First Steps into the World of Probability: Sample Space, Events, Permutations, and Combinations

Probability and combinatorics are two fundamental concepts in mathematics and statistics that help us understand and interpret many phenomena in everyday life. In this introductory post, we’ll “touch upon” the main concepts together, seeing how they can be applied in various contexts.

Probability

Probability is a mathematical measure that indicates the likelihood of an event occurring. In other words, probability tells us how many favorable cases there are relative to all possible cases.

Probability is based on two fundamental concepts: the sample space and the event.

The sample space is the set of all possible outcomes of a random experiment.
For example, if we flip a coin, the sample space is {heads, tails}. If we roll two dice, the sample space is {(1,1), (1,2), …, (6,6)}.

An event is a subset of the sample space that we’re interested in.
For instance, if we flip a coin and we’re interested in whether it will be heads or tails, the event is {heads} or {tails}. If we roll two dice and want to know if the sum of the numbers is even or odd, the event is {(2,2), (2,4), …, (6,6)} or {(1,2), (1,4), …, (5,6)}.

The probability of an event is calculated by dividing the number of favorable cases for the event to occur by the number of possible cases in the sample space.

For example:

If we have a six-sided die and want to know the probability of getting a 4 when rolling the die, we have 1 favorable case (the face with the number 4) out of 6 possible cases (the six faces of the die).
Therefore, the probability of getting a 4 is 1/6.

Other simple examples:

• The probability of getting heads when flipping a coin is 1/2
• The probability that the sum of the numbers is even when rolling two dice is 18/36 = 1/2

Probability is expressed in numbers between 0 and 1, where 0 indicates the impossibility of the event and 1 indicates the certainty of the event.

A probability value close to 0 indicates a low possibility that the event will occur, while a probability value close to 1 indicates a high possibility that the event will occur.

The Principle of Additivity of Probabilities for Incompatible Events

The principle of additivity of probabilities for incompatible events states that the probability of the union of two or more incompatible events is equal to the sum of their probabilities.

Incompatible events are events that cannot occur simultaneously, i.e., if one occurs, the other cannot occur. For example, in rolling a die, the events “rolling a 3” and “rolling a 5” are incompatible. In this case, the probability of the union of the events (i.e., rolling a 3 or a 5) is equal to the sum of their probabilities (1/6 + 1/6 = 1/3).

The Principle of Multiplication of Probabilities

The principle of multiplication of probabilities states that the probability of the intersection of two events is equal to the product of their individual probabilities, if the events are independent.

In other words, if A and B are two independent events in a probability experiment, then the probability that both occur simultaneously is given by the product of their individual probabilities:

P(A ∩ B) = P(A) x P(B).

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To calculate the probability of more complex or combined events, the rules of combinatorics are used, which study the ways in which groups of objects can be formed according to specific criteria.

Two important concepts in combinatorics are permutations and combinations.

Permutation

Permutations are the ways in which n distinct objects can be arranged in n different positions. For example:

• The permutations of the letters A, B, C are ABC, ACB, BAC, BCA, CAB, CBA
• The number of permutations of n distinct objects is calculated using the factorial n!, i.e., the product of the natural numbers from 1 to n
• The number of permutations of A, B, C is 3! = 3 x 2 x 1 = 6

Let’s see a few more examples:

How many different ways are there to arrange 4 books on a shelf?

The answer is: n! Remember that “!” denotes the factorial, i.e., the product of all positive integers from 1 to n.

Solution: 4! (4 factorial) = 24 different ways

How many permutations can I make with a set of 5 letters?

5! = 5 x 4 x 3 x 2 x 1 = 120

There are 120 possible permutations of 5 letters.

How many permutations are possible with 5 letters taken in groups of 3?

n! / (n – r)!

where “n” represents the total number of objects (in this case, the 5 letters), and “r” represents the number of objects we want to choose and arrange in a specific order (in this case, 3 letters).

So, substituting the values, we get:

5! / (5 – 3)! = 5! / 2! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 60

Therefore, there are 60 possible permutations of 5 letters taken in groups of 3.

