Essential of Probability ~ Assignment week 10

Cahaya Medina Semidang

Student Majoring in Data Science

1 Introduction

Probability is a branch of mathematics that studies the likelihood of an event occurring. This is the basis for understanding uncertainty and making data-driven predictions. The concept of probability is essential in various fields, such as statistics, finance, computer science, and social sciences. Probability measures the probability of an event occurring in the range of values between 0 and 1, where: 0 indicates that the event is unlikely to occur 1 indicates that the event must have occurred. By understanding the concept of probability, we can analyze and make decisions based on uncertainty. Probability provides a framework for understanding how events occur and how outcomes can be expected in a variety of contexts.

2 Fundamental Concept

Before entering into the discussion of the formulas and properties of probability, it is important to understand the context of this concept. By understanding the basics of probability, we can build a solid foundation for studying more complex statistical topics.

2.1 Probability

The probability of an event is related to the relationship between events that are likely to occur. Probability is a numerical measure used to describe the chances or likelihood of an event occurring. Formally, probability is defined as the ratio between the number of desired events and the number of possible events as a whole. The probability is always between 0 and 1, where 0 indicates that the event must not have happened, while 1 indicates that the event must have happened.

Probability can be define as the chance that an event will occur. This is equal to the total number of favourable outcomes divided by the total number of possible outcomes. We can properly write probability as the probability of event A equal to the number of outcomes in divided by the total number of outcomes in the sample space.

2.1.1 Basic formula of probability

$P (A) = \frac{n(A)}{n(S)}$

$n(A)$ : number of favorable outcomes
$n(S)$ : total outcomes in the sample space

Example
A fair coin is tossed once. What is the probability of getting Tail?

$P (Tail) = \frac{n (A)}{n (S)} = \frac{1}{2}$

The sample space is {Head, Tail}, so $n(S) = 2.$
There is only 1 favorable outcome {Tail}, so $n(A) = 1.$

$Therefore, P(Tail) = \frac{1}{2} = 0.5.$

2.2 Probability Rules

The probability of an event occuring always has a value between 0 and 1, inclusive. Example probability 0, means the event will never occur and probability of 1 means that event will always occur and probability of 0,5 means that event is expected to occur 50% of the time.
The probabilities of all outcomes must always add up to 1. Example if we flip a coin we know that there are two outcomes getting heads or getting tails, the probability of getting heads is 0,5 and the probability of getting tails is also 0,5. If we add these up, we get a value of 1 which satisfies this condition.

2.3 The Compliment Rule

This rule says that the probability that an event does nor occur, is equal to one minus the probability that it will occur.

The formula is

$P (A^{’}) = 1 - P (A)$

The complement of event A represents all outcomes not in A.

Example
A fair coin is tossed once. What is the probability of the complement event A?

$P (A^{’}) = 1 - P (A)$

In this experiment, the sample space is {Heads, Tails}, since the coin is fair, both outcomes are equally likely.

The event A = “Heads”, therefore:

$P (A) = \frac{1}{2}$

The complement event A’ means the coin does NOT show Heads, so A’ = “Tails”. Thus, using the complement rule:

$P (A^{’}) = 1 - \frac{1}{2} = \frac{1}{2}$

$Final Answer: The probability of the complement event
A’ is 1/2.$

3 Independent and Dependet Events

Before exploring the concepts of dependent and independent events, it is important to understand how different events within a probability experiment can relate to one another. In many real-world situations, the occurrence of one event may influence, alter, or provide information about the likelihood of another event. In other cases, an event may occur without having any effect at all on another event. Recognizing these differences is essential for accurately calculating probabilities, interpreting data, and understanding how events interact within a sample space. With this foundation in place, we can now examine the distinction between dependent and independent events more clearly.

3.1 Independent Events

Independents Events refer to the occurrence of one event not affecting the probability of another event. The outcome of one event doesn’t affect the outcome the the other event. Example: First events is Rolling a die and second events is flipping coin. The outcome of the first event doesn’t affect the outcome of the second outcome. These events are said to be independent events. In other words rolling a six doesn’t increase or decrease the probability of a coin landing on heads or tails.

