Assignment Week 10 – Essential of Probability

Basic Statistics | Week 10

1 CHAPTER 6 ~ Essentials of Probability

Basic concept of Probability

Probability is an important foundation in statistics, providing a systematic framework for understanding uncertainty and aiding in making accurate decisions. With probability, we do not rely solely on guesses or intuition but can measure the likelihood of various outcomes, recognize patterns in data, and analyze phenomena that arise from experiments or natural processes. Understanding the concept of probability well is crucial for conducting accurate data analysis, scientific research, and evidence-based practice.

2 Fundamental Concept

Summary of the explanation

Probability

Probability can be defined as the likelihood of an event occurring, which is equal to the total number of desired outcomes divided by the total number of possible outcomes. The value is in the range: \[ 0 \le P(A) \le 1 \]

  • 0 means an event is impossible

  • 1 means that the event is certain to happen

Basic Probability Formula

\[ P(A) = \frac{\text{Total number of favourable outcome}}{\text{Total number of possible outcome}} \] Relevant Example

For example, if you roll a die with 6 sides, what is the probability of getting a 4?

sample room: \[ S = \{1,2,3,4,5,6\} \] Event A = (4) \[ P(A) = \frac{1}{6} \] It means the chance of getting the number 4 is 1 out of 6 possibilities.

Sample Space

Sample Space is the set of all possible outcomes of a random experiment. Each individual outcome (for example, “getting 3”, “coin = heads”, etc.) is called a sample point, which is a member of the sample space. The sample space is usually denoted by S (or sometimes Ω).

written as: \[ S = \{\text{All Possible Outcomes}\} \] probability of two event happening together:

\[ P(A \cap B) = P(A) \times P(B) \]

Relevant Example

if we toss two coins. The possible outcomes are: \[ S = \{AA,\, AG,\, GA,\, GG\} \] If you want two images to appear, then: \[ A = \{GG\} \]
Note: probability of two event happening together \[ P(A \cap B) = P(A) \times P(B) \]

Complement Rule

The probability that an event does not occur,is equal to one minus the probability that it will occur. \[ P(A^c) = 1 - P(A) \]

  • A is an event

  • Aᶜ is the complement of A, which is the opposite event or the non-occurrence of A

  • The total probability is always 1 or 100%

So the probability of an event occurring plus the probability of the event not occurring must equal 1

Relevant Example

Example of a Complement Rule — Roll 1 Dice.

Suppose we roll one 6-sided die once. Genesis (A): gets the number 6.

Sample room: \[ S = {1,2,3,4,5,6} \]

Probability of obtaining 6: \[ P(A) = \frac{1}{6} \]

The complement of (A) (not getting 6) is: \[ A^c = {1,2,3,4,5} \]

With complementary rules: \[ P(A^c) = 1 - P(A) = 1 - \frac{1}{6} = \frac{5}{6} \] So the probability of not getting a 6 is 5/6.

Overall Conclusion:

Probability is a numerical measure that describes how likely an event is to occur. The value of probability always ranges from 0 to 1, where a value of 0 means the event is impossible, and a value of 1 means the event is certain to occur. To calculate the probability of an event, we must first know the sample space, which is the set of all possible outcomes that can occur from a random experiment. For example, in the experiment of tossing two coins, the sample space is S = {HH, HT, TH, TT}, meaning there are four possible outcomes. Once the event and the sample space are known, we can use various probability rules, one of which is the complement rule, a rule for calculating the probability that an event does not occur. If A is the event being considered, its complement Aᶜ is the occurrence of anything other than A, and its probability can be calculated using the formula P(Aᶜ) = 1 – P(A).Thus, these three concepts are interconnected: the sample space helps identify possible outcomes, probability measures the chance of an event, and the complement rule helps find the likelihood of an event not occurring more easily and efficiently.

