What is Bayes's Theorem? Can someone explain in simple terms.?
343048
Bayes's Theorem is a simple mathematical formula used for calculating conditional probabilities. It figures prominently in subjectivist or Bayesian approaches to epistemology, statistics, and inductive logic. Subjectivists, who maintain that rational belief is governed by the laws of probability, lean heavily on conditional probabilities in their theories of evidence and their models of empirical learning. Bayes's Theorem is central to these enterprises both because it simplifies the calculation of conditional probabilities and because it clarifies significant features of subjectivist position. Indeed, the Theorem's central insight — that a hypothesis is confirmed by any body data that its truth renders probable — is the cornerstone of all subjectivist methodology.\n\n\n Bayes's Theorem.\n \n PE(H) = [P(H)/P(E)] PH(E) \n\nIn an unfortunate, but now unavoidable, choice of terminology, statisticians refer to the inverse probability PH(E) as the "likelihood" of H on E. It expresses the degree to which the hypothesis predicts the data given the background information codified in the probability P.\n\n                                                               Bayes's Theorem (2nd form)\n \n PE(H) = P(H)PH(E) / [P(H)PH(E) + P(~H)P~H(E)] \n\nIn this guise Bayes's theorem is particularly useful for inferring causes from their effects since it is often fairly easy to discern the probability of an effect given the presence or absence of a putative cause. For instance, physicians often screen for diseases of known prevalence using diagnostic tests of recognized sensitivity and specificity. The sensitivity of a test, its "true positive" rate, is the fraction of times that patients with the disease test positive for it. The test's specificity, its "true negative" rate, is the proportion of healthy patients who test negative. If we let H be the event of a given patient having the disease, and E be the event of her testing positive for it, then the test's specificity and sensitivity are given by the likelihoods PH(E) and P~H(~E), respectively, and the "baseline" prevalence of the disease in the population is P(H). Given these inputs about the effects of the disease on the outcome of the test, one can use it to determine the probability of disease given a positive test.\n\nSpecial Forms of Bayes's Theorem\n\n\nBayes's Theorem can be expressed in a variety of forms that are useful for different purposes. One version employs what Rudolf Carnap called the relevance quotient or probability ratio (Carnap 1962, 466). This is the factor PR(H, E) = PE(H)/P(H) by which H's unconditional probability must be multiplied to get its probability conditional on E. Bayes's Theorem is equivalent to a simple symmetry principle for probability ratios.\n\nProbability Ratio Rule. \n PR(H, E) = PR(E, H) \n\nThe term on the right provides one measure of the degree to which H predicts E. If we think of P(E) as expressing the "baseline" predictability of E given the background information codified in P, and of PH(E) as E's predictability when H is added to this background, then PR(E, H) captures the degree to which knowing H makes E more or less predictable relative to the baseline: PR(E, H) = 0 means that H categorically predicts ~E; PR(E, H) = 1 means that adding H does not alter the baseline prediction at all; PR(E, H) = 1/P(E) means that H categorically predicts E. Since P(E)) = PT(E)) where T is any truth of logic, we can think of as telling us that.\n The probability of a hypothesis conditional on a body of data is equal to the unconditional probability of the hypothesis multiplied by the degree to which the hypothesis surpasses a tautology as a predictor of the data.\n\nAnother useful form of Bayes's Theorem is the Odds Rule. In the jargon of bookies, the "odds" of a hypothesis is its probability divided by the probability of its negation:
254656
I'm assuming you are talking about a probability problem in Bayes' theory since you haven't specified what any of your symbols mean. This is why there are no responses.\nThe answer is actually quite clear. Draw three circles in a triple intersection for the events R, S and I and label it clearly. The diagram simplifies to a certain double intersection that motivates the result.
