Wikipedia 10K Redux by Reagle from Starling archive. Bugs abound!!!
The ''binomial distribution density function'' provides the probability that X success will occur in N trials of a ''binomial experiment''. A ''binomial experiment'' has the following properties. *1). It consists of a sequence of N individual trials. *2). Two outcomes are possible for each trail, success or failure. *3). The probability of success on any trial, denoted p does not vary from trail to trial. *4). The trials are independent. One common example of a binomial experiment is a sequence of N tosses of a coin. Here: *1). This is clearly an experiment that consists of a sequence of N trials. *2). There are only two outcomes, Heads and Tails, Let call Heads a success and Tails a failure. *3). The probability of success on any one trail, denoted p, here ½, does not vary from trial to trial. *4). The trails are independent – The success of the third trial, say, does not depend on the success of the 2nd trial, etc. To find the binomial distribution function of X successes in N trials, denoted f(X), we let p denote the probability of success on any one trial. Then the probability of failure on any one trial is q=1-p. Then the probability of any particular sequence of X success (heads) out of N trials (tosses) is p^X*q^(N-X). But, there are a variety of sequences of tosses that will result in X successes out of N trials. Then, using the formula for the number of ways of obtaining X successes out of N trials, we find that the number of ways of picking X objects out of N objects that there are * (X!/(N!*(N-X)!) ways, where '''N'''!=N factorial. * We will denote (N!/(X!*(N-X)!) by: * (N) * (X). (apologies…for symbolism) Then the probability of N successes out of N trials is given by the FunctioN: F(X)= (N) * p^X*q^(N-X) = (N) * p^X*(1-p)^(N-X) (X) (X) For example, consider the experiment of tossing a die with the usual 6 sides. Suppose we want to know the probability of getting three "1"s in 5 tosses. Then p=1/5. So q=4/5. Then, using the binomial probability function for 3 successes out of 5 trials we get the probability of this occurring as: * F(2)= (5!/(3!*2!)) * (1/5)^3 * (4/5)^2=(5*2)* (1/125)*(16/25)=(10)* (16/3023) * =160/3125=.0512 ---- Your presentation is more complete, certainly, but I fail to see my error. Please look again at BinomialDistribution. Thanks. DickBeldin---- Please go to BinomialDistribution, where I have indicated what the errors were and how to see them. There errors will remain in "Previous Versions" and stand as evidence for the errors. Are they yours? Yes, I corrected them after over a week. No, there are no major errors anymore. RoseParks