Someone, without getting too specific, pointed out that I made an error.
I realized immediately that the number of combinations fitting the
requirement (i.e., the numerator)was 9 instead of 36 (I'd forgotten one
pairing determined the other, so each # of combinations had to be
divided by two).
However, the resulting probabilty now seems quite low (9/840), and I
wonder what other mistake I may have made. :-) Perhaps I shouldn't be
using this forum to be practicing probability questions from twenty
years ago!
-----Original Message-----
From: Todd E. Melander [mailto:[log in to unmask]]
Sent: Wednesday, June 18, 2003 12:14 AM
To: [log in to unmask]
Subject: RE: probability question
The answer, of course is that it is the total number of different ways
that could happen over the total number of different pairings.
Total # of pairings: Imagine the eight people line up in chairs
1,2,...,8. Now imagine that the pairings will be 1&2, 3&4, 5&6, 7&8.
How many different ways can the people sit in the chairs? Hence, how
many orderings (or pairings, with a caveat) That's just 8! (factorial).
The caveat: Obviously each pair could switch seats within the pair of
chairs and keep the same partnerships. Therefore, 8! will have to be
divided by two (or more generally, 2!). And again obviously, the same
set of pairings could occur in different orders. And there are 4! such
orders.
So that's the denominator. 8!/4!2!
The numerator is pretty simple I think: The total number of
intragender-only pairings would be the number of different pairings of
men, or 4C2=6, times the number of different pairings of women, or
4C2=6; that is,6*6=36.
I compute that to be 36/840 = .043 approximately.
In general, with X men and Y women and Z within a group (such as 2
within a pair, where X=cZ and Y=dZ (c,d being positive integers), and
letting W=X+Y, the result is:
(X choose Z)*(Y choose Z) / [(W)!/{(Z!)(W/Z)!}
I send this to the list only so that I can be corrected if I made a
mistake.
Todd
> -----Original Message-----
> From: A UK-based worldwide e-mail broadcast system mailing list
> [mailto:[log in to unmask]] On Behalf Of Isaac Dialsingh
> Sent: Tuesday, June 17, 2003 8:59 PM
> To: [log in to unmask]
> Subject: Re: probability question
>
> Can anyone assist with this question:
> A group of eight card players consist of 4 men and 4 women. If 4 pairs
are
> selected at random from the 8 players, find the probability that no
pair
> consists of a man and woman.
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