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Binomial distribution

From Wikipedia, the free encyclopedia

Binomial distribution
Probability mass function
Probability mass function for the binomial distribution
Cumulative distribution function
Cumulative distribution function for the binomial distribution
Parameters – number of trials
– success probability for each trial
Support – number of successes
Median or
Mode or
Ex. kurtosis
in shannons. For nats, use the natural log in the log.
Fisher information
(for fixed )
Binomial distribution for  p = 0.5 {\displaystyle p=0.5} with n and k as in Pascal's triangleThe probability that a ball in a Galton box with 8 layers (n = 8) ends up in the central bin (k = 4) is  70 / 256 {\displaystyle 70/256} .
Binomial distribution for
with n and k as in Pascal's triangle

The probability that a ball in a Galton box with 8 layers (n = 8) ends up in the central bin (k = 4) is .

In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q = 1 − p). A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.

The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution remains a good approximation, and is widely used.

YouTube Encyclopedic

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  • ✪ Binomial distribution | Probability and Statistics | Khan Academy
  • ✪ Finding The Probability of a Binomial Distribution Plus Mean & Standard Deviation
  • ✪ The Binomial Distribution: Crash Course Statistics #15
  • ✪ The Binomial Distribution / Binomial Probability Function
  • ✪ Lesson 11 - Binomial Distribution (Probability Tutor)