The formula is

$P (A \cap B) = P (A) \cdot P (B)$

$P(A ∩ B)$ : the probability that events A and B occur together.

$P(A)$ : the probability of event A occurring.

$P(B)$ : the probability of event B occurring.

This formula applies when events A and B are independent, meaning the occurrence of one event does not affect the probability of the other.

Example
What are the odds of getting a 5 on dice and heads on a coin?

$P (5) = \frac{1}{6}$

$P (H) = \frac{1}{2}$

$P (5 and H) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$

3.2 Dependent Events

Dependents Events refer to the occurence of one events affecting the probability of another events. The outcome of one events does influence the outcome of the other event.

The formula is

$P (A \cap B) = P (A) \cdot P (B | A)$

$P(A ∩ B)$ : the probability that events A and B occur together.

$P(A)$ : the probability that event A occurs.

$P(B | A)$ : the conditional probability of event B occurring given that A has already occurred.

This formula applies when events A and B are dependent, meaning the occurrence of A affects the probability of B.

Example

we have a box that contains 10 marbles. 7 marbles are green and three of the marbles are blue. If we randomly select two marbles from this box:

1. what is the probability of drawing a green marble and then a blue marble, without replacement?

Step 1 -> First intake: Green Marbles

$P (First Green Marbles) = \frac{7}{10}$

Step 2 -> Second intake: blue marbles, after which one green is taken

Since one marble (green) has been taken and not returned, the remaining marbles:

Green remaining: 6
Fixed blue: 3
Total marbles remaining: 9

So the probability of taking blue on the second take is:

$P (Second Blue Marbles | First Green Marbles ) = \frac{3}{9}$

Step 3 -> Combined Probability

since this is event is independent, we use

$P (A \cap B) = P (A) \cdot P (B | A)$

$P (green then blue) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90} = \frac{7}{30} \approx 0.233$

2. What is the probability of getting two green marbles in a row (no return)?

Step 1 -> First intake: Green Marbles

$P (First Green Marbles) = \frac{7}{10}$

Step 2 -> Second intake: Green Marbles again

Since one marble (green) has been taken, the remaining marbles:

Green remaining: 6
Total marbles remaining: 9

$P (Second green Marbles | First Green Marbles ) = \frac{6}{9}$

Step 3 -> Combined Probability

$P (A \cap B) = P (A) \cdot P (B | A)$

$P (Two Green marbles) = \frac{7}{10} \times \frac{6}{9} = \frac{42}{90} = \frac{7}{15}$

These two examples show the essence of the Dependent Event where the odds of the second take change because the outcome of the first take affects the number of marbles remaining.

4 Probability of Union Events

In everyday life, we are often faced with various possibilities that can occur simultaneously. For example, when we roll a die, we might want to know the probability of getting an even number or a number greater than three. To understand situations like this, the field of probability provides a very useful concept called the union of events. This concept helps us calculate the likelihood of at least one of several events occurring, enabling us to make more accurate predictions and make better decisions based on probability.

Review

In the previous sub-chapter, we have learned about the probability of events. Where the probability of an event can be calculated by:

$P (A) = \frac{total number of favourable outcomes}{total number of
possible outcomes}$

In addition, we have also learned about the sample room. Sample Space is the entire of outcomes in statistical experiment. Example if rooll a 6-sided dice, there are 6 different outcomes. This would bet he sample space for rolling one dice.

If we rolling a dice, the sample space is:

$S = {1,2,3,4,5,6}, n(S) = 6.$

if we rolling 2 dice at the same time, every dice have a 6 probability. So, total sample space:

$n(S) = 6 x 6 = 36.$

4.1 Union Event

A combination of events or unions is an event in which A occurs, or B occurs, or both occur. This concept is important because many events in real life overlap so we need a proper way to calculate the total odds without counting the same part twice. If A and B are two occurrences in the sample space S, then a combination of the two can be written:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Probability of the union of events calculate for either event occuring. We know that we are dealing with a union of events when a probability question includes the word or.