It’s essential to note that, when selecting a group of objects from a larger set, the order in which the objects are chosen matters. If the order didn’t matter, we would need to use the combination formula.

Ecco la traduzione in inglese del testo, mantenendo i tags WordPress:

The Concept of Combination

Combinations are the ways in which k objects can be chosen from n distinct objects without considering the order. For example:

• The combinations of two letters from A, B, C are AB, AC, BC
• The number of combinations of k objects from n distinct objects is calculated using the binomial coefficient C(n, k) = n! / (k! x (n-k)!)
• The number of combinations of two letters from A, B, C is C(3, 2) = 3! / (2! x (3-2)!) = 3

Let’s see a few more examples:

How many combinations are possible for a set of 10 people taken in groups of 3?

To calculate the number of combinations possible for a set of 10 people taken in groups of 3, we can use the combination formula:

n! / (k! * (n – k)!)

where “n” represents the total number of objects (in this case, the 10 people) and “k” represents the number of objects we want to choose without considering the order (in this case, 3 people).

So, substituting the values, we get:

10! / (3! * (10 – 3)!) = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / ((3 x 2 x 1) * (7 x 6 x 5 x 4 x 3 x 2 x 1)) = 120

A class consists of 12 boys and 4 girls. If three students are chosen at random, what is the probability that they are all boys?

This example is taken from the State Exam, math topic 1 (PNI, a.s. 2000-2001 – Scientific High School Course)

The probability of choosing three all-boy students can be calculated as the ratio between the number of ways to choose three boys (if we want to choose three all-boy students, we must consider all possible groups of 3 boy students that can be formed from the 12 boy students) and the total number of ways to choose three students from all sixteen.

The number of ways to choose three boys from the class of 12 boys is given by the combination of 3 elements chosen from the 12 boy students. We can calculate this number using the combination formula:

C(12, 3) = 12! / (3! * (12-3)!) = 220

The total number of ways to choose three students from the class of 16 students is given by the combination of 3 elements chosen from the 16 students.

C(16, 3) = 16! / (3! * (16-3)!) = 560

Therefore, the probability of choosing three all-boy students is:

P(three boys) = C(12, 3) / C(16, 3) = 220 / 560 = 11 / 28

The Binomial Distribution as an Application of Probability and Combinatorics

In a post specifically dedicated to probability distributions, I examined in detail the properties of the binomial distribution. I refer to the post for all the details.
In this context, I would like to briefly introduce the binomial distribution in a direct and practical way, solely to answer questions like:

• I want to know the probability of getting heads 5 times or less in 10 coin tosses.
• I want to calculate the probability of answering correctly 15 questions or more out of 20 multiple-choice questions, answering completely randomly.
• I want to find the probability of drawing less than 20 white balls in 100 draws (with replacement) from an urn containing 10 white balls and 90 black balls.

Let’s look at the first question. I want to know the probability of getting heads 5 times or less in 10 coin tosses.
Proceeding logically, I should calculate the sum of the binomial probabilities for k = 0, 1, 2, 3, 4, and 5. That is:

P(X <= 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Using the binomial probability formula and substituting n = 10 and p = 1/2, we get:

P(X <= 5) = C(10,0) x (1/2)^0 x (1/2)^10 + C(10,1) x (1/2)^1 x (1/2)^9 + … + C(10,5) x (1/2)^5 x (1/2)^5

Simplifying the calculations and using a calculator, we get:

P(X <= 5) = <0.001 + <0.01 + <0.04 + <0.12 + <0.21 + <0.25

P(X <= 5) = 0.63

So, the probability of getting heads 5 times or less in 10 coin tosses is approximately 63%.

Is there a simpler method to arrive at the correct result?

We can introduce the binomial distribution’s cumulative distribution function (CDF).

The CDF is a function that calculates the probability that the random variable X is less than or equal to a certain value k. It is denoted by F(k) and defined as:

F(k) = P(X <= k) = sum of binomial probabilities for i = 0, 1, …, k

This function can be calculated using an approximate formula or a precompiled table. For example, using an online table like this:

https://www.statisticshowto.com/tables/binomial-distribution-table/