- [Voiceover] Let's define a random variable x as being equal to the number of heads, I'll just write capital H for short, the number of heads from flipping coin, from flipping a fair coin, we're gonna assume it's a fair coin, from flipping coin five times. Five times. Like all random variables this is taking particular outcomes and converting them into numbers. And this random variable, it could take on the value x equals zero, one, two, three, four or five. And I what want to do is figure out what's the probability that this random variable takes on zero, can be one, can be two, can be three, can be four, can be five. To do that, first let's think about how many possible outcomes are there from flipping a fair coin five times. Let's think about this. Let's write possible outcomes. Possible outcomes from five flips. From five flips. These aren't the possible outcomes for the random variable, this is literally the number of possible outcomes from flipping a coin five times. For example, one possible outcome could be tails, heads, tails, heads, tails. Another possible outcome could be heads, heads, heads, tails, tails. That is one of the equally likely outcomes, that's another one of the equally likely outcomes. How many of these are there? For each flip you have two possibilities. Let's write this down. Let me... The first flip, the first flip there's two possibilities, times two for the second flip, times two for the third flip. Actually maybe we'll not use the time notation, you might get confused with the random variable. Two possibilities for the first flip, two possibilities for the second flip, two possibilities for the third flip, two possibilities for the fourth flip, and then two possibilities for the fifth flip, or two to the fifth equally likely possibilities from flipping a coin five times, which is, of course, equal to 32. This is going to be helpful because for each of the values that the random variable can take on, we just have to think about how many of these equally likely possibilities would result in the random variable taking on that value. Let's just delve into it to see what we're actually talking about. I'll do it in this light, let me do it in... I'll start in blue. Let's think about the probability that our random variable x is equal to one. Well actually, let me start with zero. The probability that our random variable x is equal to zero. That would mean that you got no heads out of the five flips. Well there's only one way, one out of the 32 equally likely possibilities, that you get no heads. That's the one where you just get five tails. So this is just going to be, this is going to be equal to one out of the 32 equally likely possibilities. Now, for this case, to think in terms of binomial coefficients, and combinatorics, and all of that, it's much easier to just reason through it, but just so we can think in terms it'll be more useful as we go into higher values for our random variable. This is all buildup for the binomial distribution, so you get a sense of where the name comes from. So let's write it in those terms. This one, this one, this one right over here, one way to think about that in combinatorics is that you had five flips and you're choosing zero of them to be heads. Five flips and you're choosing zero of them to be heads. Let's verify that five choose zero is indeed one. So five choose zero. Write it over here. Five choose zero is equal to five factorial over, over five minus zero factorial. Well actually over zero factorial times five minus zero factorial. Well zero factorial is one, by definition, so this is going to be five factorial, over five factorial, which is going to be equal to one. Once again I like reasoning through it instead of blindly applying a formula, but I just wanted to show you that these two ideas are consistent. Let's keep going. I'm going to do x equals one all the way up to x equals five. If you are inspired, and I encourage you to be inspired, try to fill out the whole thing, what's the probability that x equals one, two, three, four or five. So let's go to the probability that x equals two. Or sorry, that x equals one. The probability that x equals one is going to be equal to... Well how do you get one head? It could be, the first one could be head and then the rest of them are gonna be tails. The second one could be head and then the rest of them are gonna be tails. I could write them all out but you can see that there's five different places to have that one head. So five out of the 32 equally likely outcomes involve one head. Let me write that down. This is going to be equal to five out of 32 equally likely outcomes. Which of course is the same thing, this is going to be the same thing as saying I got five flips, and I'm choosing one of them to be heads. So that over 32. You could verify that five factorial over one factorial times five minus-- Actually let me just do it just so that you don't have to take my word for it. So five choose one is equal to five factorial over one factorial, which is just one, times five minus four-- Sorry, five minus one factorial. Which is equal to five factorial over four factorial, which is just going to be equal to five. All right, we're making good progress. Now in purple let's think about the probability that our random variable x is equal to two. Well this is going to be equal to, and now I'll actually resort to the combinatorics. You have five flips and you're choosing two of them to be heads. Over 32 equally likely possibilities. This is the number of possibilities that result in two heads. Two of the five flips have chosen to be heads, I guess you can think of it that way, by the random gods, or whatever you want to say. This is the fraction of the 32 equally likely possibilities, so this is the probability that x equals two. What's this going to be? I'll do it right over here. And actually no reason for me to have to keep switching colors. So five choose two is going to be equal to five factorial over two factorial times five minus two factorial. Five minus two factorial. So this is five factorial over two factorial times three factorial. And this is going to be equal to five times four times three times two, I could write times one but that doesn't really do anything for us. Then two factorial's just going to be two. Then the three factorial is three times two. I could write times one, but once again doesn't do anything for us. That cancels with that. Four divided by two is two. Five times two is 10. So this is equal to 10. This right over here is equal to 10/32. 10/32. And obviously we could simplify this fraction, but I like to leave it this way because we're now thinking everything is in terms of 32nds. There's a 1/32 chance x equals zero, 5/32 chance that x equals one and a 10/32 chance that x equals two. Let's keep on going. I'll go in orange. What is the probability that our random variable x is equal to three? Well this is going to be five, out of the five flips we're going to need to choose three of them to be heads to figure out which of the possibilities involve exactly three heads. And this is over 32 equally likely possibilities. And this is going to be equal to, five choose three is equal to five factorial over three factorial times five minus three factorial. Let me just write it down. Five minus three factorial, which is equal to five factorial over three factorial times two factorial. That's exactly what we had up here and we just swapped three and the two, so this also is going to be equal to 10. So this is also going to be equal to 10/32. All right, two more to go. And I think you're going to start seeing a little bit of a symmetry here. One, five, 10, 10, let's keep going. Let's keep going, and I haven't used white yet. Maybe I'll use white. The probability that our random variable x is equal to four. Well, out of our five flips we want to select four of them to be heads, or out of the five-- We're obviously not actively selecting. One way to think of it, we want to figure out the possibilities that involve out of the five flips, four of them are chosen to be heads, or four of them are heads. And this is over 32 equally likely possibilities. So five choose four is equal to five factorial over four factorial times five minus four factorial which is equal to, well that's just going to be five factorial, this is going to be one factorial right over here. That doesn't change the value, you just multiply one factorial times four factorial, so it's five factorial over four factorial, which is equal to five. So once again this is 5/32. And you could have reasoned through this because if you're saying you want five heads, that means you have one tail. There's five different places you could put that one tail. There are five possibilities with one tail. Five of the 32 equally likely. And then, and you could probably guess what we're gonna get for x equals five because having five heads means you have zero tails, and there's only gonna be one possibility out of the 32 with zero tails, where you have all heads. Let's write that down. The probability, the probability that our random variable x is equal to five. So we have all five heads. You could say this is five and we're choosing five of them to be heads. Out of the 32 equally likely possibilities. Well five choose five, that's going to be... Let me just write it here since I've done it for all of the other ones. Five choose five is five factorial over five factorial times five minus five factorial. Well this right over here is zero factorial, which is equal to one, so this whole thing simplifies to one. This is going to be one out of-- 1/32. So you see the symmetry. 1/32, 1/32. 5/32, 5/32; 10/32, 10/32. And that makes sense because the probability of getting five heads is the same as the probability of getting zero tails, and the probability of getting zero tails should be the same as the probability of getting zero heads. I'll leave you there for this video. In the next video we'll graphically represent this and we'll see the probability distribution for this random variable.



