Pi and ProbabilityDate: 12/21/97 at 19:34:02 From: Leduc Subject: Hi I'm doing a report for my math class on the relation between pi and probability. I've been looking on the Internet for information, but eveything I come across is way beyond my level. Could you send me some info? I'm an 11th grader taking Advanced Functions. Could you also send me info on the different ways to calculate pi. I'm doing a science project on this. I know so far that there is a way to calculate pi using arctangent, but I'm not sure how. Claire Date: 12/23/97 at 15:31:42 From: Doctor Rob Subject: Re: Hi Search the World Wide Web looking for "Buffon" together with "needle". I know it sounds crazy, but do it anyway. That will help you with part 1. There is a lot of information on the many, many ways to calculate Pi. You can try the following URL on the World Wide Web, and the pages linked to it: http://mathforum.org/dr.math/faq/faq.pi.html -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 01/14/98 at 16:10:30 From: Leduc Subject: Hi Could you please tell me how probability is related to the normal distribution curve or any other aspects of probability? Thanks, Claire Date: 01/14/98 at 17:14:01 From: Doctor Rob Subject: Re: Hi Thanks for asking, Claire. If you have an event with probability 1/2, like getting heads when you flip a coin, and you count the number of successes after n trials, for some positive integer n, you will find that the probability of getting exactly k successes and m failures, where k + m = n, is given by b(k; n, 1/2) = 2^(-n)*n!/(k!*m!), where r! = r*(r-1)*...*2*1 is "r factorial." Define h = 2/Sqrt[n], and x(k) = (k - n/2)*h, and then it turns out that b(k; n, 1/2) ------------ h*phi[x(k)] approaches 1 as n and k grow without bound in such a way that x(k)^3/Sqrt[n] -> 0. (This means k stays "near" n/2). Here phi(x) = e^(-x^2/2)/Sqrt(2*Pi), the normal distribution curve. This can be generalized to a situation where the probabilities of success and failure are not equal, as well. An example is rolling a die, and getting a 3, whose probability of success is 1/6, and of failure 5/6. This is so famous a theorem that it has a name: The DeMoivre-Laplace Limit Theorem. This means that for k and n satisfying those conditions, phi[x(k)] ~=~ b(k; n, 1/2)*Sqrt[n]/2. This approximation is surprisingly good, even for fairly small n's, like perhaps 50. Thus you could say that the normal distribution curve can be used to approximate the distribution of successes in coin-flipping with a large number of trials. I hope that this is what you wanted. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/