Here’s an illustration of the sample room for two dice (36 sample space):

Dice

Example 1: What is the probability of rolling two even numbers?

The even numbers on the dice are 2, 4, and 6. If there are two dice, then each dice has 3 even chances. So that the total pairs are two even digits:

$3 x 3 = 9.$

So:

$P (Two even number) = \frac{9}{36}$

Example 2: What is the probability of rolling at least one two?

We look for pairs that contain at least 2 on one of the dice. There are a total of 11 couples who meet these conditions, so that:

$P (At least one two) = \frac{11}{36}$

Example 3: What is the probability of rolling two even numbers and at least one two?

These two events overlap. If we pay attention, there are 5 couples who meet both conditions.

So:

$P (A \cap B) = \frac{5}{36}$

Example 4:What is the probability of rolling two even numbers or at least one two?

For the combination of events, we use the formula:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Substitution of values from previous calculations:

$P (A \cup B) = \frac{9}{36} + \frac{11}{36} - \frac{5}{36} = \frac{15}{36} \approx 0.4167 (41.67 %)$

Here is an illustration of a Venn diagram that illustrates the relationship between the two events:

Diagram Venn

5 Exclusive and Exhaustive Events

Two important concepts that are often used to explain the relationship between events are exclusive events and exhaustive events. Before we get into definitions and examples, let’s first understand how events in probability can be interrelated or form an entire sample space.

5.1 Exclusive Events

Two events are said to be mutually exclusive if they do not have the same outcome. In other words, both events cannot occur at the same time.

Characteristic:

There are no wedges on the Venn diagram.
Also known as disjoint events.
Mathematical condition:

$P (A \cap B) = 0$

Example

1. in one roll of the dice:

Event A: the number 3 appears
Event B: the number 5 appears

A and B cannot occur at the same time, so they are mutually exclusive events.

2. Examples of mutually exclusive incidents with Two Dice

The two-dice sample chamber contains 36 number pairs.

Event A: Appears at least one number 5

Number of results that meet: 11
Probability:

$P (A) = \frac{11}{36}$

Event B: The sum of the two dice is less than 4

Outcome that satisfies: (1,1), (1,2), (2,1)
Probability:

$P (B) = \frac{3}{36}$

Question: Are A and B exclusive events?

Event A requires the number 5 on one of the dice.
Event B only allows a total of 2 or 3.
No Result has a number of 5 and the number is < 4.

Since there are no slices:

$P (A \cap B) = 0$

So, A and B are mutually exclusive events

5.2 Exhaustive Events

A group of events is called exhaustive or comprehensive when they collectively cover the entire sample space. This means that in every experiment, there must be one of the events that occurs.

Characteristics:

In a Venn diagram, both cover the entire box of the sample space.
It is permissible to have slices, it does not have to negate each other.
Mathematical conditions:

$P (A \cup B) = 1$

Example

1. On a Single Dice:

Event A: appears an even number
Event B: appears an odd number

A and B together cover all possibilities, so they are exhaustive.

2. Example of a Comprehensive Event with Two Dice :

Event A: Appear at least one number 6

Totatl outcome : 11
Probabiity :

$P (A) = \frac{11}{36}$

Event B: The number of both dice is less than 11

Total outcome : 33
Probabbility:

$P (B) = \frac{33}{36}$

In the video, it is explained that there are 8 outcomes that become slices A and B.

Using the combined formula:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$P (A \cup B) = \frac{11}{36} + \frac{33}{36} - \frac{8}{36} = \frac{36}{36} = 1$

Conclusion: events A and B are exhaustive events

3. An example of an event that fulfills both properties at once: the sum of two even and odd dice.

Event A: The number of two even dice

Result: 18
Probability:

$P (A) = \frac{18}{36}$

Event B: Second number of odd dice

Result: 18
Probability:

$P (B) = \frac{18}{36}$

Question 1, What is an Exclusive Occurrence?

A number cannot be even and odd at the same time.
There is no slice.

$P (A \cap B) = 0$

Question 2, What is the Whole Event?

