scipy.special.bdtr#

scipy.special.bdtr(k, n, p, out=None) = <ufunc 'bdtr'>#

Binomial distribution cumulative distribution function.

Sum of the terms 0 through floor(k) of the Binomial probability density.

\[\mathrm{bdtr}(k, n, p) = \sum_{j=0}^{\lfloor k \rfloor} {{n}\choose{j}} p^j (1-p)^{n-j}\]

Parameters:

karray_like: Number of successes (double), rounded down to the nearest integer.
narray_like: Number of events (int).
parray_like: Probability of success in a single event (float).
outndarray, optional: Optional output array for the function values

Returns:

yscalar or ndarray: Probability of floor(k) or fewer successes in n independent events with success probabilities of p.

Notes

The terms are not summed directly; instead the regularized incomplete beta function is employed, according to the formula,

\[\mathrm{bdtr}(k, n, p) = I_{1 - p}(n - \lfloor k \rfloor, \lfloor k \rfloor + 1).\]

Wrapper for the Cephes [1] routine bdtr.

Array API Standard Support

bdtr has experimental support for Python Array API Standard compatible backends in addition to NumPy. Please consider testing these features by setting an environment variable SCIPY_ARRAY_API=1 and providing CuPy, PyTorch, JAX, or Dask arrays as array arguments. The following combinations of backend and device (or other capability) are supported.

Library	CPU	GPU
NumPy	✅	n/a
CuPy	n/a	✅
PyTorch	✅	⛔
JAX	✅	⛔
Dask	✅	n/a

See Support for the array API standard for more information.

References

[1]

Cephes Mathematical Functions Library, https://netlib.org/cephes/

Examples

We have a coin for which the probability of showing heads when flipped is 0.525. The coin is flipped 16 times. What is the probability that the number of heads is less than or equal to 5?

Let X be the number of heads. We want to find the probability that X <= k, given k is 5, the total number of flips n is 16 and the probability p of heads for a single flip is 0.525. This is what bdtr(k, n, p) computes:

>>> from scipy.special import bdtr

>>> bdtr(5, 16, 0.525)
np.float64(0.07281293895810999)

The following plot shows the graph of bdtr(k, 16, 0.525):

>>> import numpy as np
>>> import matplotlib.pyplot as plt

>>> n = 16
>>> p = 0.525
>>> k = np.arange(n + 1)
>>> plt.plot(k, bdtr(k, n, p), 'o')
>>> plt.grid(True)
>>> plt.xlabel('k')
>>> plt.title(f"bdtr(k, {n}, {p})")
>>> plt.show()