Generator.binomial(self, n, p, size=None)

Draws samples from a binomial distribution.

Samples are drawn from a binomial distribution with specified parameters, n trials and p probability of success where n an integer >= 0 and p is in the interval [0,1]. (n may be input as a float, but it is truncated to an integer in use)


Parameter of the distribution, >= 0. Floats are also accepted, but they will be truncated to integers.


Parameter of the distribution, >= 0 and <=1.

sizeint or tuple of ints, optional

Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn.


Drawn samples from the parameterized binomial distribution, where each sample is equal to the number of successes over the n trials.


The probability density for the binomial distribution is

P(N) = \binom{n}{N}p^N(1-p)^{n-N},

where n is the number of trials, p is the probability of success, and N is the number of successes.

When estimating the standard error of a proportion in a population by using a random sample, the normal distribution works well unless the product p*n <=5, where p = population proportion estimate, and n = number of samples, in which case the binomial distribution is used instead.

For example, a sample of 15 people shows 4 who are left handed, and 11 who are right handed. Then p = 4/15 = 27%. 0.27*15 = 4, so the binomial distribution should be used in this case.


  • If n is neither a scalar nor None : NotImplementedError occurs.

  • If p is neither a scalar nor None : NotImplementedError occurs.


Draw samples from the distribution:

>>> import nlcpy as vp
>>> rng = vp.random.default_rng()
>>> n, p = 10, .5  # number of trials, probability of each trial
>>> s = rng.binomial(n, p, 1000)

A real world example. A company drills 9 wild-cat oil exploration wells, each with an estimated probability of success of 0.1. All nine wells fail. What is the probability of that happening?

Let's do 20,000 trials of the model, and count the number that generate zero positive results.

>>> sum(rng.binomial(9, 0.1, 20000) == 0)/20000.  
array(0.38625)  #  or 38%.