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Probability ? Rate Topic: -----

#1 1123581321 


Quark
When a probability question says 'what is the chances of getting at least one or at least two', how many does it mean in accordance with its stated amount..
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#2 RealFunnyFungi 


Quark
I guess you can just simply use the fact that 1-P(at most 1) = P(at least 2) to solve the problem...
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#3 1123581321 


Quark
sorry, im not sure i really understand..

Do you mean that no greater probability than 1 can be calculated when you taken away the probability of the whole etc..
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#4 Guest_sofiarunner_*

you need to find out the probability of the event not occurring first, let it be p
now your answer will be Ans= (1-p)
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#5 csmyth3025 


Atom

View Post1123581321, on 6 May 2011 - 09:14 AM, said:

sorry, I'm not sure i really understand..

Do you mean that no greater probability than 1 can be calculated when you taken away the probability of the whole etc..

Perhaps it would be easier to think of it this way: If you toss a two-headed coin up in the air, the probability that it will come up heads is 1 (or 100%, if you prefer) and the probability that it will come up tails is 1-1=0.

If you toss a regular coin up in the air, the probability that it will come up heads is 1/2 (or 0.5, if you prefer) and the probability that it will come up tails is 1-0.5=0.5

If you toss a six-sided die into the air, the chance that the number one will come up is 1/6 (or about 0.167) and the chance that any other number will come up is 1-0.167=0.833 (about 83%).

Chris
"It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow." (Robert Goddard - from his high school graduation oration, "On Taking Things for Granted", June 1904)

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#6 khaled 


Meson
Let me detail this,

Case: What is the probability of getting 3 or more Tails when tossing a fair coin 6 times

Possible Outcomes: { Head, Tail }

P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}

P(Head) = 1/2, P(Tail) = 1/2

Conditional Probability: how likely to get an outcome based on previous outcomes

P(Head,Head,Head) =
P(Head) \times P(Head|Head) \times P(Head|Head,Head) = P(Head) \times (\frac{P(Head \cap Head)}{P(Head)}) \times (\frac{P(Head \cap Head \cap Head)}{P(Head) \times P(Head)})

But, to solve our case, we consider n = number of Heads in the Trials ...

P(n > 2) = 1 - P'(n > 2) = 1 - P(n < 3)

To evaluate a simple case, we can do this,

P(n > 2) = 1 - P(n < 3) = 1 - (P(n=0) + P(n=1) + P(n=2))

In complex cases, you have to use Probability Distribution and Mass Function

This post has been edited by khaled: 2 August 2011 - 09:46 PM

Everything is a graph

twitter: @khaledkhunaifer, Blog: KhaledKhunaifer:Blog
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#7 DrRocket 


Primate

View Postkhaled, on 2 August 2011 - 09:42 PM, said:

Let me detail this,

Case: What is the probability of getting 3 or more Tails when tossing a fair coin 6 times

Possible Outcomes: { Head, Tail }

P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}


This is not only wrong, but absurd. This formula would have the probability of an event decrease to 0 as the numbeer of outcomes increased without bound. A heurestic treatment, loosely based on the law of large numbers often presents the probability of an event A as  P(A) = \lim_{N \to \infty} \dfrac { nummber \  of \ A \ outcomes}{N} where N is the number of trials.

Quote

P(Head) = 1/2, P(Tail) = 1/2

Conditional Probability: how likely to get an outcome based on previous outcomes.


wrong again. But closer.

By definition, P(A|B) =\dfrac {P(AB)}{P(B))}

Which, if A and B are independent reduces to P(A|B)=P(A)

Quote

P(Head,Head,Head) =
P(Head) \times P(Head|Head) \times P(Head|Head,Head) = P(Head) \times (\frac{P(Head \cap Head)}{P(Head)}) \times (\frac{P(Head \cap Head \cap Head)}{P(Head) \times P(Head)})

But, to solve our case, we consider n = number of Heads in the Trials ...

P(n > 2) = 1 - P'(n > 2) = 1 - P(n < 3)

To evaluate a simple case, we can do this,

P(n > 2) = 1 - P(n < 3) = 1 - (P(n=0) + P(n=1) + P(n=2))

In complex cases, you have to use Probability Distribution and Mass Function


Since the probability of the outcomes of heads or btails in separate tosses of a fair coin are independent, conditional probabilities offer no nrw insight (see above). In short this is ridiculous and

P(Head) \times P(Head|Head) \times P(Head|Head,Head) = P(Head) \times (\frac{P(Head \cap Head)}{P(Head)}) \times (\frac{P(Head \cap Head \cap Head)}{P(Head) \times P(Head)})= P(Head)^3

There is no such thing as "probability in the complex case". Probability measures are real valued.