Source: Material 6 Essentials of Probability

https://bookdown.org/dsciencelabs/intro_statistics/06-Essentials_of_Probability.html

3 Independent And Dependent Event

Note:

Probability of Independent and Dependent Events explains the difference between independent events and dependent events.

It explains how to calculate the probability when two (or more) events occur together, using different formulas depending on whether the events are independent or not.

Independent Events

Independent events are two events that do not affect each other. This means that the outcome of the first event does not influence the likelihood of the second event occurring.

\[ P(A \cap B) = P(A) \times P(B) \]

Relevant Example

If you roll a six sided die and flip a coin, what is the probability of rolling a five and getting heads?

Determine the probability of each event

  • The probability of getting heads on a coin: \[P(getting\ heads)=\frac{1}{2}\]

  • The probability of rolling a 5 on a die: \[P(rolling\ a\ 5)=\frac{1}{6}\] Value substitution: \[P(H \cap 5)=\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\] As a result, the probability of getting heads and rolling a 5 is equal to 1/12 or 0.0833

Dependent Event

Dependent events are two events that influence each other. That is, the outcome of the first event changes the probability of the second event. \[ P(A \cap B) = P(A) \times P(B|A) \] Relevant Example

In a box, there are two types of balls:

  • 7 green balls

  • 3 blue balls

Total balls: 10 balls

If we randomly draw 2 balls without replacement, what is the probability of drawing a green ball first and then a blue ball?

Because without replacement, the probability of the second event is influenced by the outcome of the first draw → Dependent Events

  1. Chance of picking the first green ball

Number of green balls = 7 Total balls = 10 \[P(green)=\frac{7}{10}\] 2. After one green is taken

Remaining green balls = 6 Remaining total balls = 9 Number of blue balls remains = 3

Probability of taking a blue ball after a green: \[P(second\ blue \mid first\ green)=\frac{3}{9}=\frac{1}{3}\] Value substitution:

\[\frac{7}{10} \times \frac{1}{3} = \frac{7}{30}\]

Overall Conclusion:

In probability, Independent Events are two events that do not affect each other, so the outcome of the first event does not change the probability of the second event. Conversely, Dependent Events are two events that are related, so the outcome of the first event affects the probability of the second event. An example of independent events is flipping a coin and rolling a die, because the coin result does not affect the die result. Meanwhile, an example of dependent events is drawing balls consecutively without replacement, because after the first ball is drawn, the total number of balls changes, affecting the probability of the second event. Understanding the difference between these two concepts is important for determining the correct formula in calculating combined probabilities.

Source: Material 6 Essentials of Probability

https://bookdown.org/dsciencelabs/intro_statistics/06-Essentials_of_Probability.html

4 The Union of Events

Note:

Watching this video is useful for understanding how to determine the probability of the union of two or more events in a sample space. Through the explanation of concepts and example problems, this video helps us learn how to calculate the likelihood that at least one of several events can occur, whether the events are mutually exclusive or not and have an intersection.

Probability of the Union of Events

In probability, we often want to know the chances that at least one of several events may occur. For that, we use the concept of the Union of Events.

What is a Union

If there are two events A and B, then the union A ∪ B means the event that:

  • A occurs, or

  • B occurs, or

  • A and B occur simultaneously

Combined Probability Formula

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Why should we subtract
P(A∩B)?

Because the intersection part (events that are included in both) is counted twice when adding
P(A) + P(B), so it must be subtracted once to make the calculation correct.

Mutually Exclusive Events

If 𝐴 and 𝐵 cannot occur at the same time (mutually exclusive), then: \[ P(A \cup B) = P(A) + P(B) \] Because P(A ∩ B) = 0

Relevant Example

Example 1 — Events That Are Not Mutually Exclusive

In one class there are 40 students:

  • 18 students like Mathematics (A)

  • 22 students like Science (B)

  • 10 students like both (A ∩ B)

Calculate the probability that a randomly selected student likes Mathematics or Science.

