hello, zerobladex!

    I think I've reasoned correctly.
    If I'm wrong, surely someone will point it out.


    There are 13 guests at a hotel. You are one of the guests.
    There are 3 rooms: one with 2 beds, one with 3 beds, and one with 4 beds.
    If nine guests are chosen at random and placed randomly into the rooms,
    what is the probability that you will be chosen and will be in the room with 2 beds?

    There are: . {13\choose9} = 715 ways the 9 guests can be chosen.
    . . There are: . 1\cdot{12\choose8} = 495 ways that you are among the nine.
    . . Hence: . P(\text{you are chosen}) \;=\;\frac{495}{715} \:=\:\frac{9}{13}


    There are: . {9\choose2,3,4} = 1260 ways to assign the rooms.
    . . There are: . 1\cdot{8\choose1,3,4} = 280 ways that you are given the 2-bed room.
    . . Hence: . P(\text{2-bed room}) \:=\:\frac{280}{1260} \:=\:\frac{2}{9}


    Therefore: . P(\text{you are chosen }\wedge\text{ 2-bed room}) \;=\;\frac{9}{13}\cdot\frac{2}{9}\;=\;\frac{2}{13}

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  3. #3
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    Thanks a lot for the help I just have a small question about [math]
    {9\choose2,3,4} = 1260
    [/tex]


    How would you input these numbers into a calculator? So far I've only done combinations with one number underneath, so could you please write out the actual operations for this?

    Thanks again.
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  4. #4
    MHF Contributor matheagle's Avatar
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    This is rather straight forward.
    You have 9/13 chance of being selected.
    Then the CONDTIONAL chance of being in a 2 bedroom is 2/9.
    There are two equal beds in that room out of all 2+3+4=9 beds.
    So
    P(ending up in a 2 bedroom and being chosen)
    =P(Chosen) times P(ending up in a 2 bedroom|being a chosen one)
    =(9/13)(2/9)
    =2/13.
    No need to consider all 3 rooms.
    Its the 2 bedroom vs the other two rooms.
    You should only think of it as a binomial and not a trinomial.

    You can view this as 13 slots, 4 I guess sleep in the car.
    The slots are equally likely, so you have 2/13 chance of getting one of those
    two beds.
    The first 9 slots are the beds in those rooms and the remaining losers
    sleep in the car.
    Last edited by matheagle; February 17th 2009 at 11:38 PM.
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