Re: Jack , *the stripper*
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Thu, 29 Jun 2006 19:29:01 EDT
*** Even if you accept the binomial expansion as
being correct, the j=0 term is 1, not (1-p)^k.
Jack ***
Had you notice that YOU ARE COMPLETELY WRONG?
___(a + b)^k = kC0 * a ^ k * b^0 + …
___________= a ^ k +...
___a = 1 - p , b = p
Not 1, NEVER EVER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
(when you attend High School did you failed the
lesson to go to skate? - You slipped majestically,
nose to the ground).
(cft. what I wrote on Jun. 28, 2006 5:27 AM).
BE MORE CAREFUL.
_______licas (Luis A. Afonso)
LAA --> ( (1-p )+ p)) ^ k =
_LAA -->_______sum (kCj* (1-p)^j )______ 0<= j <=k
LAA -->Therefore
LAA -->1 = (1-p) ^k + sum (kCj* (1-p)^j )____ 1<= j <=k
Luis, set j=0 for the first term in your sum. You should get kC0*(1-p)^0 = 1*1 = 1.
Jack
.
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