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
1008113
The Russian mathematician P. L. Chebyshev (1821- 1894) discovered that the fraction of observations falling between two distinct values, whose differences from the mean have the same absolute value, is related to the variance of the population. Chebyshev's Theorem gives a conservative estimate to the above percentage.\n\nFor any population or sample, at least (1 - (1 / k)2) of the observations in the data set fall within k standard deviations of the mean, where k ³ 1.\n\nUsing the concept of z scores, we can restate Chebyshev's Theorem to say that for any population or sample, the proportion of all observations, whose z score has an absolute value less than or equal to k, is no less than \n(1 - (1 / k2 )). For k = 1, this theorem states that the fraction of all observations having a z score between -1 and 1 is (1 - (1 / 1))2 = 0; of course, this is not a very helpful statement. But for k ³ 1, Chebyshev's Theorem provides a lower bound to the proportion of measurements that are within a certain number of standard deviations from the mean. This lower bound estimate can be very helpful when the distribution of a particular population is unknown or mathematically intractable.
937552
You can do this two ways...\n\n1. Find P(x=1) + P(x=2) + P(x=3).\n\n2. An easier way: find P(x >= 1) by taking P(x = 0) and subtracting that from 1.\n\nEither way, from the table, you should get the same thing.\n\nedit: That rule of thumb will work for the binomial distribution because it is discrete.  It would not for, say, continuous distributions.  A better rule of thumb might be this...\n\nP(X >= x) = 1 - P(X < x).\n\nThis will always work.  Hopefully you understand the reason behind it; the probability of an event is equal to one minus the probability of the opposite of that event.
407360
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.
554599
4.2\n\nfor a binomial distribution the mean is the number * the probability
887177
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
480069
It takes only one example to prove the statement to be false, and I provided one:\n\n(Now, while you may never naturally encounter the set of numbers I used, the definitions of mean and median fit, so the statement is false.)
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
176780
I have a different point of view to the post above. If the probability of the given event occuring is 0, the the probability P(A|B) is a nonsense sentence and undefined, in the same manner as division by 0. There's nothing wrong with the existence of meaningless statements.\nThe aforementioned probability being zero is only enforced if you change the probability space to that of P(A AND B) which can easily be argued to be zero.
114803
If you have two events "A" and "B"\n\nThe probability that\n\na) Both would occur = P(A intersection B)\nb) Only one would occur = P (A intersection B') + P (B intersection A')\nc) At least one of them would occur = P (A union B)\nd) None of them would occur = P (A' intersection B')\n\nTry this site for a good explanation on probability\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
1000630
Pythagoras Theorem asserts that for a right triangle with short sides of length a and b and long side of length c \na2 + b2 = c2 \n\nor c2-a2 = b2 \n\nhere is a link with all the necessary thing you have to know about this theorem . I hope it helps , just read it carefully and everything is going to be fine :)
906595
C.\nThis is a binomial probability problem.\nThe binomial formula says \nP(r out of n) = (nCr)(p^r)(q^(n-r)), where n is the number of trials (here 6 rolls), r is the number of successes (here 0, 1, 2, and 3 twos), p is the probability of success (here 1/6), and q is the probability of failure (here 5/6)\nYou need to plug in 0, 1, 2, and 3 for "r" into this formula.  If you do, you get:\nP(0 twos) = .3348979767\nP(1 two) = .401877572\nP(2 twos) = .200938786\nP(4 twos) = .05358367627\nYou add up these answers to get your final answer, about .9913
540053
In the experiment of tossing a coin ten times.\nTotal number of possible choices = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 \n\n      = 1024 \n => n = 1024\n\nLet A be the event of getting number of heads is double the number of tails.\n\nFor event A\n------------\nNumber of favourable choices  =  Number of chances of getting heads is double the number of tailes.\n\n                  = No chances for occurance of event A\n                  = 0\n           =>  m  = 0\n\nProbablity of occurance of an event A =  Number of favourable choices / total number of possible choices\n     => P(A) =  m/n\n     => P(A) =  0/1024\n     => P(A) =  0\nTherefore the probability is zero.\n\n\n\nFor all about probabilty:\n\nhttp://www.futureaccountant.com
427277
Maybe you will find this article useful. It explains the concept of information entropy.