probability

问题 I was curious to know, how do I implement probability in Java? For example, if the chances of a variable showing is 1/25, then how would I implement that? Or any other probability? Please point me in the general direction. 回答1: You'd use Random to generate a random number, then test it against a literal to match the probability you're trying to achieve. So given: boolean val = new Random().nextInt(25)==0; val will have a 1/25 probability of being true (since nextInt() has an even probability

问题 Using R, it is trivial to calculate the quantiles for given probabilities in a sampled distribution: x <- rnorm(1000, mean=4, sd=2) quantile(x, .9) # results in 6.705755 However, I can't find an easy way to do the inverse—calculate the probability for a given quantile in the sample x . The closest I've come is to use pnorm() with the same mean and standard deviation I used when creating the sample: pnorm(5, mean=4, sd=2) # results in 0.6914625 However, because this is calculating the

问题 i've been looking around, but i'm not sure how to do it. i've found this page which, in the last paragraph, says: A simple generator for random numbers taken from a Poisson distribution is obtained using this simple recipe: if x 1 , x 2 , ... is a sequence of random numbers with uniform distribution between zero and one, k is the first integer for which the product x 1 · x 2 · ... · x k+1 < e -λ i've found another page describing how to generate binomial numbers, but i think it is using an

问题 Is there a direct way of getting the Nth combination of an ordered set of all combinations of nCr? Example: I have four elements: [6, 4, 2, 1]. All the possible combinations by taking three at a time would be: [[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]]. Is there an algorithm that would give me e.g. the 3rd answer, [6, 2, 1], in the ordered result set, without enumerating all the previous answers? 回答1: Note you can generate the sequence by recursively generating all combinations with the

问题 In some code I want to choose n random numbers in [0,1) which sum to 1 . I do so by choosing the numbers independently in [0,1) and normalizing them by dividing each one by the total sum: numbers = [random() for i in range(n)] numbers = [n/sum(numbers) for n in numbers] My "problem" is, that the distribution I get out is quite skew. Choosing a million numbers not a single one gets over 1/2 . By some effort I've calculated the pdf, and it's not nice. Here is the weird looking pdf I get for 5

问题 I am very new to SAS and trying to predict probabilities using logistic regression in SAS. I got the code below from SAS Support web site: data vaso; length Response $12; input Volume Rate Response @@; LogVolume=log(Volume); LogRate=log(Rate); datalines; 3.70 0.825 constrict 3.50 1.09 constrict 1.25 2.50 constrict 0.75 1.50 constrict 0.80 3.20 constrict 0.70 3.50 constrict 0.60 0.75 no_constrict 1.10 1.70 no_constrict 0.90 0.75 no_constrict 0.90 0.45 no_constrict 0.80 0.57 no_constrict 0.55 2

问题 I have been running this code for almost ten hours with no love: while ( true ) { int r = rand(); assert( r != 0 ); } I expect rand() to eventually roll a zero and thus trigger the assert. Am I doing something wrong or does rand() never return zero? Or have I not been waiting long enough to expect to see it? I'm on 2012-vintage 2GHz i7. 回答1: If this link is the definition of Mac OS X rand() , then 0 will never be produced. It's not a very good pseudo random number generator, IMHO. Amongst its

问题 This question already has answers here : How do I generate random integers within a specific range in Java? (65 answers) Closed 6 years ago . I'm currently working through an exercise where I am required to generate a random number which can be one of 4 values. (note I'm only allowed to use Math.random()) 0 1 2 or 3 Currently I am using this: randno = (int) (Math.random()*4); // range 0-3 However, the outcome MUST have equal probability. My tests so far (although the method is lacking) shows

问题 I admit I am being lazy here but is anyone aware of a free math .net library which gives me the output of a normal distribution given a mean vector and a covariance matrix? Thanks. 回答1: Math.NET might work for you. http://www.mathdotnet.com/ http://numerics.mathdotnet.com/probability-distributions/ Multivariate Distributions Dirichlet Inverse Wishart Matrix Normal Multinomial NormalGamma Wishart Function Reference: http://api.mathdotnet.com/Numerics/MathNet.Numerics.Distributions/MatrixNormal

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed last year . How could you use the prob argument in the sample command in R so that it covers conditional probability as well? For example, the letters of the alphabet: the probability of generating the letter 'b' depends on the previous letter generated. Thanks! 回答1: The question is still a bit difficult to