Binomial distribution formula?
953152
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).
764469
X follows a Binomial distribution with n = 10 and probability of success p = 1/6
371621
The Binomial and Poisson probability distributions are both discrete probability distributions for modelling events.\n\nThe Poisson is useful for modeling arrivals per unit time. If you are running a customer support call center, a fast-food restaurant, an amusement park, or the toilet facilities at a venue, you might try to use the Poisson to estimate the rate at which people arrive in 10 minutes, or in an hour.  The Poisson tells you that while the average rate might be 20 per hour, the peak arrival rate for the 99th percentile might actually be as high as 31 per hour.  For capacity planning purposes, you might actually add more servers or more toilets so that people have to wait less time.\n\nThe Binomial can be used to model probability that someone does or does not do something.  Let's say 3 out of 10 people in one trial respond yes in a consumer survey.  Now let's say another survey has 4 out of 10 people say yes.  Does this mean that people were more likely to say use in the second survey?  The answer is probably not.  The Binomial allows you to model the the probability that a larger sample or population will say yes given the mean behavior you've measured.  This will allow you to access whether there is a true difference or if the difference is in the expected random variation.\n\nThe use of these distributions and others is often called Operations Research (OR) in business and military settings.\nDoing a Yahoo Search for operations research binomial may help give you more applications:\nhttp://search.yahoo.com/search?ei=utf-8&fr=slv1-&p=binomial%20operations%20research
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
1180637
I was answering another similar question, and I came across a website that has a downloadable sample spreadsheet for doing statistical distributions.  Play with this and see if it helps you out.
982580
Use binomial coefficients to solve this.\n\n| n |           n!\n|    |   =  ----------\n| k |        (n-k)!k!\n\nn is the number of possibilities and k is the number of unordered outcomes.\n\nCombination of six numbers from ten possibilities.\n\n| 10 |       10!          3628800         \n|      | = ------------ = --------------- = 210\n|   6 |    (10-6)!6!     (24)(720)\n\nCombination of four numbers from ten possibilities.\n\n| 10 |\n|      |\n|   4 |\n\nYou should be able to figure out the second one on your own.\nHere's a link on factorials, just in case you don't know what they are.\n\nhttp://mathworld.wolfram.com/Factorial.html
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
92833
The Poisson distribution is commonly applied to a count of events happening within some time interval or spatial area.  Examples include the number of accidents happening on a certain stretch of road in a month, the number of purchases of a given product in a store in a day, the number of flaws in a bolt of fabric, the number of weeds in a 100 square foot plot of land.\n\nThe binomial distribution describes the number of events that occur out of a fixed number of trials.  Examples include, count the number of heads in 100 coin flips, the number of correct answers when someone guesses on every question of a multiple choice test, etc.\n\nThe difference between them is that in the Poisson applications, we don't really know how many events are possible.  Practically speaking, there is an upper limit (such as, we can't sell more of a product than is on the shelf).  However, it's usually applied to cases where the events are pretty rare, in which case it's an excellent approximation to the real behavior.\n\nMeanwhile, the Binomial distribution assumes a fixed number of attempts or trials (each coin toss, or each test question).\n\nPossibly adding to the confusion is this fact.  A binomial distribution with a very large number of trials is practically indistinguishable from a Poisson distribution.  For example, the number of red-heads in a stadium full of people could either be modeled as a Poisson distribution, or as a Binomial.\n\nWhen the number of trials is very large, the Binomial distribution probabilities are difficult to compute.  In such cases, people tend to either approximate it with a Poisson distribution, or use the continuous Normal (bell curve) distribution.\n\nIf it's a count of fairly uncommon events occuring in a specific period of time or spatial area, then use the Poisson.  If it's a count of events out of a known, limited number of trials or opportunities, use the Binomial.\n\nI hope this helps.
341707
The RAND() function is not the appropriate way to solve this.\n\nThe number of cars can be determined a couple of ways.\n\n1.  One is to use NORMINV(.99,2600000,100955) = 2834856\n2.  Or you could use the standard normal distribution:\n\n    x= 2600000 + 100955 NORMSINV(.99)\n     = 2600000 + 100955(2.326348) = 2834856\n\nRAND() just gives you a random number between 0 and 1.  I don't see how that would help you in this problem.\n\nThe second problem is also easily solved using the the standard normal distribution:\n\nFirst, the 2.4 million is 200000 below the mean, \nso z=-200000/100955= -1.98108\n\n2.7 million is 100000 above the mean, so \nz=100000/100955=.99054\n\nSo the probability the number of cars will be between 2.4 and 2.7 million is\n\n= NORMSDIST(.99054) - NORMSDIST(.023791)\n=.839045 - .023791 = .815254
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.
584200
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
242656
This one is easy-assume a 50% chance each time, so its 1/2 X 1/2 X 1/2 X 1/2, so 1 chance in 16.  There is another way of figuring this, using combinations, like sn777 did, but sn777 made a logical error.  There are these possible combinations: 4 m, 3m 1F, 2m, 2F, 1m 3F, and 4F. So just looking at that, you would think its 25%. BUT, there is only one combination of 4M, while there are 4 combinations that make 3m 1F, and 6 combinations that make 2m and 2F, etc-so the total possible combinations of 4 children is 16, and 4M is only one of those.
304995
1{p+5}/u-6\n\nThe 1 can be distributed into the {p+5}\n\n{p+5}/u-6
460739
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
192872
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]
1017859
Well, the equation is : (p^2) + (2pq)+(q^2)= 1....where (p^2) and (q^2) represent the frequency of homozygous alleles and (pq+qp=2pq) represent the frequency of heterozygous alleles.  \n\nThe way to use this formula is for example : " if you have a plant population that consists of 84% plants with Red flowers (RR, Rr -alleles) and 16% White flowers ( rr-alleles)," you use the formula above as follows:\n\n(q^2) = .16= White flowered plants (rr trait)\n(p^2)+ (2pq)= .84= Red flowered plants (RR, or Rr traits)\n\n\n\nNow there is another equation : p+q=1\n\nSo, take the square root of (q^2)=.16, q=.4 and then plug that into p+q=1. Now solve for p,  p should be .6.  Now  note that q=.4 and p=.6 and with that said plug it into the original equation in order to find the frequency. \n\n(p^2) +(2pq)+(q^2)=1\n\n(p^2) = (.6) (.6) = .36  or 36%  for the homozygous dominant trait (RR)\n\n(q^2)= (.4)(.4)= .16 or 16% for the homozygous recessive trait- (rr)\n\n(2pq) = (2)(.6)(.4)= .48 or 48% for the heterozygous trait - (Rr).\n\nAll of this should equall 100%.  So 36%+16%+48%=100%.\n\nSo, basically the Hardy-Weinberg equation is to see the frequency of the alleles (RR,Rr, and rr) in a population.
729913
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.