Use previous equation to compute the probabilities with N = 100 (population size), K = 5 (number of success in population), n = 2 (sample size):

• Find exactly one issue:

p(k = 1) = (5·95)/(100·99/2) = (10·95)/(100·99) = 0.096

• Find at least one issue is to find: P(x ≥ 1). Observe that p(x = 0) + p(x = 1) = 1. Then p(x = 0) + P(x ≥ 1) = 1 leads to P(x ≥ 1) = 1 − p(x = 0). Now, using equation to compute p(k = 0):

p(k = 0) = (95·94/2)/(100·99/2) = (95·94)/(100·99) = 0.9020

Finally: P(x ≥ 1) = 1 − 0.9020 = 0.098.

from scipy.stats import hypergeom

import matplotlib.pyplot as plt

import numpy as np

#creating an array of values between

#0 to n+1 with a difference of 1

# K = Pop. size, K = success cases in pop., n = sample size.

[N, K, n] = [20, 7, 12]

x = np.arange(0, n+1)

y = hypergeom(N, K, n).pmf(x)

plt.plot(x,y,'r-',x, y,'bo')

plt.show()

from scipy.stats import hypergeom

import matplotlib.pyplot as plt

import numpy as np

# computing P(X) for each X.

x = [0, 1, 2]

N = 100

K = 5

n = 2

y = hypergeom(N, K, n).pmf(x)

print('  X  |',x)

print('P(X) |',y)

   X | [0, 1, 2] 

P(X) | [0.9020202 0.0959596 0.0020202]