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<math>\sum\limits_{k=0}^N {N \choose k} (x)^{N-k}y^k[1+(-1)^k] </math> Eq. 3 | <math>\sum\limits_{k=0}^N {N \choose k} (x)^{N-k}y^k[1+(-1)^k] </math> Eq. 3 | ||
| − | + | By comparing equations 1 and 3 you can derive a condensed equation in the form of <math>(x+y)^n + (x-y)^n</math> where <math>x = 1-P</math> and <math>y=P</math> : | |
<math> ((1-P)+P)^n + ((1-P)-P)^n = 1^n + (1-2P)^n </math> Eq. 4 | <math> ((1-P)+P)^n + ((1-P)-P)^n = 1^n + (1-2P)^n </math> Eq. 4 | ||
Note that even though the hint gives away x = 1-P and y = P, we use the binomial theorem to prove the correlation between equations 1 and 3. | Note that even though the hint gives away x = 1-P and y = P, we use the binomial theorem to prove the correlation between equations 1 and 3. | ||
Revision as of 11:52, 23 September 2008
The first step for this problem is to map out what the probability that x is even would be:
$ P[X is Even]= \sum\limits_{k=0,even}^N P[x=k]= \sum\limits_{k=0}^N {N \choose k} P^k(1-P)^{N-k}[1+(-1)^k] $ Eq. 1
Next we must expand $ (x+y)^n + (x-y)^n $ using the binomial thereom:
$ (x+y)^n + (x-y)^n = \sum\limits_{k=0}^N {N \choose k} y^kx^{N-k} + \sum\limits_{k=0}^N {N \choose k} (-y)^kx^{N-k} $ Eq. 2
This simplifies to:
$ \sum\limits_{k=0}^N {N \choose k} (x)^{N-k}y^k[1+(-1)^k] $ Eq. 3
By comparing equations 1 and 3 you can derive a condensed equation in the form of $ (x+y)^n + (x-y)^n $ where $ x = 1-P $ and $ y=P $ :
$ ((1-P)+P)^n + ((1-P)-P)^n = 1^n + (1-2P)^n $ Eq. 4
Note that even though the hint gives away x = 1-P and y = P, we use the binomial theorem to prove the correlation between equations 1 and 3.
