Probability question

  • Thread starter Dell
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  • #1
Dell
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, 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
 

Answers and Replies

  • #2
tiny-tim
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Hi Dell! :smile:

(I'm not sure that's the quickest way to do it, but anyway …)
3rd place (A'/C/C') => 3 options - say C is in 3rd place...

3rd place (A'/C/C') => 3 options - say A' is in 3rd place...

No, in the first line, there are only 2 options, and in the second line only 1 option. :wink:
 
  • #3
Dell
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how so? could i not choose any of the 3??

how would you have solved the problem?
 
  • #4
tiny-tim
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how so? could i not choose any of the 3??

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:
 
  • #5
Dell
590
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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
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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how would you have gona about solving the problem

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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