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Probability question

  1. Nov 13, 2009 #1
    , what is the probability that no husband and wife are standing next to each other if:
    there are 2 couples are standing in a line?
    there are 3 couples are standing in a line?
    -------------------------------------------------------
    for 2 couples, my options are A,A',B,B' (where A and A' are a couple)

    1st place (A/A'/B/B') => 4 options - say A is in 1st place...
    2nd place (B/B') => 2 options - say B is in 2nd place...
    3rd place (A') => 1 option
    4th place (B') => 1 option

    4*2*1*1=8 options that no husband and wife are standing next to one another.
    4!=24 ways to place 4 people in a line

    8/24=1/3
    P(no couples)=1/3
    ------------------------------------------------------------
    now for the 2nd case where there are 3 couples

    1st place (A/A'/B/B'/C/C') => 6 options - say A is in 1st place...
    2nd place (B/B'/C/C') => 4 options - say B is in 2nd place...
    3rd place (A'/C/C') => 3 options - say C is in 3rd place...
    4th place (A'/B') => 2 options - say A' is in 4th place...
    5th place (B'/C') => 2 options - say B' is in 5th place...
    6th place (C') => 1 option

    6*4*3*2*2=288 options

    <<OR>>

    1st place (A/A'/B/B'/C/C') => 6 options - say A is in 1st place...
    2nd place (B/B'/C/C') => 4 options - say B is in 2nd place...
    3rd place (A'/C/C') => 3 options - say A' is in 3rd place...
    4th place (C/C') => 2 options - say C is in 4th place...
    5th place (B') => 1 option
    6th place (C') => 1 option

    6*4*3*2= 144 options

    there are 6! ways to arrange the 6 people in the line

    P(no couples)= (288+144)/6!=3/5


    BUT THE CORRECT ANSWER IS ALSO MEANT TO BE 1/3
     
  2. jcsd
  3. Nov 13, 2009 #2

    tiny-tim

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    Hi Dell! :smile:

    (I'm not sure that's the quickest way to do it, but anyway …)
    No, in the first line, there are only 2 options, and in the second line only 1 option. :wink:
     
  4. Nov 14, 2009 #3
    how so? could i not choose any of the 3??

    how would you have solved the problem?
     
  5. Nov 14, 2009 #4

    tiny-tim

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    No, because by doing that, you've counted those 3 ways twice

    you can only count things once!

    you must count 2 of them the first way, and the other 1 the second way. :wink:

    You need to convince yourself of this! :smile:
     
  6. Nov 14, 2009 #5
    okay, i see what you are saying, all together there are 3 options, and counted them as 6,

    how would you have gona about solving the problem
     
  7. Nov 14, 2009 #6

    tiny-tim

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    dunno :redface:

    but it looks unnecessarily long, so if I had the time, I'd try to find something neater. :smile:
     
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