Probability mass function

In general, if the random variable X follows the binomial distribution with parameters n and p ∈ [0,1], we write X ~ B(np). The probability of getting exactly k successes in n trials is given by the probability mass function:

for k = 0, 1, 2, ..., n, where

is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows. k successes occur with probability pk and n − k failures occur with probability (1 − p)n − k. However, the k successes can occur anywhere among the n trials, and there are different ways of distributing k successes in a sequence of n trials.

In creating reference tables for binomial distribution probability, usually the table is filled in up to n/2 values. This is because for k > n/2, the probability can be calculated by its complement as

Looking at the expression f(knp) as a function of k, there is a k value that maximizes it. This k value can be found by calculating

and comparing it to 1. There is always an integer M that satisfies[1]

f(knp) is monotone increasing for k < M and monotone decreasing for k > M, with the exception of the case where (n + 1)p is an integer. In this case, there are two values for which f is maximal: (n + 1)p and (n + 1)p − 1. M is the most probable outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the mode.

Cumulative distribution function

The cumulative distribution function can be expressed as:

where is the "floor" under k, i.e. the greatest integer less than or equal to k.

It can also be represented in terms of the regularized incomplete beta function, as follows:[2]

Some closed-form bounds for the cumulative distribution function are given below.


Suppose a biased coin comes up heads with probability 0.3 when tossed. What is the probability of achieving 0, 1,..., 6 heads after six tosses?



If X ~ B(n, p), that is, X is a binomially distributed random variable, n being the total number of experiments and p the probability of each experiment yielding a successful result, then the expected value of X is:[4]

For example, if n = 100, and p = 1/4, then the average number of successful results will be 25.

Proof: We calculate the mean, μ, directly calculated from its definition

and the binomial theorem:

It is also possible to deduce the mean from the equation whereby all are Bernoulli distributed random variables with ( if the ith experiment succeeds and otherwise). We get:


The variance is:

Proof: Let where all are independently Bernoulli distributed random variables. Since , we get:

Higher moments

The first 6 central moments are given by


Usually the mode of a binomial B(n, p) distribution is equal to , where is the floor function. However, when (n + 1)p is an integer and p is neither 0 nor 1, then the distribution has two modes: (n + 1)p and (n + 1)p − 1. When p is equal to 0 or 1, the mode will be 0 and n correspondingly. These cases can be summarized as follows:

Proof: Let

For only has a nonzero value with . For we find and for . This proves that the mode is 0 for and for .

Let . We find


From this follows

So when is an integer, then and is a mode. In the case that , then only is a mode.[5]


In general, there is no single formula to find the median for a binomial distribution, and it may even be non-unique. However several special results have been established:

  • If np is an integer, then the mean, median, and mode coincide and equal np.[6][7]
  • Any median m must lie within the interval ⌊np⌋ ≤ m ≤ ⌈np⌉.[8]
  • A median m cannot lie too far away from the mean: |mnp| ≤ min{ ln 2, max{p, 1 − p} }.[9]
  • The median is unique and equal to m = round(np) when |m − np| ≤ min{p, 1 − p} (except for the case when p = 1/2 and n is odd).[8]
  • When p = 1/2 and n is odd, any number m in the interval 1/2(n − 1) ≤ m ≤ 1/2(n + 1) is a median of the binomial distribution. If p = 1/2 and n is even, then m = n/2 is the unique median.

Covariance between two binomials

If two binomially distributed random variables X and Y are observed together, estimating their covariance can be useful. The covariance is

In the case n = 1 (the case of Bernoulli trials) XY is non-zero only when both X and Y are one, and μX and μY are equal to the two probabilities. Defining pB as the probability of both happening at the same time, this gives

and for n independent pairwise trials

If X and Y are the same variable, this reduces to the variance formula given above.

Related distributions

Sums of binomials

If X ~ B(np) and Y ~ B(mp) are independent binomial variables with the same probability p, then X + Y is again a binomial variable; its distribution is Z=X+Y ~ B(n+mp):

However, if X and Y do not have the same probability p, then the variance of the sum will be smaller than the variance of a binomial variable distributed as

Ratio of two binomial distributions

This result was first derived by Katz et al. in 1978.[10]

Let X ~ B(n,p1) and Y ~ B(m,p2) be independent. Let T = (X/n)/(Y/m).

