Determine probabilities

In summary, the conversation was about a probability theory problem involving sets and summation. The solution involved using the formula for the sum of a geometric progression and noting that P({n}) is different from P({1,2,...,n}). After some discussion and clarification, the correct answer was found to be the negative of the given answer.
Physics news on Phys.org
  • #2
Shackleford said:
I'm not sure how to start this problem. I've been doing probability theory homework all day, so I'm a bit worn out. Plus, it's almost midnight.

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
VietDao29 said:
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(Ec)

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

P(E) = P(S) - P(Ec) = 1 - 2-k
 
  • #4
Shackleford said:
That line of reasoning did cross my mind right before I went to bed.

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

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

P(E) = P(S) - P(Ec) = 1 - 2-k

How did you know that P(Ec) = 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), and E is the set of all integers from 1 to n. Then, how can you calculate P(E)?

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

Oh, and by the way, P({n}) is totally different from P({1; 2; ...; n}). One is a set with only one element, and the other one is a set with elements from 1 to n.
 
Last edited:
  • #5
VietDao29 said:
How did you know that P(Ec) = 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), and E is the set of all integers from 1 to n. 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
Shackleford said:
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 totally different from P({1; 2; ...; n}). One is a set with only one element, 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 learned in the past?
 
  • #7
VietDao29 said:
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 totally different from P({1; 2; ...; n}). One is a set with only one element, 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 learned in the past?

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

Sum from k = 0 to n of rhok = ( 1 - rhon+1 ) / ( 1 - rho )
 
  • #8
Shackleford said:
The professor gave us a few sum equations we would have to use.

Sum from k = 0 to n of rhok = ( 1 - rhon+1 ) / ( 1 - rho )

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

The Geometric Sequence is a sequence, in which, the later (following) terms is formed by multiplying the former (previous) term by a constant r (call common 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; ar2; ar3; ar4; ...

The sum of the first n terms of a GP can be calculated by:
[tex]\sum_{i = 1} ^ n (a r ^ {i - 1}) = a\frac{1 - r ^ n}{1 - r}[/tex]

Where a is the first term of the GP, r is the common ratio, and n is th number of terms.

Now, apply the above formula to see if you can get the result. :)
 
Last edited by a moderator:
  • #9
VietDao29 said:
Yup, that's the sum of the first n terms in a http://en.wikipedia.org/wiki/Geometric_progression" [Broken] (abbr: GP) (or you can call it Geometric Sequence).

The Geometric Sequence is a sequence, in which, the later (following) terms is formed by multiplying the former (previous) term by a constant r (call common 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; ar2; ar3; ar4; ...

The sum of the first n terms of a GP can be calculated by:
[tex]\sum_{i = 1} ^ n (a r ^ {i - 1}) = a\frac{1 - r ^ n}{1 - r}[/tex]

Where a is the first term of the GP, r is the common ratio, and n is 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.
 
Last edited by a moderator:
  • #10
Shackleford said:
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
VietDao29 said:
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-n

Sum from k = 0 to n of rhok = ( 1 - 2n+1) / ( 1 - 2 )
 
  • #12
Shackleford said:
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 rhok = ( 1 - 2n+1) / ( 1 - 2 )

rho is not 2. Please look again closely at your sum, it's:
[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-k to somethingk?

And by the way, I think you should use the sum of the first n terms in a GP, instead of your professor's one. But if you still want to use the equation given by your professor, then you should do some adjustment to the index of the series. Hint: your professor's equation has one extra term, right? So, what can we do to it, to make it vanish?
 
  • #13
Nevermind. I figured out (a). I'll try the rest of them.
 
Last edited:
  • #14
What am I doing wrongly here?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110209_192854.jpg?t=1297301586 [Broken]
 
Last edited by a moderator:
  • #15
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?
 
Last edited:
  • #16
Unit said:
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.
 

1. What is the definition of probability?

Probability is the measure of the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

2. How do you calculate probability?

To calculate probability, divide the number of favorable outcomes by the total number of possible outcomes. This is known as the classical probability formula.

3. What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely. Experimental probability is based on actual trials or observations and may vary from the theoretical probability.

4. How is probability used in real life?

Probability is used in many real-life situations, such as predicting the weather, analyzing risk in insurance and finance, and making decisions in sports and gambling.

5. Can the probability of an event be greater than 1?

No, the probability of an event cannot be greater than 1. This would indicate that the event is certain to occur, which is not possible. The probability of an event can never be negative either.

Similar threads

  • Calculus and Beyond Homework Help
Replies
15
Views
3K
  • Calculus and Beyond Homework Help
Replies
14
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
878
  • Calculus and Beyond Homework Help
Replies
12
Views
6K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top