Probability

 
| Experiment       | Outcomes    | Sample Space      |
|------------------|-------------|-------------------|
| Roll a die       | 1,2,3,4,5,6 | S = {1,2,3,4,5,6} |
| Tooss a coin once| Head, Tail  | S = {Head, Tail}  |

 
Simple event: For example, the head in flipping the coin - E = {heads}

Compount event: For example, the roll of a dice which events are less than 4 - E = {1,2,3,4}

 
import pandas as pd
from pandas import DataFrame

df = DataFrame({
    'in_favor': [15, 4], 
    'against': [45, 36]
}, index=['male', 'female'])

df_total = DataFrame({
    'in_favor': df['in_favor'].sum(), 
    'against': df['against'].sum()
},index=['total'])

df_table = pd.concat([df, df_total])
df_table['total'] = df_table.sum(axis=1)

#            against  in_favor  total
#    male         45        15     60
#    female       36         4     40
#    total        81        19    100

 
P(in_favor|male) = male_in_favor / total_males = 15 / 60 = .25

 
P(A|B) = P(A) or P(B|A) = P(B)

 
From the same table before what's the probability of a random selected person be male and in favor?

P(A) = 60 / 100 = .60
P(B|A) = P(in_favor|male) = 15 / 60 = .25
P(A and B) = P(A) * P(B|A) = .60 * .25 = .15

 
Again from the table above find the probability of the random selected person be a female OR against.

P(A) = 40 / 100 = .40
P(B) = 81 / 100 = .81

P(against|female) = 36 / 40 = 0.9
P(A and B) = .40 * .9 = 0.36
(.40 + .81) - 0.36 = 0.85

Or 40 (total_fem) + 45 (other against) = 85

 
For example, in a groups of 5000 adults, 3500 are in favor of stricter gun control law, 1200 are against such laws, and 300 have no opinion. A is the event that this adult is in favor of stricter gun control law. What is the complementary event of A?

P(A) = 3500/5000 = .70
~P(A) = 1 - .70 = .30

 
A, B = set([1,2]), set([1,3])
# Union
print(A & B)
{1}

# Intersection
print(A | B)
{1, 2, 3}

# Complement
space = {1,2,3,4,5,6}
print(space - A)
{3, 4, 5, 6}

 
import math
from itertools import permutations

print(list(permutations(['a','b','c'])))
[('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'), ('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')]

print(math.factorial(3))
6

 
from itertools import combinations

list(combinations(['a', 'b', 'c'], 3))
[('a', 'b', 'c')]

 
The calculation of permutation for this is nPk = n! / (n-k)!. 
The calculation for combinations for this is nCk = n! / k!(n -k)!

 
print(list(combinations(['a', 'b', 'c'], 2)))
[('a', 'b'), ('a', 'c'), ('b', 'c')]

print(list(permutations(['a', 'b', 'c'], 2)))
[('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'c'), ('c', 'a'), ('c', 'b')]

n = 3
k = 2

# Permutation
def calc_perm(n, k):
    return math.factorial(n) / math.factorial(n - k)
print(calc_perm(3, 2))
6.0

# Combination
def calc_comb(n, k):
    return math.factorial(n) / (math.factorial(k) * math.factorial(n - k))

print(calc_comb(3, 2))
3.0

 
We want p(Bowl 1|vanilla) not p(vanilla|Bowl 1), the answer for this second is 30/40, and the answer would be What is the probability of get a vanilla cookie from Bowl 1. p(A|B) != p(B|A). To find the answer we can go to the Baye's theorem, find out more on the book:

P(A|B) = P(A) * p(B|A) / P(B)

P(A) = 1/2 (random bowl) = .5
P(B|A) = 30/40
P(B) = 50 / 80 (probability of get it from both bowls)

P(A|B) = (.5 * .75) / 0.625 = 0.6 for bowl 1 and 1 - .6 = .4 for bowl 2

 
import numpy as np
from sklearn.naive_bayes import GaussianNB

# Suppose vanilla cookie is represented by 1 and chocolate by 2
# lets fill the bowls
    
bowl_1 = list(map(lambda x: [x], ([1] * 30) + ([2] * 10)))
bowl_2 = list(map(lambda x: [x], ([1] * 20) + ([2] * 20)))

# To create the target list we fill a list with B1 for bowl_1
# and B2 for bowl_2
data = np.array(bowl_1 + bowl_2)
target = ['b1'] * 40 + ['b2'] * 40

# Training the predictor
pred = GaussianNB()
pred.fit(data, target)

# Where's my vanilla cookie came from?
print(pred.predict(1))
['b1']

# What are the probabilities?
print(pred.classes_)
['b1' 'b2']

print(pred.predict_proba(1))
[[ 0.61708042  0.38291958]]

 
# 20C5
n = 20
k = 5
print(calc_comb(n, k))

15504.0

 
import math
import scipy.stats

# Given an example, 10% of all machines manufactured by a large 
# eletronic company are defective. A quality control inspetor randonmly 
# selects three machines from the production line. 
# What is the probability that exactly ONE of the THREE machines is
# defective?

p = .10  # p(success) - get a defective machine
q = .90  # 1 - p
n = 3 #  number of trials - machines
x = 1 # number of success

three_comb_1 = math.factorial(n) / (math.factorial(x) * math.factorial(n - x))

p_x_1 = three_comb_1 * (p ** x) * (q ** (n - x))
print(p_x_1)
0.24300000000000005

# Using ScipPy
print(round(scipy.stats.binom.pmf(x, n, p), 4))
0.243

 
# For the other example, we have a test's question with 5 alternatives, 
# just ONE is correct, which is the probability of miss TWO questions?

n = 4
x = 2
p = 4 / 5
q = 1 / 5

four_comb_2 = 6

p_x_2 = four_comb_2 * (p**2) * (q**(n-x))
print(p_x_2)
0.15360000000000007

# Using ScipPy
print(round(scipy.stats.binom.pmf(x, n, p), 4))
0.1536

 
%matplotlib inline

import numpy
import matplotlib.pyplot as plt

# Creating standard normal distribution data
mu, sigma = 0, 1
s = np.random.normal(mu, sigma, 1000)

count, bins, ignored = plt.hist(s, 30, normed=True)

# Display histogram of distribution and PDF
plt.plot(bins, 1/(sigma * np.sqrt(2 * np.pi)) *
         np.exp( - (bins - mu)**2 / (2 * sigma**2) ))

 
# Exercises

# 1. A delivery company have measured that mean time to delivery is
# 21 minutes and the standard deviation is 7 minutes. For a random
# delivery, what is the probability of the time be in 21 - 28 minutes.

from scipy.stats import norm
mu, sigma = 21, 7

z1 = norm(mu, sigma).cdf(21)
z2 = norm(mu, sigma).cdf(28)
print("Probability %", (z2 - z1) * 100)

Probability % 34.1344746069

 
# 2. For a mean of 24 years and a standard deviation of 6 years, what
# is the probability of a machine long between 17.52 and 29.7 years.

mu, sigma = 24, 6

z1 = norm(mu, sigma).cdf(17.52)
z2 = norm(mu, sigma).cdf(29.7)

print((z2 - z1) * 100)
68.8872783603

 
mu, sigma = 126, 10
# 3. Calculate the zcore for 116, 136, 131, 141

def zscore(x, mu, sigma):
    return (x - mu) / sigma

for x in [116, 136, 131, 141]:
    print(zscore(x, mu, sigma))

-1.0
1.0
0.5
1.5

# What is the probability of get someone with score higher than 136.
print( norm(mu, sigma).cdf(136) - 0.5)
0.341344746069