Then log(T) is approximately normally distributed with mean log(p1/p2) and variance ((1/p1) − 1)/n + ((1/p2) − 1)/m.

Conditional binomials

If X ~ B(np) and Y | X ~ B(Xq) (the conditional distribution of Y, given X), then Y is a simple binomial random variable with distribution Y ~ B(npq).

For example, imagine throwing n balls to a basket UX and taking the balls that hit and throwing them to another basket UY. If p is the probability to hit UX then X ~ B(np) is the number of balls that hit UX. If q is the probability to hit UY then the number of balls that hit UY is Y ~ B(Xq) and therefore Y ~ B(npq).


Since and , by the law of total probability,

Since the equation above can be expressed as

Factoring and pulling all the terms that don't depend on out of the sum now yields

After substituting in the expression above, we get

Notice that the sum (in the parentheses) above equals by the binomial theorem. Substituting this in finally yields

and thus as desired.

Bernoulli distribution

The Bernoulli distribution is a special case of the binomial distribution, where n = 1. Symbolically, X ~ B(1, p) has the same meaning as X ~ Bernoulli(p). Conversely, any binomial distribution, B(np), is the distribution of the sum of n Bernoulli trials, Bernoulli(p), each with the same probability p.[11]

Poisson binomial distribution

The binomial distribution is a special case of the Poisson binomial distribution, or general binomial distribution, which is the distribution of a sum of n independent non-identical Bernoulli trials B(pi).[12]

Normal approximation

Binomial probability mass function and normal probability density function approximation for n = 6 and p = 0.5
Binomial probability mass function and normal probability density function approximation for n = 6 and p = 0.5

If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(np) is given by the normal distribution

and this basic approximation can be improved in a simple way by using a suitable continuity correction. The basic approximation generally improves as n increases (at least 20) and is better when p is not near to 0 or 1.[13] Various rules of thumb may be used to decide whether n is large enough, and p is far enough from the extremes of zero or one:

  • One rule[13] is that for n > 5 the normal approximation is adequate if the absolute value of the skewness is strictly less than 1/3; that is, if
  • A stronger rule states that the normal approximation is appropriate only if everything within 3 standard deviations of its mean is within the range of possible values; that is, only if
This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above.

The rule is totally equivalent to request that

Moving terms around yields:

Since , we can apply the square power and divide by the respective factors and , to obtain the desired conditions:

Notice that these conditions automatically imply that . On the other hand, apply again the square root and divide by 3,

Subtracting the second set of inequalities from the first one yields:

and so, the desired first rule is satisfied,

  • Another commonly used rule is that both values and must be greater than or equal to 5. However, the specific number varies from source to source, and depends on how good an approximation one wants. In particular, if one uses 9 instead of 5, the rule implies the results stated in the previous paragraphs.

Assume that both values and are greater than 9. Since , we easily have that

We only have to divide now by the respective factors and , to deduce the alternative form of the 3-standard-deviation rule:

The following is an example of applying a continuity correction. Suppose one wishes to calculate Pr(X ≤ 8) for a binomial random variable X. If Y has a distribution given by the normal approximation, then Pr(X ≤ 8) is approximated by Pr(Y ≤ 8.5). The addition of 0.5 is the continuity correction; the uncorrected normal approximation gives considerably less accurate results.

This approximation, known as de Moivre–Laplace theorem, is a huge time-saver when undertaking calculations by hand (exact calculations with large n are very onerous); historically, it was the first use of the normal distribution, introduced in Abraham de Moivre's book The Doctrine of Chances in 1738. Nowadays, it can be seen as a consequence of the central limit theorem since B(np) is a sum of n independent, identically distributed Bernoulli variables with parameter p. This fact is the basis of a hypothesis test, a "proportion z-test", for the value of p using x/n, the sample proportion and estimator of p, in a common test statistic.[14]

For example, suppose one randomly samples n people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of n people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion p of agreement in the population and with standard deviation

Poisson approximation

The binomial distribution converges towards the Poisson distribution as the number of trials goes to infinity while the product np remains fixed or at least p tends to zero. Therefore, the Poisson distribution with parameter λ = np can be used as an approximation to B(n, p) of the binomial distribution if n is sufficiently large and p is sufficiently small. According to two rules of thumb, this approximation is good if n ≥ 20 and p ≤ 0.05, or if n ≥ 100 and np ≤ 10.[15]

Concerning the accuracy of Poisson approximation, see Novak,[16] ch. 4, and references therein.

