What is the formula for finding the expected value in a probability distribution?
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Discrete Case:\nE[X] = Sigma_{i=1}^n x_i*p(X = x_i) \n\nwhere Sigma_{i=1}^n p(X = x_i) = 1\n\nContinuous Case:\nE[X] = int(x*p(X = x)dx,-infty,infty)\n\nThink of it as a weighted *average*.
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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.
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When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
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You must look at the formula for standard deviation of the sample distribution of (I'm thinking) binomial distributions. The proportion "p" is a parameter of the formula. You should check it out but, if I remember correctly, it's:\ns.d. of the sample distribution = sqrt[p*(p-1)/n]
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(A) The probability that a wait will be necessary is the same as the probability that a car comes within the next T seconds: 1 - exp(-lambda T). So the probability of no wait is:\n\nP = exp(-lambda*T)\n\n(B) Let Q be the expected waiting time. If t is the time until the next car arrives (t < T), the expected additional wait time will be Q, giving an expected total wait time of t + Q. In other words the expectation value of the total wait time is:\n\nQ = &#92;integral_0^T (lambda exp(-lambda*t))(t + Q) dt\n= [-exp(-lambda t)(Q + 1/lambda + t]_0^T\n= Q + 1/lambda - exp(-lambda T)(Q + 1/lambda + T)\n\nSolving for Q gives the answer:\n\nQ = (exp(+lambda T) - 1)/lambda - T
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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
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Here is a Bayes Theorem Calculator\nhttp://psych.fullerton.edu/mbirnbaum/bayes/BayesCalc.htm\nhere is another one\nhttp://members.aol.com/johnp71/bayes.html\n\nHere is a Bayes theorem tutorial\nhttp://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutorialsf3/unit6_6.html
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Do you mean E(X)? That would be the mean value of X, assuming we have a normal (Gaussian) distribution.
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There will be a table in your statistics textbook that gives probabilities for a normal (Gaussian) distribution.  Usually it is for a "standard normal" where the mean is 0, and the standard deviation is 1.  The table will show upper tail probabilities for values of x ranging from 0. to 3. (or higher), where the it gives the probability that random sample xi > x.  For general samples with a mean different than 0 and a standard deviation different than 1, you have to apply a conversion to use this table, as follows:\n\nLet m = the mean of a normal distribution\nq = the standard deviation (usually called by the greek letter sigma), yi = boundary of upper tail distribution.\nIn this case, m = 92500, q = 8400, yi = 95000.  The value of x to look up in the table is \n(1) x = (yi-m)/q.  \nThis will tell you the probability that ONE random element, xi > x.\n\nHowever, if you take a sample of N elements, if the standard deviation of the entire sample is q, the standard deviation on the MEAN of the sample is \n(2) q'=q/sqrt(N).\nSo for the sample of 64 families, the standard deviation on the mean is 8400/sqrt(64) = 1050.  This is the value to use for q in equation (1).  Equation (1) then gives x = (95000-92500)/1050 = 2.38.\n\n  So the probability is the same as the probability that x > 2.38 for a standard normal distribution.  From the table (or from an online calculator like the one at\nhttp://bayes.math.montana.edu/cgi-bin/Rweb/buildModules.cgi\nI find that the probability = 0.008656319 , less than 1%
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Suppose P(k) is the probability that k of the N pots are occupied. Considering the process to begin on the second throw, there are N possible states (k=1,2,3...N). The initial probability distribution (after the first rock is thrown) is:\n\nP(1)=1, P(2)=P(3)=...=P(N)=0\n\nEach time a rock is thrown at random, the probability distribution undergoes a constant linear transformation (this is a Markov process):\n\nP --> P'(k) = (k/N)*P(k) + ((k-1)/N)*P(k-1)\n\nTo answer the question it's necessary to calculate P(N) after r transitions, equate it to .95 and solve for r. That can be done in principle for various specific values of N. For example for N=2, it's not to hard to see that:\n\nP(1) = (1/2)^r\nP(2) = 1 - (1/2)^r\n\nSolving (1/2)^r = 1-.95 = .05 gives about r = 4 (i.e. 5 actual rocks).\n\nI can also solve the case N=3 exactly. In this case, the transition P -> P' caused by each throw is: \n\nP'(1) = (1/3)P(1)\nP'(2) = (2/3)P(1) + (2/3)P(2)\nP'(3) = P(3) + (1/3)P(2)\n\nThis has solution (P(k) after r transitions):\n\nP(1) = (1/3)^r\nP(2) = (2^(r+1) - 2)/3^r\nP(3) = 1-P(1)-P(2)\n\nSo the exact answer in this case is the solution of:\n\nP(1) + P(2) = (2^(r+1)-1)/3^r = (2^(r+1)-1)/3^r = .05\n\n==> r is between 7 and 8 (i.e. between 8 and 9 actual throws).\n\nIf there's an expression for P(N) for general values of N, it's probably relatively complicated. \n\nHowever for large N it might be a pretty good approximation to treat the occupation states of each pot as independent (even though they're not really). Taking r as the actual number of rocks thrown, the probability that any given pot is occupied is exactly:\n\nP = 1 - ((N-1)/N)^r\n\nIf these events were independent, the probability that all pots are occupied would be:\n\nP^N = [1-((N-1)/N)^r]^N = .95\n\n==>\n\nr = ln(1- .95^(1/N))/(ln(N-1)-ln(N))\n\nFor N=3 this gives r = 10 which is a bit larger than the exact value, but it's probably more accurate for larger N.
