- #1

Shackleford

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http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png?t=1296970739

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- Thread starter Shackleford
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- #1

Shackleford

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http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png?t=1296970739

- #2

VietDao29

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http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png?t=1296970739

It's pretty easy. Maybe you've overlooked the problem.

- Firstly, you need to note that:

[tex]\mbox{If } A \cap B = \emptyset \mbox{, then } P(A \cup B) = P(A) + P(B)[/tex] - Secondly, reading the problem, we also know that: [tex]P(\{ n \}) = 2 ^ {-n}[/tex]

So, if we have: [tex]E = \{ 1, 2, 3, ..., n \}[/tex], then what's [tex]P(E) = P(\{ 1, 2, 3, ..., n \})[/tex]?

- #3

Shackleford

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It's pretty easy. Maybe you've overlooked the problem.

- Firstly, you need to note that:

[tex]\mbox{If } A \cap B = \emptyset \mbox{, then } P(A \cup B) = P(A) + P(B)[/tex]- Secondly, reading the problem, we also know that: [tex]P(\{ n \}) = 2 ^ {-n}[/tex]

So, if we have: [tex]E = \{ 1, 2, 3, ..., n \}[/tex], then what's [tex]P(E) = P(\{ 1, 2, 3, ..., n \})[/tex]?

That line of reasoning did cross my mind right before I went to bed.

P(S) = P(E) + P(E

where E

P(E) = P(S) - P(E

- #4

VietDao29

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That line of reasoning did cross my mind right before I went to bed.

P(S) = P(E) + P(E^{c})

where E^{c}is the sum from k (>n) to infinity [tex]P(\{ k \}) = 2 ^ {-k}[/tex]

P(E) = P(S) - P(E^{c}) = 1 - 2^{-k}

How did you know that P(E

Don't over complicate the problem. You know that: [tex]P(\{ k \}) = 2 ^ {-k}[/tex], i.e (P({1}) = 2

-----------------------

Oh, and by the way, P({

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- #5

Shackleford

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How did you know that P(E^{c}) = 2^{-k}?

Don't over complicate the problem. You know that: [tex]P(\{ k \}) = 2 ^ {-k}[/tex], i.e (P({1}) = 2^{-1}, P({2}) = 2^{-2}, P({3}) = 2^{-3},... and so on), andEis the set of all integers from 1 ton. Then, how can you calculate P(E)?

You would sum it up, but would you have to reorder the limits of summation to get the given answer?

- #6

VietDao29

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Would you have to reorder the limits of summation?

No, you don't need to reorder anything, you just need to stick to the fact that:

[tex]\mbox{If } A \cap B = \emptyset \mbox{, then } P(A \cup B) = P(A) + P(B)[/tex]

and this:

P({n}) is

-------------------

Yup, 'sum it up' is the correct way to go. Now, look at the sum, does it looks like something you've learnt in the past?

- #7

Shackleford

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No, you don't need to reorder anything, you just need to stick to the fact that:

[tex]\mbox{If } A \cap B = \emptyset \mbox{, then } P(A \cup B) = P(A) + P(B)[/tex]

and this:

P({n}) istotally differentfrom P({1; 2; ...; n}). One is a set withonly oneelement, and the other one is a set with elements from 1 to n.

-------------------

Yup, 'sum it up' is the correct way to go. Now, look at the sum, does it looks like something you've learnt in the past?

The professor gave us a few sum equations we would have to use.

Sum from k = 0 to n of rho

- #8

VietDao29

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The professor gave us a few sum equations we would have to use.

Sum from k = 0 to n of rho^{k}= ( 1 - rho^{n+1}) / ( 1 - rho )

Yup, that's the sum of the first

The Geometric Sequence is a sequence, in which, the later (following) terms is formed by multiplying the former (previous) term by a

An example of a Geometric Progression is

- 2; 4; 8; 16; ... (common ratio
*r*= 2). - -1; 3; -9; 27; ... (common ratio
*r*= -3).