You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#8 khaled 


Meson
Dr. Rocket .. with all regards, neither my post nor yours is wrong

So, just because I mentioned the simple situation doesn't mean it's wrong or absurd, and my definition
of conditional probability is not wrong in my point of view ...

So, your equation to calculate P(A) looks complex to me, can you show how to use it ?
Everything is a graph

twitter: @khaledkhunaifer, Blog: KhaledKhunaifer:Blog
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#9 DrRocket 


Primate

View Postkhaled, on 6 August 2011 - 01:15 AM, said:

Dr. Rocket .. with all regards, neither my post nor yours is wrong

So, just because I mentioned the simple situation doesn't mean it's wrong or absurd, and my definition
of conditional probability is not wrong in my point of view ...

So, your equation to calculate P(A) looks complex to me, can you show how to use it ?


You are precisely half right. My post is correct.

You have no idea what you are talking about. Read my post again.

Better yet read Loeve's Probability Theory.

You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#10 Bignose 


Icon
Maths Expert
Khaled, DrRocket is right. Some of your equations are quite wrong.

View Postkhaled, on 2 August 2011 - 09:42 PM, said:

P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}


For example, this equation is very naive in its assumptions.

This is only right if the probability of every event is exactly equal, but that is hardly necessary. Consider an unfair coin, where heads will come up much more often than one half. Your equation above is wrong, because even though the coin is weighted to be unfair, there are still two outcomes, yet the probability of heads is already stated to not be 1/2.

Or, consider rolling 2 fair six sided dice, and call the result the total sum of the pips on the top sides. Your equation above predicts that each sum, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 are equal because those are the possible outcomes. But, the chances of rolling a sum of 2 is only 1/36 while rolling a sum of 7 is far more likely. 6/36 in fact.

View Postkhaled, on 6 August 2011 - 01:15 AM, said:

and my definition of conditional probability is not wrong in my point of view ...


Also, there is a very precise definition of conditional probability that is in exceptionally common use, it probably isn't a good idea to redefine it willy nilly...
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#11 khaled 


Meson
Still, I have no idea how to use this equation  P(A) = \lim_{N \to \infty} \frac{nummber of A outcomes}{N}

isn't \lim_{N \to \infty} \frac{C}{N} = \lim \frac{C}{\infty} = 0 \;\; for \;\; 0 \leq C < \infty ..?

This post has been edited by khaled: 6 August 2011 - 11:21 AM

Everything is a graph

twitter: @khaledkhunaifer, Blog: KhaledKhunaifer:Blog
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#12 DJBruce 


Molecule

View Postkhaled, on 6 August 2011 - 11:18 AM, said:

Still, I have no idea how to use this equation  P(A) = \lim_{N \to \infty} \frac{nummber of A outcomes}{N}

isn't \lim_{N \to \infty} \frac{C}{N} = \lim \frac{C}{\infty} = 0 \;\; for \;\; 0 \leq C < \infty ..?


The numerator is a function of N.
"To give anything less than your best is to sacrifice the Gift."

"A lot of people run a race to see who is fastest. I run to see who has the most guts, who can punish himself into exhausting pace, and then at the end, punish himself even more."
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#13 DrRocket 


Primate
In a rigorous treatment of the theory of probability, the probability of elementary events is given, not calculated. The starting point for probability theory is a Set, a sigma algebra of subsets and a positive measure of total mass 1 on that sigma algebra.

Probabilities based on relative frequency of occurrence are really estimates based on the law of large numbers. Probabilities based on combinatorics are really definitions of a probability measure, and an assumption that some given class of events are of equal probability. This sort of treatment is usually found in very elementary and non-rigorous treatments of probability theory that attempt to give an overview of the subject while avoiding the measure theory on which rigorous probability has been based since the work of Kolmogorov.

The heuristic "definition" in terms of relative frequencies is conceptually useful but is not a practical way to determine actual probabilities. It is not strictly speaking correct, hence my qualification of it as heuristic, but rather is roughly a converse to the law of large numbers, modulo some loose language as to the sense in which things are meant to converge (see "convergence in probability" or "convergence in measure".)

The only way to do this correctly is to use the general theory of measure and integration. For that see the book of Loeve. Probability is the most misused and incorrectly presented branch of mathematics. A good deal of what one finds in engineering, physics and introductory mathematics texts is not strictly correct.

You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#14 amanda more 


Baryon

View PostDrRocket, on 6 August 2011 - 10:47 PM, said:

In a rigorous treatment of the theory of probability, the probability of elementary events is given, not calculated. The starting point for probability theory is a Set, a sigma algebra of subsets and a positive measure of total mass 1 on that sigma algebra.

