m (4.2b Steve Streeter moved to 4.1b Steve Streeter: Improperly named originally) |
|||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 2: | Line 2: | ||
| − | <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>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> Eq. 1 |
| + | |||
| + | //comment// | ||
| + | |||
| + | Is this true? Won't the probability double because when k is even <math>[1+(-1)^k] = 2</math>? Shouldn't you put a 1/2 somewhere because of this?? Please advise. Thanks, | ||
| + | |||
| + | Ken Pesyna | ||
| + | |||
| + | //end comment// | ||
| + | |||
Next we must expand <math>(x+y)^n + (x-y)^n</math> using the binomial thereom: | Next we must expand <math>(x+y)^n + (x-y)^n</math> using the binomial thereom: | ||
| − | <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> | + | <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> Eq. 2 |
This simplifies to: | This simplifies to: | ||
| − | <math>\sum\limits_{k=0}^N {N \choose k} (x)^{N-k}y^k[1+(-1)^k] </math> | + | <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 | ||
| + | |||
| + | //Comment// | ||
| + | |||
| + | I might be wrong, but 1^n + (1-2P)^n should be greater than 1. If so, P[X is even] > 1, which doesnt look reasonable. | ||
| − | + | //End of Comment// | |
| − | + | 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. | |
Latest revision as of 07:42, 15 October 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
//comment//
Is this true? Won't the probability double because when k is even $ [1+(-1)^k] = 2 $? Shouldn't you put a 1/2 somewhere because of this?? Please advise. Thanks,
Ken Pesyna
//end comment//
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
//Comment//
I might be wrong, but 1^n + (1-2P)^n should be greater than 1. If so, P[X is even] > 1, which doesnt look reasonable.
//End of Comment//
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.
