Now that you know what sets are, we can go on and work with two sets that are of key importance when talking about probability: the event space and the sample space. These two concepts are foundational for calculating probabilities when assuming each event in the event space has the same probability of happening. Typical examples are rolling a dice (if the dice is "fair", the chance of throwing each number between 1 and 6 is 1/6) and flipping a coin (1/2 heads vs tails). You'll get a better sense of how all of this works in this lab.

You will be able to:

• Calculate probabilities by using relative frequency of outcomes to event space
• Use Python to demonstrate the axioms of probability

First, create a set roll_dice that holds the sample space of rolling a 6-sided dice once.

roll_dice = None

Now, let's assume that the event space is defined by "throwing a number higher than 4". This means that we consider the outcome "successful" if a 5 or a 6 is thrown. Create a set that holds these values.

event = None

Now use the formula $P(E) = \dfrac{|E|}{|S|}$ (This formula is called "Laplace's formula" and strongly related to the law of relative frequency) to calculate the probability.

prob_5_6 = None
prob_5_6  # 0.3333333333333333

Using this formula, it should be clear that the answer is 1/3 or 0.3333....

The law of relative frequency can be used to prove certain probabilities. But how does this work exactly? You're about to find out!

$$P(E) = \lim_{n\rightarrow\infty} \dfrac{S{(n)}}{n}$$

As you can see in the formula, the law states that when repeating an experiment $n$ times, where $n$ is very big, and you divide the number of "good" outcomes by the sample space (here we call it event E), you get to the probability of the event E. It should be clear that we get a more accurate number for P(E) when $n$ grows.

Let's see how this works. First, let's randomly generate values between 1 and 6. You can use numpy (imported as np) to generate random integers between 1 and 6 by setting the correct arguments. The np.random module is a very useful tool for this. We helped you with the code here, but you'll get more practice and a thorough explanation later on!

import numpy as np
np.random.randint(1,7) # you will get a random value between 1 and 6. See how it changes when you rerun

Now, let's repeat this experiment 10 times, then 1000 times, then 1 million times, then 100 million times. You can do this by specifying the argument size within the numpy function used above. Store the values in the pre-defined variables below.

np.random.seed(12345) # to ensure reproducibility of results

dice_10 = np.random.randint(1,7,size= None)
dice_1k = np.random.randint(1,7,size= None)
dice_1m = np.random.randint(1,7,size=None)
dice_100m = np.random.randint(1,7,size=None)

Next, let's count the number of "events". Remember that an event here is defined as throwing a 5 or a 6. Store them in the values below.

event_10 = None
event_1k = None
event_1m = None
event_100m = None

Next, you'll divide the number of events for each $n$ by the respective values for $n$. What do you see?

prob_10 = None
prob_1k = None
prob_1m = None
prob_100m = None
prob_10, prob_1k, prob_1m, prob_100m  # 0.5 0.331 0.333657 0.33329752

You see that the probability converges to 0.3333333... for higher values of $n$.

You're working at the United Nations, and want to get a better sense of the world population.

You come across some numbers and find the list of probabilities of being an inhabitant for each of the seven continents (rounded up to 3 digits):

• P(Africa) = 0.161
• P(Antarctica) = 0.000
• P(Asia) = 0.598
• P(Europe) = 0.10
• P(North-America) = 0.079
• P(Australia) = 0.005
• P(South-America) = 0.057

store these values using the variable names below:

P_afr = None
P_ant = None
P_as = None
P_eur = None
P_na = None
P_aus = None
P_sa = None

Now create the sample space set names continents. Store the sample space in a numpy array.

continents = np.array([P_afr, P_ant, P_as, P_eur, P_na, P_aus, P_sa])
print(continents)

We want to make sure that the three probability axioms are fulfilled because they assure us that $(\Omega,E,P)$ is a probability space:

• if we have a sample space $S$ (or $\Omega$)
• if we have an event space $E$ and a probability measure $P$,
• and the three probability axioms introduced in the previous lesson are fulfilled,

The third axiom is fairly ad hoc, and you will basically have to deduce from the context whether individual events are independent. It is fairly straightforward, however, that people cannot be inhabitants of two continents at the same time, so for now, we will assume that we're good for axiom three.

