|  |  | 
  
    |  |  | 
 
  |   |  | 
  
    |  |  | 
  
    |  |  | 
  
    |   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
_______________________________________________
Xen-ppc-devel mailing list
Xen-ppc-devel@xxxxxxxxxxxxxxxxxxx
http://lists.xensource.com/xen-ppc-devel
 | 
 |  | 
  
    |  |  |