All numbers are definitely even or odd.
Together they cover the entire sample space.

$P (A \cup B) = 1$

Conclusion: A and B are events that negate each other while complementing each other.

6 Binomial Experiment and Binomial Formula

Some experiments have special characteristics, where each experiment has only two possible outcomes—for example, successful or fail, win or lose, present or absent. This kind of experiment is known as a binomial experiment. Having understood the characteristics of these experiments, we can use the binomial formula to calculate the probability of a certain number of successes in a certain number of experiments.

6.1 Binomial Formula

When we talk about the binomial probability distribution we are referring to the probability of a “success” or a “failure” in an experiment that is repeated multiple times. We can easily remember this by paying attention to the prefix “Bi-“ which literally means two (success or failure).

$P (k) = (\binom{n}{k}) p^{k} (1 - p)^{n - k}$

Where:

$n$ : total experiment
$k$ : total succes
$p$ : probability of success on each attempt
$1 -
p (or q)$ : probability of failure on each attempt

$(\binom{n}{k}) = \frac{n!}{k! (n - k)!}$ : combination

Binomial Setting

An experiment can be modeled with a binomial distribution if it meets the following four conditions:

The number of trials (n), mus be fixed
There are only two possible outcomes for each trial; success or failure
The probability (p) of success must be constant for every trial
Each trial must be independent

6.2 Example

Example 1: What is the probability of getting exactly one head if we flip a regular coin three time? And is this a binomial experiment?

Step 1: check binomial setting

Is there a fixed number of trials? Yes -> (n = 3)
Are the two possible outcomes a success and a failure? Yes -> (Success = H, Failure = T)
Is the probability of success constans for each trial? Yes -> (P = 5/10=0,5)
Are trials independent of each other? Yes

Because it meets 4 conditions, the experiment is a binomial experiment

Step 2: The probability

Illustration of throwing coin three time

Diagram Venn

For example, we consider:

A = H
G = T

so, There are 3 possible sequences:

Because P(H) = 0.5 and P(T) = 0.5, So for each sequence:

$0.5 x 0.5 x 0.5 = 0.125$

Total probability for three outcomes:

$0.125 + 0.125 + 0.125 = 0.375$

Calculation with formula

$n = 3, k = 1, p = 0.5$

$P (1) = (\binom{3}{1}) (0.5)^{1} (0.5)^{2} = 3 \times 0.125 = 0.375$

Example 2:

We have 10 marbles in a box, there:

3 pink marbles
2 green marbles
5 blue marbles

Question: if we pick out five marbles with replacement, what is the probability of drawing exactly two green marbles, andi s this a binomial experiment?

Step 1: Check binomial setting

Is there a fixed number of trials? Yes -> (n = 5)
Are the two possible outcomes a success and a failure? Yes -> (Success = green marbles, Failure = not getting a green marbles)
Is the probability of success constans for each trial? Yes -> (P = 2/10=0,2)
Are trials independent of each other? Yes

Because it meets 4 conditions, the experiment is a binomial experiment

Step 2: The probability

example:

G = Green
- = Not green

Outcome	Trial1	Trial2	Trial3	Trial4	Trial5
1	G	–	–	–	–
2	–	G	–	–	–
3	–	–	G	–	–
4	–	–	–	G	–
5	–	–	–	–	G
6	G	–	–	–	–
7	–	G	–	–	–
8	–	–	G	–	–
9	–	–	–	G	–
10	–	–	–	–	G

Each sequence has:

2 success (G) = $p^{2} = {0.2}^{2}$
3 failure (-) = $p^{3} = {0.8}^{3}$

So, probability each outcome:

${0.2}^{2} \times {0.8}^{3} = 0.02048$

Total 10 outcome:

$10 x 0.02048 = 0.2048$

Calculation with formula

$n = 5, k = 2, p = 0.2$

$P (2) = (\binom{5}{2}) (0.2)^{2} (0.8)^{3} = 0.2048$

The result is the same as calculating the numeracy of 10 outcomes

7 Visualizing the Binomial Distribution

The understanding of the concept of binomials is not only limited to the calculation of numbers, but also to how the data can be visualized to make it easier to understand. The binomial visualization provides a way to see the probability distribution of an experiment that has two possible outcomes, such as “successful” or “fail”. With the help of graphs or diagrams, distribution patterns, odds of each event, and overall trends can be recognized faster than just looking at numerical values. Therefore, having known and understood binomial calculations and formulas, it is important to understand how visual representations can strengthen our intuition towards the probability and distribution of data.

