1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple probability problem

  1. Nov 18, 2009 #1
    Please help with this probability question,
    , 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 18, 2009 #2
    Re: probability

    Apperently your first case with 3 couples, is the case where the first 3 chosen persons are all from different couples. There will be only 2 persons to choose from in third place

    Your second case with 3 couples is the case where the first 3 chosen persons are from only 2 different couples. There's only 1 person to choose from in third place
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple probability problem
Loading...