Start with the population of o's.  If the probability of formation is equal, regardless of concentration (!), then for each o, it's 50/50 whether it combines with another o or an x.  So, 7.5 o-o pairs, and 15 xo pairs.  Remaining 55 x's form x-x pairs, since there ain't no o's left.

IF this were chemistry and not some silly statistics text problem, then LeChatlier's principle would lead to to believe that the abundance of x's would tend to favor formation of xo's more than oo's.  There are treatments of this in chemistry textbooks, with discussions of the equilibrium constants that define the final concentrations.

btrueblood - There's something about your answer that bugs me.  I'll try and put it into words when i've had a play.  

Cheers

Greg Locock

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I know, Greg.  The problem is in the statement

"If the probability of formation is equal, regardless of concentration (!)"

It does depend on the concentration (if I pick and remove two o's from the bag, the number of o's to be picked from the bag is diminished on the next trial), which is why the "equal probability for each pair formation" statement is too loose, and the reason for my ().  Ok, I admit, I'm playing games here.  The OP's problem also needs to define whether pairs are removed from play or not.

Oh, and there is a treatment of the pair-selection problem in one of my wife's probability texts...will have to go dig it out of storage...someday when I'm not so busy.

Herman,

Let Nxx, Noo and Nxo represent the number of pairs XX, the number of pairs OO and the number of pairs XO, respectively. Then, the three numbers are random but their sum is always 50.

The range for Nxx is (20, 35).
The range for Noo is (0, 15).
The range for Nx0 is (0, 30).

Theoretically, the three numbers have a joint distribution. Its derivation requires a good amount of computations. However, all the sixteen possibilities of (Nxx, Noo, Nxo) are:

(20, 0, 30)
(21, 1, 28)
(22, 2, 26)
(23, 3, 24)
     .
     .
     .
 
(34, 14, 2)
(35, 15, 0).

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