So, the terms of a GP can be written as:

The sum of the first

[tex]\sum_{i = 1} ^ n (a r ^ {i - 1}) = a\frac{1 - r ^ n}{1 - r}[/tex]

Where

Now, apply the above formula to see if you can get the result. :)

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- #9

Shackleford

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Yup, that's the sum of the firstnterms in a http://en.wikipedia.org/wiki/Geometric_progression" [Broken] (abbr:GP) (or you can call itGeometric Sequence).

The Geometric Sequence is a sequence, in which, the later (following) terms is formed by multiplying the former (previous) term by aconstantr(callcommon ratio).

An example of a Geometric Progression is

- 2; 4; 8; 16; ... (common ratio
r= 2).- -1; 3; -9; 27; ... (common ratio
r= -3).

So, the terms of a GP can be written as:

a;ar;ar^{2};ar^{3};ar^{4}; ...

The sum of the firstnterms of a GP can be calculated by:

[tex]\sum_{i = 1} ^ n (a r ^ {i - 1}) = a\frac{1 - r ^ n}{1 - r}[/tex]

Whereais the first term of the GP,ris the common ratio, andnis th number of terms.

Now, apply the above formula to see if you can get the result. :)

Well, I worked it out yesterday before I left home, and I got the negative of the answer given.

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- #10

VietDao29

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Well, I worked it out yesterday before I left home, and I got the negative of the answer given.

Well, that's strange. The answer provided is correct. Can you show us the work so that we can check it, and tell you where you went wrong? :)

- #11

Shackleford

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Well, that's strange. The answer provided is correct. Can you show us the work so that we can check it, and tell you where you went wrong? :)

Well, your sum goes from 1 to n. The equation I have goes from 0 to n. Since E goes from 1 to n, should I adjust the limits of summation? Can't believe I'm this rusty on this stuff. -_-

P({n}) = 2

Sum from k = 0 to n of rho

- #12

VietDao29

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Well, your sum goes from 1 to n. The equation I have goes from 0 to n. Since E goes from 1 to n, should I adjust the limits of summation? Can't believe I'm this rusty on this stuff. -_-

Yes, if you use the equation your professor gave you, then you should adjust the index a little bit.

P({n}) = 2^{-n}

Sum from k = 0 to n of rho^{k}= ( 1 - 2^{n+1}) / ( 1 - 2 )

rho is

[tex]\sum_{k = 1} ^ n 2 ^ {-k}[/tex]

And the equation your professor gave is:

[tex]\sum_{k = 0} ^ n \rho ^ k = \frac{1 - \rho ^ {n + 1}}{1 - \rho}[/tex]

How can you change 2

And by the way, I think you should use the sum of the first

- #13

Shackleford

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Nevermind. I figured out (a). I'll try the rest of them.

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- #14

Shackleford

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What am I doing wrongly here?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110209_192854.jpg?t=1297301586 [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110209_192854.jpg?t=1297301586 [Broken]

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- #15

Unit

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For (b), you wrote,

[tex] E = \{n+1, n+2, n+3, ...\} \Rightarrow P(\{2k+1\}) [/tex]

That is not true. Like VietDao said, if A intersect B is empty, then P(A U B) = P(A) + P(B). Consider the fact that E = union of all the following: {n+1}, {n+2}, {n+3}, ..., noting that each set is disjoint from every other set. What can you say about P(E) in this light?

[tex] E = \{n+1, n+2, n+3, ...\} \Rightarrow P(\{2k+1\}) [/tex]

That is not true. Like VietDao said, if A intersect B is empty, then P(A U B) = P(A) + P(B). Consider the fact that E = union of all the following: {n+1}, {n+2}, {n+3}, ..., noting that each set is disjoint from every other set. What can you say about P(E) in this light?

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- #16

Shackleford

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For (b), you wrote,

[tex] E = \{n+1, n+2, n+3, ...\} \Rightarrow P(\{2k+1\}) [/tex]

That is not true. Like VietDao said, if A intersect B is empty, then P(A U B) = P(A) + P(B). Consider the fact that E = union of all the following: {n+1}, {n+2}, {n+3}, ..., noting that each set is disjoint from every other set. What can you say about P(E) in this light?

Okay. I'll take a guess. For P({E}), I could sum up P({n+1}), P({n+2}), P({n+3}), etc., since each of the sets is disjoint.

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