1123581321 Posted May 5, 2011 Share Posted May 5, 2011 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.. Link to comment Share on other sites More sharing options...

RealFunnyFungi Posted May 5, 2011 Share Posted May 5, 2011 I guess you can just simply use the fact that 1-P(at most 1) = P(at least 2) to solve the problem... Link to comment Share on other sites More sharing options...

1123581321 Posted May 6, 2011 Author Share Posted May 6, 2011 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.. Link to comment Share on other sites More sharing options...

Guest sofiarunner Posted May 6, 2011 Share Posted May 6, 2011 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) Link to comment Share on other sites More sharing options...

csmyth3025 Posted July 13, 2011 Share Posted July 13, 2011 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 Link to comment Share on other sites More sharing options...

khaled Posted August 2, 2011 Share Posted August 2, 2011 (edited) 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 } [math]P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}[/math] P(Head) = 1/2, P(Tail) = 1/2 Conditional Probability: how likely to get an outcome based on previous outcomes [math]P(Head,Head,Head) = [/math] [math]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)})[/math] But, to solve our case, we consider n = number of Heads in the Trials ... [math]P(n > 2) = 1 - P'(n > 2) = 1 - P(n < 3)[/math] To evaluate a simple case, we can do this, [math]P(n > 2) = 1 - P(n < 3) = 1 - (P(n=0) + P(n=1) + P(n=2))[/math] In complex cases, you have to use Probability Distribution and Mass Function Edited August 2, 2011 by khaled Link to comment Share on other sites More sharing options...

DrRocket Posted August 3, 2011 Share Posted August 3, 2011 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 } [math]P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}[/math] 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 [math] P(A) = \lim_{N \to \infty} \dfrac { nummber \ of \ A \ outcomes}{N}[/math] where [math]N[/math] is the number of trials. 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, [math]P(A|B) =\dfrac {P(AB)}{P(B))}[/math] Which, if [math]A[/math] and [math]B[/math] are independent reduces to [math]P(A|B)=P(A)[/math] [math]P(Head,Head,Head) = [/math][math]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)})[/math] But, to solve our case, we consider n = number of Heads in the Trials ... [math]P(n > 2) = 1 - P'(n > 2) = 1 - P(n < 3)[/math] To evaluate a simple case, we can do this, [math]P(n > 2) = 1 - P(n < 3) = 1 - (P(n=0) + P(n=1) + P(n=2))[/math] 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 [math]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[/math] There is no such thing as "probability in the complex case". Probability measures are real valued. Link to comment Share on other sites More sharing options...

khaled Posted August 6, 2011 Share Posted August 6, 2011 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 ? Link to comment Share on other sites More sharing options...

DrRocket Posted August 6, 2011 Share Posted August 6, 2011 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. Link to comment Share on other sites More sharing options...

Bignose Posted August 6, 2011 Share Posted August 6, 2011 Khaled, DrRocket is right. Some of your equations are quite wrong. [math]P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}[/math] 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. 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... Link to comment Share on other sites More sharing options...

khaled Posted August 6, 2011 Share Posted August 6, 2011 (edited) Still, I have no idea how to use this equation [math] P(A) = \lim_{N \to \infty} \frac{nummber of A outcomes}{N} [/math] isn't [math]\lim_{N \to \infty} \frac{C}{N} = \lim \frac{C}{\infty} = 0 \;\; for \;\; 0 \leq C < \infty[/math] ..? Edited August 6, 2011 by khaled Link to comment Share on other sites More sharing options...

DJBruce Posted August 6, 2011 Share Posted August 6, 2011 Still, I have no idea how to use this equation [math] P(A) = \lim_{N \to \infty} \frac{nummber of A outcomes}{N} [/math] isn't [math]\lim_{N \to \infty} \frac{C}{N} = \lim \frac{C}{\infty} = 0 \;\; for \;\; 0 \leq C < \infty[/math] ..? The numerator is a function of N. Link to comment Share on other sites More sharing options...

DrRocket Posted August 6, 2011 Share Posted August 6, 2011 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. Link to comment Share on other sites More sharing options...

amanda more Posted August 7, 2011 Share Posted August 7, 2011 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? Link to comment Share on other sites More sharing options...

DrRocket Posted August 7, 2011 Share Posted August 7, 2011 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. Link to comment Share on other sites More sharing options...

khaled Posted August 7, 2011 Share Posted August 7, 2011 I usually use this: [math]P(A) = (\frac{1}{number of A outcomes})^{N}[/math], which for large N leads to small numbers that are rounded as zero But, if I use this: [math]\frac{number of A outcomes}{N}[/math], this would approach zero slower Laplace smothering used logarithm and exp functions to work that problem .. I've worked on it regarding Markov Models -2 Link to comment Share on other sites More sharing options...

amanda more Posted August 7, 2011 Share Posted August 7, 2011 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? Link to comment Share on other sites More sharing options...

