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| robjay... |
 Posted: Mon Oct 26, 2009 3:36 am |
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Guest
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Hi
I am having problems generalising the following problem.
Say you have five (1,2,3,4,5) people which are selecting colours that
they like from the possible choices of Red, Green or Blue. They may
select more than one colour, but they must select at least one colour.
It's easy to answer questions like, Given a person has selected two
colours, what is the probability that they have Red in their selection
(2/3).
If we are then given extra information, then it can make it much
harder to answer this same question. If for example we are told that 4
people liked Red, 4 People liked Green, but only one person liked
Green, then answering the above question becomes much harder.
By considering the possible combinations I have managed to calculate
that given a person who says they like two colours, the probability
that they like Red is 14/15, Blue is 14/15 and Green is 2/15.
My problem is that I can't figure out how to derive a general result,
when there are C colours, n people, and some given numbers for the
number of people who liked a particular colour.
Any help that anyone may be able to give me would be gratefully
received.
Cheers
James |
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| robjay... |
| Posted: Mon Oct 26, 2009 8:42 am |
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Guest
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Sorry, but there is a mistake in the above, it should read:
4 people liked Red, 4 People liked Blue, but only one person liked
Green. |
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| Ray Koopman... |
| Posted: Mon Oct 26, 2009 9:01 am |
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Guest
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On Oct 25, 6:03 pm, robjay <james.robertso... at (no spam) yahoo.co.uk> wrote:
[quote]Hi
I am having problems generalising the following problem.
Say you have five (1,2,3,4,5) people which are selecting colours that
they like from the possible choices of Red, Green or Blue. They may
select more than one colour, but they must select at least one colour.
It's easy to answer questions like, Given a person has selected two
colours, what is the probability that they have Red in their selection
(2/3).
If we are then given extra information, then it can make it much
harder to answer this same question. If for example we are told that 4
people liked Red, 4 People liked Green, but only one person liked
Green, then answering the above question becomes much harder.
By considering the possible combinations I have managed to calculate
that given a person who says they like two colours, the probability
that they like Red is 14/15, Blue is 14/15 and Green is 2/15.
My problem is that I can't figure out how to derive a general result,
when there are C colours, n people, and some given numbers for the
number of people who liked a particular colour.
Any help that anyone may be able to give me would be gratefully
received.
Cheers
James
[/quote]
I don't think there is a nice solution to the general problem.
Think of an N x C matrix in which rows correspond to people,
columns correspond to colors, and each element is a 1 iff that
person liked that color, and is a 0 otherwise. If all you know
are the column totals, plus the fact that no row is all 0's,
that leaves quite a bit of freedom for placing 1's and 0's,
even if you consider two matrices equivalent if they can be
made equal by permuting the rows. |
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| Allen McIntosh... |
| Posted: Tue Oct 27, 2009 6:05 am |
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Guest
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Ray Koopman wrote:
[quote]On Oct 25, 6:03 pm, robjay <james.robertso... at (no spam) yahoo.co.uk> wrote:
I am having problems generalising the following problem.
[Original problem statement snipped]
I don't think there is a nice solution to the general problem.
Think of an N x C matrix in which rows correspond to people,
columns correspond to colors, and each element is a 1 iff that
person liked that color, and is a 0 otherwise. If all you know
are the column totals, plus the fact that no row is all 0's,
that leaves quite a bit of freedom for placing 1's and 0's,
even if you consider two matrices equivalent if they can be
made equal by permuting the rows.
[/quote]
A mathematical programming package can sometimes get specific solutions
to specific questions for *small* instances of problems like these if
you pose them properly. Probably should direct followup to appropriate
newsgroup for that. |
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