Probabilities based on relative frequency of occurrence are really estimates based on the law of large numbers. Probabilities based on combinatorics are really definitions of a probability measure, and an assumption that some given class of events are of equal probability. This sort of treatment is usually found in very elementary and non-rigorous treatments of probability theory that attempt to give an overview of the subject while avoiding the measure theory on which rigorous probability has been based since the work of Kolmogorov.

The heuristic "definition" in terms of relative frequencies is conceptually useful but is not a practical way to determine actual probabilities. It is not strictly speaking correct, hence my qualification of it as heuristic, but rather is roughly a converse to the law of large numbers, modulo some loose language as to the sense in which things are meant to converge (see "convergence in probability" or "convergence in measure".)

The only way to do this correctly is to use the general theory of measure and integration. For that see the book of Loeve. Probability is the most misused and incorrectly presented branch of mathematics. A good deal of what one finds in engineering, physics and introductory mathematics texts is not strictly correct.


I am unsure how to explain this. I want nonscientists to understand the absolute basics of probabiliy.

"Assume a state takes in 100 million dollars. 50 million is kept by the government ." "I understand that> I know the lottery makes money for the state. Then I ask so "when a $1 lottery ticket is purchased 50 cents goes to the government." 9 out of 10 lottery players find this goes over their heads. Why? Am I actually using some kind of sophisticated theory here?



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#15 DrRocket 


Primate

View Postamanda more, on 7 August 2011 - 04:48 AM, said:

I am unsure how to explain this. I want nonscientists to understand the absolute basics of probabiliy.

"Assume a state takes in 100 million dollars. 50 million is kept by the government ." "I understand that> I know the lottery makes money for the state. Then I ask so "when a $1 lottery ticket is purchased 50 cents goes to the government." 9 out of 10 lottery players find this goes over their heads. Why? Am I actually using some kind of sophisticated theory here?





No, what you are observing is both correct and very simple.

But you are talking to lottery players, and that is a group that accepts a very poor bet from an economic perspective, some for the entertainment value of a minor expense, but many from a position of abject ignorance.

You will find that almost any logical argument goes over the heads of a great many people. Probability is often not intuitive and goes over the heads of many more.

You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#16 khaled 


Meson
I usually use this: P(A) = (\frac{1}{number of A outcomes})^{N}, which for large N leads to small numbers that are rounded as zero

But, if I use this: \frac{number of A outcomes}{N}, this would approach zero slower

Laplace smothering used logarithm and exp functions to work that problem .. I've worked on it regarding Markov Models
Everything is a graph

twitter: @khaledkhunaifer, Blog: KhaledKhunaifer:Blog
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#17 amanda more 


Baryon

View PostDrRocket, on 7 August 2011 - 05:29 AM, said:

No, what you are observing is both correct and very simple.

But you are talking to lottery players, and that is a group that accepts a very poor bet from an economic perspective, some for the entertainment value of a minor expense, but many from a position of abject ignorance.

You will find that almost any logical argument goes over the heads of a great many people. Probability is often not intuitive and goes over the heads of many more.


Any books on that?
There are many very odd things happening now. In anyones personal life, I don't find that intense logic helps very well in day to day life. A well developed EQ appears to provide an easier time of it.

I have found the very disturbing gamemanship in Congress to have had no logical basis.

This stuff must be a symptom of an underlying malaise.

Could fear cause people to lose their heads? Is there an example to use where people can find probability more intuitive?

There are the scientists/technologists and the everybody else. The "everybody else" runs this country.

They are ignorant of even the most basic "words" in the language of science - math. Could this lack be one of the root causes of what appears to be crazy behavior? They can't understand arithmetic and so cannot understand the effects of their behavior?
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#18 John Cuthber 


Icon
Chemistry Expert
http://en.wikipedia....with_Statistics
What's this signature thingy then? Did you know Santa only brings presents to people who click the + sign? -->
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#19 DrRocket 


Primate

View Postkhaled, on 7 August 2011 - 12:38 PM, said:

I usually use this: P(A) = (\frac{1}{number of A outcomes})^{N}, which for large N leads to small numbers that are rounded as zero

But, if I use this: \frac{number of A outcomes}{N}, this would approach zero slower

Laplace smothering used logarithm and exp functions to work that problem .. I've worked on it regarding Markov Models


This just plain wrong. Obviously wrong. Patently ridiculous.

You need to learn some mathematics. Probability theory is just a small part of the obvious lack.

You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#20 amanda more 


Baryon

View PostJohn Cuthber, on 7 August 2011 - 06:00 PM, said:



I guess I am trying to show how to tell the truth with statistics. At least with the lottery example.

Is it because those exposed to numbers in this country are suspect about any math and rightly so? Is that why they cannot speak even the equivalent of a pidgin style of math?

I appreciate the efforts to explain here. I'll source the book.
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