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xen-ppc-devel
Re: [XenPPC] [PATCH] Flush the ERAT early for secondary CPUs
p = probability of success = .997 (897 / 900)
q = probability of failure = .003 (1.0 - .997)
n = number of trials = 2323
X = number of successes = 2323
Applying these to the binomial probability formula, we get:
P(2323) = 2323! / ((2323 - 2323)! * 2323!) * .997**2323 * .003**
(2323-2323)
= .0009307922
So we conclude that the probability that our trials with this patch
applied achieved exactly 2323 successes because of chance alone is .
0009.
Not prematurely rounding p to a useless precision gives you
p**2323 ~ 0.000428
even. And that just calculates the chance that 2323 tries all
succeed given that the chance for one to succeed is 897/900; it
doesn't compare two hypotheses at all.
Segher
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