@shallow-blue said
Of course - since I did it arithmetically - but I'd still like to know if there is a solution based on principles rather than mere calculation.
Unfortunately, when it comes to probabilities, there often isn't. (Knot theory is even worse.)
I'm not sure what principles you are looking for, but they are all implicit in your calculation.
The set can either be matched
or not matched. The probabilities of each event are independent, and thus additively form the space of all possible outcomes.
P ( mismatched ) = 1 - P( matched )
You can compute the LHS directly ( as you have) or indirectly from the RHS.
You might instead use combinatorial argument:
WWWWWW BBBB
To calculate P ( matched )
P( matched ) = ( C(6,2) + C(4,2) ) / C(10,2)
= ( 15 + 6 )/45
= 7/15
Then it follows that:
P ( mismatched ) = 1 - P( matched )
= 1 - 7/15
= 8/15
However, there is nothing wrong with treating the draws as chronologically ordered as you have done, as I believe simultaneously drawing two things is a probability zero event anyhow.
Hope that is what you are looking for, if not your going to have to elaborate a bit further.
P.S. if you are interested, there are still some unsolved puzzle in the lineup below.
Topic: Probabilty
https://www.redhotpawn.com/forum/posers-and-puzzles/joe-should-go-to-bed.188751
Topic: Algebra
https://www.redhotpawn.com/forum/posers-and-puzzles/one-train-two-train-red-train-blue-train.188801