Bayes  (statistics),  any help would be greatly appreciated?
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a) P(D+|K+) = P(D+ and K+)/P(K+)\n17/30 = P(D+ and K+)/.6\nP(D+ and K+) = .34\n\nb) P(D- | K+) = P(D- and K+)/P(K+)\n=P(D- and K+)/.6\n=(P(K+) - P(D+ and K+))/.6\n=.26/.6\n= 13/30
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1. probabilty of getting a value that extreme in truth if the null hypothesis is true\n2. 1. null hypothesis is false \n    2. null hypothesis is true and there's another factor at work\n3. conclusion: it is likely that you should reject the null hypothesis\n4. wtf
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If you have a computer program that can compute binomial probabilities, you can use that.\n\nIf you are doing this by hand, however, I would suggest using the normal approximation to the binomial here.\n\nn = number of trials = 200\np = prob. of success = 1/6 if we consider getting a five to be a success.\n\nThen np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal.  This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27.\n\nNow for the continuity correction.  The rectangle over 40 goes from 39.5 to 40.5.  Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included.\n\nz = (39.5-33.33)/5.27 = 1.17\nArea to the right of 1.17 on the standard normal is 0.121.
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By reading your question, it appears that you are first and foremost seeking to "define" mathmatically what it is that you are attempting to do to reach your results-- presumably so that you can study that area of statistics further and apply it to your experiment.\n\nWith that in mind, what you are dealing with in your experiment is called "Event Probability."  In Event Probability, you are dealing with concepts of "conditional probability", "frequency", and "relative frequency."  For instance, what you are doing is very similar to the flow charts that appear in many family's medical books with a "Decision Tree" that allows you to answer "Yes" and "No" questions along a Flow Chart to reach an ultimate probability that you have a particular illness.  In other words, you are creating an algorithm by measuring the "relative frequency" of events and compound events and drawing a conclusion from that information.  People who perform these experiments are called "frequentists."  For all intents and purposes, this is what actuaries do at insurance companies and financial firms.  I'd read about those areas of statistics first which should prove very helpful in providing you with a structure for reaching the most helpful and accurate conclusions from your experiment.
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yea, multiplying the single event probabilities of two independent events to get the prob of the independent events conjunction is correct. In other words (to avoid you going to grab an Advil already): p(A&B)=p(A)*P(B) assuming A and B two independent events. That's BASIC stats.\n\nIn excel, use the function POWER. First argument the probability, second argument the number of occurence in a row.
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Try this website for help with your statistics homework:\n\nhttp://www.statsoft.com/textbook/stathome.html
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Start from www.statistics.com or search "statistics books" in Yahoo Answers. Greetings.
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since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
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1. Jessica M Utts and Robert F Heckard (2006) Statistical Ideas and Methods, Thomson Brookes/Cole \n*** (3 stars)\n2. Gerald Keller and Brian Warrack, Statistics for Management and Economics –6th Edition, Duxbury. This textbook comes with a CD-ROM comprising applets for interactive examples, a student study guide and data files.\n****(4 stars) - quite good
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plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
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Hi! I'm a statistician and I use http://www.statsoft.com as well as cheap textbooks from Half Priced Books in my area. Both seem to help. If you have a specific question, feel free to email me. Don't know how much good I'll be because I use fancy software for my analysis, but I'll try.
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use the formula for confidence interval.. its found on most statistics book.. i would type the formula, but that wouldn't be helping you to be a better student..
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try this site\n\n\nhttp://www.futureaccountant.com/probability/
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This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
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As to part one, I can not recommend any acceptable, known test for such data (perhaps what I've written below might work).\n\nAs to part two, this is a horrid approximation since the midrange as the mean is a major assumption (even more far-reaching than assuming something about the populations' probability distributions).\n\nFor part 3 and also part 1:\n\nThe first problem you run into is that you do not know anything about the nature of the population distribution (normal, uniform, etc).  Thus, a nonparametric test would be required.  I know of no nonparametric tests for any of the statistics you have (minimum, maximum or range).\n\nIf it is possible to obtain the actual values for each of the samples in each set, the Kruskal-Wallis test should suffice.  It is nonparametric, so there are no problems with the requirement of assumptions about population distributions.\n\nFurther, its job is to determine if two samples are far enough apart such that the null hypothesis that the population distributions are the same can be rejected.  This would give you what you desire.\n\nAs far as your problem is concerned, the only statistics you have are minimum, maximum and range.   Perhaps one could derive a test to test the minima or maxima of each set, but one still runs into the problem that there can be no parameter placed on the population distribution.\n\nIf you can make an assumption about the distributions of the populations of interest (normal or whatnot), then it would be possible to derive such a test.  This would be a tedious task, however, as it would require thorough knowledge of order statistics and distribution functions.\n\nOne would need to use a test statistic based of off the difference in the two statistics that you use (I would use the maxima since they are further apart than the minima and probability distributions for ranges are uncharted territory unless you're a Ph.D.).  Thus, your test statistic is based off of the difference between the maxima (Y(n) - Z(n) where the (n) are subscripts, which is 2 in this case).\n\nTo perform this test, you would need to derive the probability distribution of the difference of the two maxima and use that to find the probability of the difference being two or greater.  This is quite a task for anyone and requires a large amount of knowledge in mathematical statistics.  I believe I could perform such a task, but it would likely take some time to complete.\n\nRemember though that you must make a critical assumption about the population before you can even do this.  I would recommend using the assumption of uniformity if you are unsure of the populations' distributions.  This is because the derived test would then be least likely to give a type I error, where you reject the data improperly, because the test is less likely to reject.
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Note: they aren't asking just for the chance that at least 2 children have blue eyes.  They are asking for the probability, given that you know at least one already does.\n\nThis is conditional probability.\n\nP(B | A) = P(A and B) / P(A)\n\nSo the probability that you have at least 2, given you have at least 1 is the probability you have at least 2 divided by the probability you have at least 1.\n\nFor that part though, you would use binomal distribution.\n\nFirst, what is the probability that at least one child has blue eyes?\nThis would be 1 minus the probability that nobody has blue eyes.  \nP ( blue >= 1 ) = 1 - (0.75)^5 \nP ( blue >= 1 ) = 0.762695312 \n\nNow the probability that at least *two* children have blue eyes:\nP( blue = 2 ) + P ( blue = 3 ) + P ( blue = 4 ) + P ( blue = 5 ).  Again you can figure the negative statement and subtract from 1.  So this would be equivalent to:\n1 - P ( blue = 0 ) - P ( blue = 1 ).\n\nThis would be:\nP ( blue >= 2 ) = 1 - (0.75)^5 - (0.25)(0.75)^4 * C(5,1)\nThe C(5,1) means the ways to choose 1 person from 5, which is just 5.\nP ( blue >= 2) = 1 - (0.75)^5 - (0.25)(0.75)^4 * 5 \nP ( blue >= 2) = 0.3671875\n\nSo putting it all together:\nP ( blue >= 2 | blue >= 1 ) = P ( blue >= 2 ) / P ( blue >= 1 )\n= 0.3671875 / 0.762695312\n= 0.481434059\n≈ 48.1%
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These are my steps; it might be different for different people.\n\nCritical value approach:\n\n1. State null and alternative hypothesis.\n2. Compute the test statistic.\n3. Find the critical value and rejection region.\n4. Make you conclusion.\n\nP-value approach:\n\n1. State null and alternative hypothesis.\n2. Compute the test statistic.\n3. Find the p-value.\n4. Make your conclusion.