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P (X>=5) \n= 1 - [  P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)] \n\nRandam sample of 100 days from the past 10 yrs... \nC(360*10, 100)
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Here are the formulas:\nIf P(E) = a/b, then the odds for E are a to (b-a)\nTo find the odds against E just reverse the order.\nOdds against E are (b-a) to a\n\nIf the odds of E are a to b, then P(E) = a/(a+b)\n\nFor example,\nif P(R) = 1/10, then the odds for R are\n1 to (10-1) or\n1 to 9.  \nThus, the odds against R are 9 to 1.\n\nAnother example,  say the odds of R are 5 to 3.  \nThe probability of R is 5/(5+3) = 5/8.\nThe probability of not R would then be \n1 - 5/8\n= 3/8
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The binomial distribution is given by:\n\n    N!\n----------- x (p)^n x (1-p)^(N-n)\nn!(N-n!)\n\nWhere N is the number of trials, n is the number of successes and p is the probability of success\n\nIn this question, n, the "number of successes" is the number of failures caused by operator error. p, the "probability of success" is the probability that the failure is caused by operator error (i.e. 30%).\n\nSo, in the questions you need to substitute the following for each variable:\n\ni) N=20, p=0.3\n\nAs this question uses the phrase "at least", you will need to use the equation more than once and add all the results. Repeat the calucation for each n for n=10 to n=20 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=10, 0.03081708\nn=11, 0.01200665\nn=12, 0.00385928\nn=13, 0.00101783\nn=14, 0.00021811\nn=15, 0.00003739\nn=16, 0.00000501\nn=17, 0.0000005\nn=18, 0.00000004\nn=19, 0.00000000\nn=20, 0.00000000\n\nTotal: 0.0479619 or 4.80% chance that at least 10 are caused by operator error\n\nii) N=20, p=0.3\n\nThis question is similar to q1, however you need to repeat the calculation for each n for n=0 to n=4 and add the results.\n\nI got the following values for each n (rounded to 8 d.p)\n\nn=0, 0.00079792\nn=1, 0.00683934\nn=2, 0.02784587\nn=3, 0.07160367\nn=4, 0.13042097\n\nTotal: 0.23750778 or 23.75% chance that no more than 4 are caused by operator error\n\niii) In the random sample, only 5 are errors. Work out the probability of this being the case using N=20, n=5, p=0.3. If it seems reasonable, then the 30% figure is probably right, and the sample just contains a low number of failures due to operation error. If the probability is impossibly small, then the 30% is not correct.\n\nThe probability I calculated for n=5 was 0.178863051. This indicates a 17.89% chance that there would be 5 errors in the sample. This is not so outrageous to suggest that 30% is an incorrect percentage of operator errors. It's just that in this sample, the number of actual errors was less than the number of expected errors (which is 30% x 20, or 6 errors).
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I've found that the Wikipedia articles on probability distributions are top notch and contain enough information for you to solve this problem on your own.\n\nHint: solve the algebraic expressions that relate mean and variance so you can express p or n as algebraic relationship of one another and then do some substitions.\n\nAt first I thought you needed to know p or n up-front however I realize that you have all the info you need to solve this problem algebraically.\n\nGood luck with your homework.
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try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
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P(Last is W) = P{(2nd is W and 1st is W) or (2nd is B and 1st is W)}\n= P(2nd is W and 1st is W) + P(2nd is B and 1st is W) (since these are mutually exclusive events)\n= P(2nd is W)P(1st is W|2nd is W) + P(2nd is B)P(1st is W|2nd is B) (both use the general multiplication rule)\n= (6/10)(7/11) + (4/10)(6/11)\n= (42+24)/110\n= 66/110\n= 0.6\n\nWhich is the probability of getting a white from either box, which I think makes sense, since both have the same proportion of whites.
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try this site...\n\nhttp://www.futureaccountant.com/probability/