John Cuthber Posted August 7, 2011 Share Posted August 7, 2011 http://en.wikipedia.org/wiki/How_to_Lie_with_Statistics Link to comment Share on other sites More sharing options...

DrRocket Posted August 7, 2011 Share Posted August 7, 2011 I usually use this: [math]P(A) = (\frac{1}{number of A outcomes})^{N}[/math], which for large N leads to small numbers that are rounded as zero But, if I use this: [math]\frac{number of A outcomes}{N}[/math], 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. -1 Link to comment Share on other sites More sharing options...

amanda more Posted August 7, 2011 Share Posted August 7, 2011 http://en.wikipedia....with_Statistics 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. Link to comment Share on other sites More sharing options...

DrRocket Posted August 7, 2011 Share Posted August 7, 2011 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? Not sure what kind of books you are interested in. Two of the best probability books are Loeve's Probability Theory and Feller's An Introduction to Probability Theory and its Applications. Probability and Measure by Billingsley is also very good. For statistics I like van der Waerden's Mathematical Statistics and Cramer's Mathematical Methods of Statistics. However your observation that if the government pays out 50% on a lottery then the government also keeps 50% should not require a book, but should in fact be obvious to the casual observer. The reality that this is not obvious to some people is merely a comment on what is meant by "average intelligence". Link to comment Share on other sites More sharing options...

amanda more Posted August 8, 2011 Share Posted August 8, 2011 Not sure what kind of books you are interested in. Two of the best probability books are Loeve's Probability Theory and Feller's An Introduction to Probability Theory and its Applications. Probability and Measure by Billingsley is also very good. For statistics I like van der Waerden's Mathematical Statistics and Cramer's Mathematical Methods of Statistics. However your observation that if the government pays out 50% on a lottery then the government also keeps 50% should not require a book, but should in fact be obvious to the casual observer. The reality that this is not obvious to some people is merely a comment on what is meant by "average intelligence". Thank you. I'll either dig out my own or check out one of these. "Probability is often not intuitive and goes over the heads of many more. " I think people do use probability but don't think of it in math terms. I think if they did think of it in math terms their odds would be wildly off. Yet, they have a feel for risk to get through life. Crossing the street. Necessary but needs vigilance. Crossing a sidewalk. Not as much vigilance etc. In fact some streetwise kid can beat out an absentminded professor every time. They can pass 8th grade probability. So majority intelligence is good enough for that. As I am finding today, in our economy everyone is blinded by money and decisions about money. Lawyers and journalists are by and large not stupid. Yet they generally can't handle arithmetic. Why? Link to comment Share on other sites More sharing options...

DrRocket Posted August 8, 2011 Share Posted August 8, 2011 (edited) Lawyers and journalists are by and large not stupid. Yet they generally can't handle arithmetic. Why? I know of one former mathematics professor who is now a lawyer. I have encountered some very stupid lawyers. I taught college algebra to a local anchor man. He was one of the better students in the class. I have encountered reporters who could not count their fingers. Generalizations are usually wrong. A lot of people cannot handle arithmetic. A lot can. Edited August 8, 2011 by DrRocket Link to comment Share on other sites More sharing options...

amanda more Posted August 8, 2011 Share Posted August 8, 2011 I know of one former mathematics professor who is now a lawyer. I have encountered some very stupid lawyers. I taught college algebra to a local anchor man. He was one of the better students in the class. I have encountered reporters who could not count their fingers. Generalizations are usually wrong. A lot of people cannot handle arithmetic. A lot can. Scientists and technologists can. In fact they can speak math. I'm not saying all should do that. Instead of us having to put things in language, English why can't they (the other half) at least learn to understand the baby speech of probability? You also made a comment about logical thinking. So a tea partyer say who explains calmly some weird party line feels as if he is being logical. He is calmly explaining and seeks to give that impression. The impression and comraderie is so important he recoils at being called on it. He finds -logical thinking- a hindrance. Hence if logical thinking impedes goals then that is why many avoid it? Or are their brains set up so they can't do it? Have you met any over 160 IQ types who doesn't do logical thinking? I'm thinking there have been dictators who are smart. What comes out of their mouths may not make logical sense. I assumed they are being duplicitous. But maybe not. So, is there a way to explain to a 3 year old my example of a lottery that could be understood? I started to ask around about the lottery for the basis of a book I want to write. If I can't explain this simple example, I'm stuck. Link to comment Share on other sites More sharing options...

DrRocket Posted August 8, 2011 Share Posted August 8, 2011 [math]P ( Outcome_{i} ) = \frac{1}{Number of Outcomes}[/math] Khaled, DrRocket is right. Some of your equations are quite wrong. 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. It is only right if one in addition to the assumption that all elementary events are equally likely one makes the unconventional interpretation of "outcomes" as "possible outcomes" rather than "observed occuences of A". That may be what Khaled means, but it is contrary to normal useage. If one uses the conventional interpretation of "outcomes" then his equation would result in the probability of A being inversely proportional to the relative frequency of occurrence of A, which is backwards. Link to comment Share on other sites More sharing options...

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