To answer the previous question, the Hypergeometric distribution can be used with the following parameters: with N = 1000 (population size), K = 5 (number of success in population), n = 2 (sample size), and compute: 

• Happen exactly zero failures in inspection: P(X = 0). 

• More than zero failures in inspection: P(X >= 1) = 1 - P(X = 0). 

• What happens with an increase in the sample size ranging from 0 to 200?

Instead of employing completed equations, the answers to these two questions will be addressed through Python code.

from scipy.stats import hypergeom

import matplotlib.pyplot as plt

import numpy as np

# computing P(X) for each X.

x = [0]

N = 1000

K = 5

n = 2

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

print('  X  |',x)

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

   X | [0] 

P(X) | [0.99002002] 

from scipy.stats import hypergeom

import matplotlib.pyplot as plt

import numpy as np

# computing P(X) for each X.

x = [0]

N = 1000

K = 5

ns = list(range(0,201))

y = []

for n in ns:

  res = hypergeom(N, K, n).pmf(x)

  y.append(1-res)

plt.plot(ns,y,'-r')

plt.xlabel('n (Sample Size)')

plt.ylabel('Probability find issue (P(X>=1))')

plt.grid()

plt.show()