However, we can use the numpy array continents to verify if axiom 1 and 2 are fulfilled. Create a function check_axioms with a single input, sample_space, that returns the message "We're good!" if both axiom 1 and 2 are fulfilled, and "Not quite!" if that's not the case.

def check_axioms(sample_space):
    None

Now test your newly created function out on the sample space continents:

check_axioms(continents)

You want to make sure your test returns "Not quite!" for the following numpy arrays. Go ahead and test away!

test_1 = np.array([0.05, 0.2, 0.3, 1.01])
test_2 = np.array([0.05, 0.5, 0.6, -0.15])
test_3 = np.array([0.043,0.05,.02,0.3,0.2])

None
None
None

Great! We tested it and seems like our set continents is a true probability space.

In this exercise, we'll look at possible outcomes when throwing a dice twice. For your convenience, we created the NumPy array below.

Next, we'll compute a couple or probabilities associated with doing this.

import numpy as np
sample_dice = np.array([(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
              (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
              (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
              (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
              (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
              (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)])

Look at the shape of the array to reassure we haven't made any mistakes.

None # should be equal to (36,2)

Use Python to obtain the following probabilities:

First, use sample_dice to get "True" values for each time a 5 occurs.

set_5 = None # Your output should be an array of shape (36, 2) with booleans instead of numbers

Next, make sure that you get a value True for each pair where at least one 5 was thrown.

true_5 = None # Your output should be an array of shape (36,) and have booleans. 
              # You should obtain "True" if at least one of the previous pairs was true. 
              # Hint: Use np.any()
print(true_5)

Applying the sum() function to this result you can get to the total number of items in the event space. Divide this number by the total number in the sample space to obtain the probability of throwing a 5 at least once.

prob_5 = None
print(prob_5)

set_5 = None
set_6 = None

set_5_6 = None

set_any_5_6 = None
print(set_any_5_6) 

prob_5_6 = None
print(prob_5_6)

sum_dice = None
sum_8 = None

prob_sum_8 = None
print(prob_sum_8)

A teaching assistant is holding office hours so students can make appointments. She has 6 appointments scheduled today, 3 by male students, and 3 by female students.

Create a NumPy array of all possible orders of three male and three female students in which the teaching assistant can see students by appointment (please note: only consider gender when creating event space). Create this NumPy array in the same way as we did in the "throwing a dice twice" exercise. It will be quite a bit of typing, as your resulting NumPy array will have a shape (20,6)!

sample_mf= None

None # get the shape of sample_mf

sample_length= None
print(sample_length)

First, select the first 3 appointment slots and check for "F".

first_3_F = None
None

num_F = None
print(num_F)

F_2plus = None
print(F_2plus)

prob_F_2plus = None
print(prob_F_2plus)

None

You noticed that coming up with the sample space was probably the most time-consuming part of the exercise, and it would really become unfeasible to write this down for say, 10 or, even worse, 20 appointments in a row. You'll learn about methods that make this process easier soon!

At a supermarket, we randomly select customers, and make notes of whether a certain customer owns a Visa card (event A) or an American Express credit card (event B). Some customers own both cards. You can assume that:

• P(A) = 0.5
• P(B) = 0.4
• both A and B = 0.25.

1. Compute the probability that a selected customer has at least one credit card.

2. Compute the probability that a selected customer doesn't own any of the mentioned credit cards.

3. Compute the probability that a customer only owns VISA card.

(You can use Python here, but you don't have to)

In this lab, you got to practice your knowledge on the foundations of probability through working on problems regarding the law of relative frequency, the probability axioms, and the addition law of probability.