Formula Binomial

$P (k) = (\binom{n}{k}) p^{k} (1 - p)^{n - k}$

Where:

$n$ : total experiment
$k$ : total succes
$p$ : probability of success on each attempt
$1 -
p (or q)$ : probability of failure on each attempt

$(\binom{n}{k}) = \frac{n!}{k! (n - k)!}$ : combination

Example
If we throwing coins 2 times, so:

$n = 2$
$p =
0.5$
$k = 0,1 or 2$
$The result of the calculation of the binomial formula is :$

P(0)=0.25, P(1)=0.50, P(2)=0.25

These values are then visualized in a bar chart to show the probability of each number of successes.

When the probability value of success (p) and the number of experiments (n) are entered into the formula, the probability values for each k are depicted as a bar graph. In the case of coins (p = 0.5, n = 2), the chart is symmetrical because the chances of success and failure are the same. In other words, a binomial graph can be created simply by using the binomial formula for each value k.

Binomial Distribution Parameters

If a random variable X follows a binomial distribution, then

Parameter	Formula	Description
Mean (μ)	μ = n · p	The expected average number of successes.
Variance (σ²)	σ² = n · p · (1 - p)	A measure of the spread of the data from the mean.
Standard Deviation (σ)	σ = √(n · p · (1 - p))	The square root of the variance.

In the video, it is also shown how the value P affects the form of distribution:

Condition	Example	Description
P = 0.5	0.5	Symmetric, bell-like (fair coin).
P < 0.5	0.1	Right-skewed (few successes expected).
P > 0.5	0.8	Left-skewed (many successes ecpected).

In the video, it is also shown how the value of n (number of attempts) affects the form of the distribution:

Condition	Example	Description
Small	n = 10	The distribution is still wide and less smooth.
Medium	n = 20	The distribution is starting to look smoother and more centered around the expected value (np).
Big	n = 50	The distribution becomes much smoother, narrower, and more symmetrical.

From the diagram it can be concluded that when n increases, then the binomial distribution:

Smoother
Increasingly similar to normal distribution
Data collected around mean np

Question: When can the binomial distribution be approached normal? In the video, a rough guidline is explained to determine whether the binomial distribution can be considered close to the normal distribution. The binomial distribution can be approached normal if: -

n p \geq 10

- $n (1 - p) \geq 10$

Both must be fulfilled. However, there are some that replace the number 10 with 5

8 References

Efgivia, Mohammad Givi (2024). Widina Media Utama, https://repository.penerbitwidina.com/media/publications/584551-statistik-dan-probabilitas-4124f579.pdf
Sumarlin, Tantik. S.Kom.,M.Si (2023). Yayasan Prima Agus Teknik, https://penerbit.stekom.ac.id/index.php/yayasanpat/article/view/418/441

Outcome	Trial1	Trial2	Trial3	Trial4	Trial5
1	G	–	–	–	–
2	–	G	–	–	–
3	–	–	G	–	–
4	–	–	–	G	–
5	–	–	–	–	G
6	G	–	–	–	–
7	–	G	–	–	–
8	–	–	G	–	–
9	–	–	–	G	–
10	–	–	–	–	G

Outcome	Trial1	Trial2	Trial3	Trial4	Trial5
1	G	–	–	–	–
2	–	G	–	–	–
3	–	–	G	–	–
4	–	–	–	G	–
5	–	–	–	–	G
6	G	–	–	–	–
7	–	G	–	–	–
8	–	–	G	–	–
9	–	–	–	G	–
10	–	–	–	–	G

Essential of Probability ~ Assignment week 10