Limiting distributions

approaches the normal distribution with expected value 0 and variance 1.[citation needed] This result is sometimes loosely stated by saying that the distribution of X is asymptotically normal with expected value np and variance np(1 − p). This result is a specific case of the central limit theorem.

Beta distribution

The binomial distribution and beta distribution are different views of the same model of repeated Bernoulli trials. The binomial distribution is the PMF of k successes given n independent events each with a probability p of success. The beta distribution is the PDF for the probability of success p given n independent events with k observed successes. [17] Mathematically, when 𝛼=𝑘+1 and 𝛽=𝑛−𝑘+1, the beta distribution and the binomial distribution are related by a factor of n+1:

Beta distributions also provide a family of prior probability distributions for binomial distributions in Bayesian inference:[18]

Confidence intervals

Even for quite large values of n, the actual distribution of the mean is significantly nonnormal.[19] Because of this problem several methods to estimate confidence intervals have been proposed.

In the equations for confidence intervals below, the variables have the following meaning:

  • n1 is the number of successes out of n, the total number of trials
  • is the proportion of successes
  • is the quantile of a standard normal distribution (i.e., probit) corresponding to the target error rate . For example, for a 95% confidence level the error  = 0.05, so  = 0.975 and  = 1.96.

Wald method

A continuity correction of 0.5/n may be added.[clarification needed]

Agresti–Coull method


Here the estimate of p is modified to

Arcsine method


Wilson (score) method

The notation in the formula below differs from the previous formulas in two respects:[22]

  • Firstly, zx has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the xth quantile of the standard normal distribution', rather than being a shorthand for 'the (1 − x)-th quantile'.
  • Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use to get the lower bound, or use to get the upper bound. For example: for a 95% confidence level the error  = 0.05, so one gets the lower bound by using , and one gets the upper bound by using .


The exact (Clopper–Pearson) method is the most conservative.[19]

The Wald method, although commonly recommended in textbooks, is the most biased.[clarification needed]

Generating binomial random variates

Methods for random number generation where the marginal distribution is a binomial distribution are well-established.[24][25]

One way to generate random samples from a binomial distribution is to use an inversion algorithm. To do so, one must calculate the probability that Pr(X = k) for all values k from 0 through n. (These probabilities should sum to a value close to one, in order to encompass the entire sample space.) Then by using a pseudorandom number generator to generate samples uniformly between 0 and 1, one can transform the calculated samples into discrete numbers by using the probabilities calculated in the first step.

Tail bounds

For knp, upper bounds for the lower tail of the distribution function can be derived. Recall that , the probability that there are at most k successes.

Hoeffding's inequality yields the bound

and Chernoff's inequality can be used to derive the bound

Moreover, these bounds are reasonably tight when p = 1/2, since the following expression holds for all k ≥ 3n/8[26]

However, the bounds do not work well for extreme values of p. In particular, as p 1, value F(k;n,p) goes to zero (for fixed k, n with k < n) while the upper bound above goes to a positive constant. In this case a better bound is given by [27]

where D(a || p) is the relative entropy between an a-coin and a p-coin (i.e. between the Bernoulli(a) and Bernoulli(p) distribution):

Asymptotically, this bound is reasonably tight; see [27] for details. An equivalent formulation of the bound is

Both these bounds are derived directly from the Chernoff bound. It can also be shown that,

This is proved using the method of types (see for example chapter 11 of Elements of Information Theory by Cover and Thomas [28]).

We can also change the in the denominator to , by approximating the binomial coefficient with Stirling's formula.[29]


This distribution was derived by James Bernoulli. He considered the case where p = r/(r + s) where p is the probability of success and r and s are positive integers. Blaise Pascal had earlier considered the case where p = 1/2.

