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Determine probability

warwick

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Jan 27, 2006
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311
I need to sum the probabilities from k=1 to n of 2[sup:1ljuawnd]-k[/sup:1ljuawnd].

The professor gave us this geometric sum equation:

Sum from k=0 to n of p[sup:1ljuawnd]k[/sup:1ljuawnd] = [1 - p[sup:1ljuawnd]n+1[/sup:1ljuawnd]] / [1 - p]

Apparently, I need to get my sum into that form. Also, how do you generally change the limits of summation? I haven't played with this stuff in years. I'm ashamed.

http://i111.photobucket.com/albums/n149 ... 1296970739
 

Subhotosh Khan

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Jun 18, 2007
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18,089
warwick said:
I need to sum the probabilities from k=1 to n of 2[sup:3qgm4ey4]-k[/sup:3qgm4ey4].

The professor gave us this geometric sum equation:

Sum from k=0 to n of p[sup:3qgm4ey4]k[/sup:3qgm4ey4] = [1 - p[sup:3qgm4ey4]n+1[/sup:3qgm4ey4]] / [1 - p]

Apparently, I need to get my sum into that form. Also, how do you generally change the limits of summation? I haven't played with this stuff in years. I'm ashamed.

http://i111.photobucket.com/albums/n149 ... 1296970739
In your problem p = 1/2

Are you saying you have not seen this equation (sum of geometric progression) before????!!!!

Aren't you a Math (or Physics) major???
 

warwick

Full Member
Joined
Jan 27, 2006
Messages
311
Subhotosh Khan said:
warwick said:
I need to sum the probabilities from k=1 to n of 2[sup:7mm1kpw7]-k[/sup:7mm1kpw7].

The professor gave us this geometric sum equation:

Sum from k=0 to n of p[sup:7mm1kpw7]k[/sup:7mm1kpw7] = [1 - p[sup:7mm1kpw7]n+1[/sup:7mm1kpw7]] / [1 - p]

Apparently, I need to get my sum into that form. Also, how do you generally change the limits of summation? I haven't played with this stuff in years. I'm ashamed.

http://i111.photobucket.com/albums/n149 ... 1296970739
In your problem p = 1/2

Are you saying you have not seen this equation (sum of geometric progression) before????!!!!

Aren't you a Math (or Physics) major???
No. I said I haven't played with it in a long time. Apparently, I'm a bit rusty on it.

I figured the constant factor had to be 1/2.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,686
I will do part d) for you. It the probability of odd numbers.
\(\displaystyle \sum\limits_{k = 0}^\infty {\frac{1}{{2^{2k + 1} }}} = \frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{1}{{4^k }}} = \frac{1}{2}\frac{1}{{1 - \frac{1}{4}}} = ~?\)
 

warwick

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Jan 27, 2006
Messages
311
If p = 1/2, in order to change the lower limit of summation from 0 to 1, I subtract the p[sub:d0tuagpo]0[/sub:d0tuagpo] term, which is 1, from the new sum.

I end up with

2 - 2(1/2)(1/2)[sup:d0tuagpo]n[/sup:d0tuagpo] - 1 = 1 - (1/2)[sup:d0tuagpo]n[/sup:d0tuagpo] = 1 - 2[sup:d0tuagpo]-n[/sup:d0tuagpo] since 1[sup:d0tuagpo]n[/sup:d0tuagpo] for all n.
 

warwick

Full Member
Joined
Jan 27, 2006
Messages
311
pka said:
I will do part d) for you. It the probability of odd numbers.
\(\displaystyle \sum\limits_{k = 0}^\infty {\frac{1}{{2^{2k + 1} }}} = \frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{1}{{4^k }}} = \frac{1}{2}\frac{1}{{1 - \frac{1}{4}}} = ~?\)
How did you get to that point? The first one was a bit easier, write 2[sup:1b6zz3s7]-k[/sup:1b6zz3s7] in the form of p[sup:1b6zz3s7]k[/sup:1b6zz3s7] with n being positive integers.
 
C

ceabDachaft

Guest
You probably very clever? or cunning?
 

Subhotosh Khan

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Staff member
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Jun 18, 2007
Messages
18,089
warwick said:
pka said:
I will do part d) for you. It the probability of odd numbers.
\(\displaystyle \sum\limits_{k = 0}^\infty {\frac{1}{{2^{2k + 1} }}} = \frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{1}{{4^k }}} = \frac{1}{2}\frac{1}{{1 - \frac{1}{4}}} = ~?\)
How did you get to that point? The first one was a bit easier, write 2[sup:2hbyacqz]-k[/sup:2hbyacqz] in the form of p[sup:2hbyacqz]k[/sup:2hbyacqz] with n being positive integers.
Any odd number n can be expressed as 2k+1
 

warwick

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Jan 27, 2006
Messages
311
Okay. You're just using induction to set these up. I tend to over-complicate problems. Thanks.
 

Subhotosh Khan

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warwick said:
Okay. You're just using induction to set these up. I tend to over-complicate problems. Thanks.
No - this is not induction. This is straight-forward deduction.
 

warwick

Full Member
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Jan 27, 2006
Messages
311
Subhotosh Khan said:
warwick said:
Okay. You're just using induction to set these up. I tend to over-complicate problems. Thanks.
No - this is not induction. This is straight-forward deduction.
Well, yeah, I guess you're right. You're given the odd or even sets.

The even set probability sum would be 2[sup:ybkmpbwr]-2k[/sup:ybkmpbwr], but it would go from 1 to n.
 

warwick

Full Member
Joined
Jan 27, 2006
Messages
311
Subhotosh Khan said:
warwick said:
pka said:
I will do part d) for you. It the probability of odd numbers.
\(\displaystyle \sum\limits_{k = 0}^\infty {\frac{1}{{2^{2k + 1} }}} = \frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{1}{{4^k }}} = \frac{1}{2}\frac{1}{{1 - \frac{1}{4}}} = ~?\)
How did you get to that point? The first one was a bit easier, write 2[sup:1zsoqm18]-k[/sup:1zsoqm18] in the form of p[sup:1zsoqm18]k[/sup:1zsoqm18] with n being positive integers.
Any odd number n can be expressed as 2k+1
I'm not seeing how you get 1/4[sup:1zsoqm18]k[/sup:1zsoqm18] in the sum. I see why the (1/2) is brought out front. 2[sup:1zsoqm18]2k[/sup:1zsoqm18] = 2[sup:1zsoqm18]k[/sup:1zsoqm18]2[sup:1zsoqm18]k[/sup:1zsoqm18] = (2[sup:1zsoqm18]2[/sup:1zsoqm18])[sup:1zsoqm18]k[/sup:1zsoqm18] = 4[sup:1zsoqm18]k[/sup:1zsoqm18]?

In the GS formula, what happened to the n for (c) and (d)?

The professor gave us this equation: sum from k = 0 to n of p[sup:1zsoqm18]k[/sup:1zsoqm18] = ( 1 - p[sup:1zsoqm18]n+1[/sup:1zsoqm18] ) / ( 1 - p )

Are you taking the limit of (1/4)[sup:1zsoqm18]n+1[/sup:1zsoqm18] to be zero? Is it appropriate to the k=0 to infinity when the sum is k=1 to n? I guess n doesn't necessarily imply a finite set in this case given n is an element of S, which can be assumed infinite.
 
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