Bioequivalence and Bioavailability Forum

Hmmmmmm.......

❝ I wonder how I can derive the exact probability given the sample size N, number of chunks M (yes, it would be handy if N can be divided by M) and the number oddballs Q. The probability P can be calculated exactly but that's one I will save for a sleepless night.

Dammit, won't be able to sleep until I look a little at this shit:

Let K(A,B) be the coefficient A!/((A-B)!B!).
P(N,M,Q) is the probability that all the Q observations are in the last chunk.

The Q oddballs can be permuted among the N subjects in K(N,Q) ways. The Q oddballs can be permuted among the N/M extreme subjects in K(N/M, Q) ways.

Prob(Q oddballs in last chunk out of M chunks given N subjects)
= K(N/M, Q) / K(N,Q)

........hang on....That absolutely doesn't look right, does it???

Let's take a handy number like 60 subjects:
60 subjects, 3 chunks, 6 oddballs: P= 0.0008
Increase to 4 chunks: P=10^-4 (it is lower, so better sensitivity in this case.)

But I have a feeling my combinatolophystic capabilities are betraying me here. Where's Dr. Rotter?