Get Binomial Distribution essential facts below. View Videos or join the Binomial Distribution discussion. Add Binomial Distribution to your PopFlock.com topic list for future reference or share this resource on social media.
Probability mass function
Cumulative distribution function
– number of trials – success probability for each trial
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.
Probability mass function
In general, if the random variableX follows the binomial distribution with parameters n?N and p ? [0,1], we write X ~ B(n, p). The probability of getting exactly k successes in n independent Bernoulli 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(k, n, p) 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
f(k, n, p) 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.
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. The probability of seeing exactly 4 heads in 6 tosses is
Expected value and variance
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:
This follows from the linearity of the expected value along with fact that X is the sum of n identical Bernoulli random variables, each with expected value p. In other words, if are identical (and independent) Bernoulli random variables with parameter p, then and
This similarly follows from the fact that the variance of a sum of independent random variables is the sum of the variances.
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:
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.
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.
Any median m must lie within the interval ?np? m np?.
A median m cannot lie too far away from the mean: }.
The median is unique and equal to m = round(np) when |m - np| p, 1 - p} (except for the case when p = and n is odd).
When p = 1/2 and n is odd, any number m in the interval (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.
For k np, upper bounds can be derived for the lower tail of the cumulative distribution function , the probability that there are at most k successes. Since , these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for k >= np.
When estimating p with very rare events and a small n (e.g.: if x=0), then using the standard estimator leads to which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators. One way is to use the Bayes estimator, leading to: ). Another method is to use the upper bound of the confidence interval obtained using the rule of three: )
Even for quite large values of n, the actual distribution of the mean is significantly nonnormal. 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
The notation in the formula below differs from the previous formulas in two respects:
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 .
This result was first derived by Katz et al. in 1978.
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.
If X ~ B(n, p) and Y | X ~ B(X, q) (the conditional distribution of Y, given X), then Y is a simple binomial random variable with distribution Y ~ B(n, pq).
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(n, p) 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(X, q) and therefore Y ~ B(n, pq).
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.
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(n, p), is the distribution of the sum of nBernoulli trials, Bernoulli(p), each with the same probability p.
If n is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to B(n, p) 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. 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 is that for 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 X. If Y has a distribution given by the normal approximation, then Pr(X Y
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
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 ? = 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 n >= 100 and np
Concerning the accuracy of Poisson approximation, see Novak, ch. 4, and references therein.
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.
Mathematically, when ? = k + 1 and ? = n − k + 1, the beta distribution and the binomial distribution are related by a factor of n + 1:
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.
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.
^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.
^ abR. Arratia and L. Gordon: Tutorial on large deviations for the binomial distribution, Bulletin of Mathematical Biology 51(1) (1989), 125-131 .
^Robert B. Ash (1990). Information Theory. Dover Publications. p. 115.
^Matou?ek, J, Vondrak, J: The Probabilistic Method (lecture notes) .
^Razzaghi, Mehdi. "On the estimation of binomial success probability with zero occurrence in sample." Journal of Modern Applied Statistical Methods 1.2 (2002): 41. url