See also


  1. ^ Feller, W. (1968). An Introduction to Probability Theory and Its Applications (Third ed.). New York: Wiley. p. 151 (theorem in section VI.3).
  2. ^ Wadsworth, G. P. (1960). Introduction to Probability and Random Variables. New York: McGraw-Hill. p. 52.
  3. ^ Hamilton Institute. "The Binomial Distribution" October 20, 2010.
  4. ^ See Proof Wiki
  5. ^ See also Nicolas, André (January 7, 2019). "Finding mode in Binomial distribution". Stack Exchange.
  6. ^ Neumann, P. (1966). "Über den Median der Binomial- and Poissonverteilung". Wissenschaftliche Zeitschrift der Technischen Universität Dresden (in German). 19: 29–33.
  7. ^ Lord, Nick. (July 2010). "Binomial averages when the mean is an integer", The Mathematical Gazette 94, 331-332.
  8. ^ a b Kaas, R.; Buhrman, J.M. (1980). "Mean, Median and Mode in Binomial Distributions". Statistica Neerlandica. 34 (1): 13–18. doi:10.1111/j.1467-9574.1980.tb00681.x.
  9. ^ Hamza, K. (1995). "The smallest uniform upper bound on the distance between the mean and the median of the binomial and Poisson distributions". Statistics & Probability Letters. 23: 21–25. doi:10.1016/0167-7152(94)00090-U.
  10. ^ Katz D. et al.(1978) "Obtaining confidence intervals for the risk ratio in cohort studies". Biometrics 34:469–474
  11. ^ Taboga, Marco. "Lectures on Probability Theory and Mathematical Statistics". Retrieved 18 December 2017.
  12. ^ Wang, Y. H. (1993). "On the number of successes in independent trials" (PDF). Statistica Sinica. 3 (2): 295–312. Archived from the original (PDF) on 2016-03-03.
  13. ^ a b Box, Hunter and Hunter (1978). Statistics for experimenters. Wiley. p. 130.
  14. ^ NIST/SEMATECH, "7.2.4. Does the proportion of defectives meet requirements?" e-Handbook of Statistical Methods.
  15. ^ a b NIST/SEMATECH, " Counts Control Charts", e-Handbook of Statistical Methods.
  16. ^ Novak S.Y. (2011) Extreme value methods with applications to finance. London: CRC/ Chapman & Hall/Taylor & Francis. ISBN 9781-43983-5746.
  17. ^
  18. ^ MacKay, David (2003). Information Theory, Inference and Learning Algorithms. Cambridge University Press; First Edition. ISBN 978-0521642989.
  19. ^ a b Brown, Lawrence D.; Cai, T. Tony; DasGupta, Anirban (2001), "Interval Estimation for a Binomial Proportion", Statistical Science, 16 (2): 101–133, CiteSeerX, doi:10.1214/ss/1009213286, retrieved 2015-01-05
  20. ^ Agresti, Alan; Coull, Brent A. (May 1998), "Approximate is better than 'exact' for interval estimation of binomial proportions" (PDF), The American Statistician, 52 (2): 119–126, doi:10.2307/2685469, JSTOR 2685469, retrieved 2015-01-05
  21. ^ Pires M. A. Confidence intervals for a binomial proportion: comparison of methods and software evaluation.
  22. ^ Wilson, Edwin B. (June 1927), "Probable inference, the law of succession, and statistical inference" (PDF), Journal of the American Statistical Association, 22 (158): 209–212, doi:10.2307/2276774, JSTOR 2276774, archived from the original (PDF) on 2015-01-13, retrieved 2015-01-05
  23. ^ "Confidence intervals". Engineering Statistics Handbook. NIST/Sematech. 2012. Retrieved 2017-07-23.
  24. ^ Devroye, Luc (1986) Non-Uniform Random Variate Generation, New York: Springer-Verlag. (See especially Chapter X, Discrete Univariate Distributions)
  25. ^ Kachitvichyanukul, V.; Schmeiser, B. W. (1988). "Binomial random variate generation". Communications of the ACM. 31 (2): 216–222. doi:10.1145/42372.42381.
  26. ^ Matoušek, J, Vondrak, J: The Probabilistic Method (lecture notes) [1].
  27. ^ a b R. Arratia and L. Gordon: Tutorial on large deviations for the binomial distribution, Bulletin of Mathematical Biology 51(1) (1989), 125–131 [2].
  28. ^ Theorem 11.1.3 in Cover, T.; Thomas, J. (2006). Elements of Information Theory (2nd ed.). Wiley. p. 350. ISBN 9781118585771.
  29. ^ "Sharper Lower Bounds for Binomial/Chernoff Tails". Stack Exchange. December 7, 2015.
  30. ^ Mandelbrot, B. B., Fisher, A. J., & Calvet, L. E. (1997). A multifractal model of asset returns. 3.2 The Binomial Measure is the Simplest Example of a Multifractal

Further reading

External links

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