well-known probability distribution?
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Binomial Distribution, \nPoisson Distribution, \nNormal Distribution\n\nhttp://www.futureaccountant.com/probability/
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A few minutes of research and a basic understanding of probability should convince you that "messages" of any kind can be found in any sufficiently long piece of text using any kind of systematic text sampling technique.\n   By any kind of message, I mean just that. For any given statement, a statement that exactly contradicts the first statement is likely to be found along, as always, with a mind-boggling mountain of pure, random gibberish.\n\nAlthough it's best to determine these things for oneself, if an appeal to authority is convincing then here is a link to Caltech's mathematics department where 55 mathematicians have signed a document called, "Mathematicians' Statement on the Bible Codes"
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You mean a distribution of credit scores?
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The easiest way to generate random number from a given distribution is to generate a uniform (0,1) random number; then compute the inverse cummulative probability function from that number:\nLet u be a uniform random number\nthen, x=Pinv(u,l) distributes Poisson if Pinv stands for the inverse function of the cummulative poisson distribution.\nIt's not relevant what language you use for programming
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Of course. The pattern is well known, and is manifested as the binomial distribution and Pascal's triangle.\n\nIt is well formulated mathematically. There is a formula that can tell you, for example, what the 18th term is in the expansion of (a+b)^30.
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http://en.wikipedia.org/wiki/Cumulative_distribution_function
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The Weibull distribution is a statistical distribution developed by Wallodi Weibull to predict the "life" of roller bearings.  It has since been used in the field of reliability and life data analysis.
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Use Binomial distribution:\n\n(n nCR x) p^x (1-p)^(n-x)\n\nn nCR x = n!/(x!(n-x)!)\np=probabiliy (.10)\nn=patients (10)\nx= desired\n\na) P(X<=2) = P(X=0) + P(X=1) + P (X=2)\nb) 1- P (X<=2)... 1-prevoius answer
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Theoretical probability is finding the probability of events that come from an equiprobable sample space or, in other words, a sample space of known equally likely outcomes. \nFor example, \nSay you roll a dice.\nThe probability of rolling a 1 is the same if you roll a 2, 3, 4, 5, or 6.\nThat is, each number is equally likely.\nSince there are 6 numbers in the sample space, the probability of each event is 1/6\n\nAnother example:  Say you flip a fair coin.\nYou have two possiblities.  It could land heads, or it could land tails.  Each event is equally likely, so the probability of landing heads is the same as the probability of landing tails (which is 1/2)
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Actually, statistics books are notorious for being written in such a way that no normal human can understand them. The trick, generally, is trying to find a way to relate the problems to something you're interested in. For example, if you like basketball: one possible sampling distribution would be the number of free throws individual players make over a season. You could take a sample of, say, 10 of the players and find the mean and standard deviation. You could compute a 95% confidence interval with that information, which would allow you to say that you're 95% confident that the average number of free throws made by NBA players in a season will be between this number and that number. You could find the probability that a given player would make at least a certain number of free throws in a season by finding the z-score and looking it up on the normal probability distribution table, etc. If you're really struggling with the terminology, I recommend visiting your math help center on campus (most of them have one), or contacting a tutor.
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http://www.mathsrevision.net/alevel/\n\nhttp://www.revision-notes.co.uk/A_Level/Maths/Statistics_and_Probability/index.html\n\nhttp://www.mathsnet.net/links/revision.html
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This probability problem can be used using Binomial Distribution.  \n\nGiven a number of 'x' success, the probability in a binomial experiment is:\n\nP(x) = [n choose x]*p^x*q^(n-x)\n\nwhere:\n\nn choose x is a combination problem: n!/[(n-x)!x!]\np = probability of success\nq = probability of failure\nn = total selection\nx = successes\n\nIn this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers).  In essence, we are solving for:\n\nP(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).  \n\nHowever it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability.  This would be:\n\n1 - P(x=0) is what we are looking for.  Now all we need is to calculate the probabilty.\n\nFor P(x=0), we have the following:\n\nn = 5\nx = 0\np = (9/26) = probability of success\nq = (17/26) = probability of failure\n\nPlugging into P(x), we get:\n\nP(x=0) = (5!/5!)(9/26)^0(17/26)^5\n\nSolving, you get P(x=0) = 0.12\n\nTherefore, \n\n1 - P(x=0) is ------>\n\n1- 0.12 = 0.88\n\nBelow is a website devoted to Binomial Distribution exercises\n---------\n\nHope this helps
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try this site\n\nhttp://www.futureaccountant.com/probability/study-notes/
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Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.
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Random chances are the same as probabilities. If you are so interested, you should look into Quantum Theory. It describes it in depth.
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Find yourself a z-statistic table and look up critical value in the table.  Check your stats book on how to interpret the table.\nhttp://www.math2.org/math/stat/distributions/z-dist.htm
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For any random variable X, if a is a positive number, it is clear that the probability that X will be larger than a goes to 0 as a grows without limit. This is because a probability distribution has integral 1, and therefore the area under the probability density curve (continuous distribution) to the right of a, or the sum of the probabilities for values of X larger than a (discrete distribution) have to go to 0 as a goes to infinity.\nHope this helps -