Re: At least j successes (out of k Binomial trials)
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Sun, 02 Jul 2006 18:48:06 EDT
Depending on the values of k and j one prefers
to use
___p (at least j successes) =
= p(j) + p(j+1) + ... + p(k)_____________[1]
___ p (at least j successes) =
___1 - (p(0) + p(1) + ... + p(j-1))_______ [2]
A particularly simple method is used for the case j=1
(of at least one success) because it is the
complementary of no successes among k trials.
Because
___1 = ((1- p) + p) ^ k =
___ = sum kCj * (1-p) ^(k-j) * k^j ____0<=j<=k
___= (1-p) ^k + sum kCj * (1-p) ^(k-j) * k^j
_______________________________1<=j<=k
This equation is incorrect. For example, set p=0 and k=1. Then the right side is
(1-0)^1 + Sum[1C1*(1-0)^(1-1)*1^1] =
1+(1*1*1) = 2, which is not the same as 1.
Jack
.
We have ( the generic term of this sum is the
probability to occur EXACTTLY j successes or,
what is the same thing, to occur exactly k-j
failures):
___1 - (1-p)^k = p(at least 1 success)
_____licas (Luis A. Afonso)
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