(New page: The first step for this problem is to map out what the probability that x is even: <math>P{X is Even}= ∑_█(k=0@k even)^N▒P[x=k] = ∑_█(k=0@k even)^n▒〖(n¦k) x^k a^(n-k) 〗</...) |
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| − | The first step for this problem is to map out what the probability that x is even: | + | The first step for this problem is to map out what the probability that x is even would be: |
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| − | + | <math>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]</math> | |
| − | <math>(x+y)^n | + | Next we must expand <math>(x+y)^n + (x-y)^n</math> using the binomial thereom: |
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| + | <math>(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}</math> | ||
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| + | This simplifies to: | ||
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| + | <math>\sum\limits_{k=0}^N {N \choose k} (x)^{N-k}y^k[1+(-1)^k] </math> | ||
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| + | If you compare P{X is Even} with the binomial expansion just above then 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>: | ||
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| + | <math> ((1-P)+P)^n + ((1-P)-P)^n = 1^n + (1-2P)^n </math> | ||
Revision as of 10:40, 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] $
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} $
This simplifies to:
$ \sum\limits_{k=0}^N {N \choose k} (x)^{N-k}y^k[1+(-1)^k] $
If you compare P{X is Even} with the binomial expansion just above then 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 $
