>>11039329No your formula is not correct. Let's just look at an example:
Let's say the wheel is equally divided in 4 sections "A", "B", "C", and "D".
And we want to know the probability of getting exactly one "A" and exactly one "B" in two spins.
So the variables are: n=2, a=1, b=1, p=1/4, q=1/4.
Let's list all the possible outcomes of the two spins (these outcomes all have equal probability):
AA, AB, AC, AD, BA, BB, BC, BD, CA, CB, CC, CD, DA, DB, DC, DD
As you can see 2 of the 16 outcomes have exactly one "A" and exactly one "B". So the probability is 2/16 = 1/8.
But if you plug the variables into your formula, you get 3/32, which is wrong.
While if you plug the variables into my formula, you get 1/8, which is correct.
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>the P(A=a) part is already ""checking for"" there being no more of the first result in the remaining n?a observationsthat's precisely the problem
the second part of your formula is calculating as if you haven't already checked for this, while my formula corrects for this