Solution:

\[ P(A) = \frac{18}{40}, \quad P(B) = \frac{22}{40}, \quad P(A \cap B) = \frac{10}{40} \]

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

\[ P(A \cup B) = \frac{18}{40} + \frac{22}{40} - \frac{10}{40} = \frac{30}{40} = \frac{3}{4} \] It means: the probability that a selected student likes Mathematics or Science is 3/4.

Example 2 — Mutually Exclusive Events

Rolling 1 die:

  • A = the number 1 appears

  • B = the number 6 appears

Because both numbers cannot appear at the same time:

\[ P(A) = \frac{1}{6}, \quad P(B) = \frac{1}{6} \]

\[ P(A \cup B) = P(A) + P(B) \]

\[ P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \] It means: the probability of getting a 1 or 6 is 1/3

Source:

5 Exclusive and Exhaustive Events

Mutually Exclusive Events

Two events (or more) are said to be mutually exclusive if they cannot occur at the same time in a single trial. This means that if event A occurs, then event B cannot occur at the same time. Therefore, the probability of both events occurring is zero \[ P(A \cap B) = 0 \] Since there is no overlap between the two events, the formula for the combined probability of two mutually exclusive events is: \[ P(A \cup B) = P(A) + P(B) \]

Exhaustive Events

(sometimes called collectively / jointly exhaustive), a set of events is said to be exhaustive if together they cover all possible outcomes of an experiment (sample space S). This means that every time the experiment is conducted, one (or more) of the events in the set must occur.

Unlike mutually exclusive events, if both events are complete, then the combined probability of both events is 1 \[ P(A \cup B) = 1 \]

Characteristics or Differences:

Aspect / Feature Mutually Exclusive Events Exhaustive Events
Relationship among events Events cannot occur at the same time; if A occurs, B cannot occur. Events can overlap; the main requirement is that the union of all events covers the entire sample space.
What is guaranteed per trial At most one event from the set can occur (or none). At least one event from the set must occur.
Set notation / probability \(A \cap B = \emptyset \Rightarrow P(A \cap B) = 0\) \(\bigcup E_i = S \Rightarrow\) at least one \(E_i\) occurs
Combination with other properties Can be combined with exhaustiveness → “Mutually Exclusive & Exhaustive” (MECE). Exhaustive events are not necessarily mutually exclusive; events may overlap.
Practical implication Allows simple addition rule: \(P(A \cup B) = P(A) + P(B)\) if events are mutually exclusive. Ensures all possible outcomes are considered, useful for law of total probability and complements.

Example:

Scenario / Trial Event Set Mutually Exclusive? Exhaustive? Explanation
Coin toss A = “Heads”, B = “Tails” Yes Yes Only one outcome can occur per toss, and together they cover all possible outcomes.
Dice roll A = “Roll a 1”, B = “Roll a 4” Yes No A and B cannot occur together, but outcomes 2, 3, 5, 6 are not covered.
Dice roll A = “Prime number” {2,3,5}, B = “Even number” {2,4,6}, C = “Odd number” {1,3,5} No Yes Some events overlap (2 in A & B, 3 & 5 in A & C), but together they cover all outcomes.
Dice roll A = “Even number” {2,4,6}, B = “Odd number” {1,3,5} Yes Yes A and B cannot occur together, and together they cover all possible outcomes (MECE).
Coin toss A = “Heads”, B = “Tails” Yes Yes Same as first row; mutually exclusive and exhaustive for coin toss.

Source:

6 Binominal Experiments and Binomial Formula

The Binomial Experiment

A binomial experiment is a series of trials carried out multiple times, where each trial can only result in one of two outcomes, the probability of success remains the same, and each trial does not affect the others.

An experiment is called binomial if it meets the following conditions:

  • There is a fixed number of trials (n).

  • Each trial has only two possible outcomes — for example, “success” or “failure” (just two outcomes).

  • Each trial is independent — the result of one trial does not affect another trial.

  • The probability of success (p) is the same for each trial — it does not change throughout the trials.

Relevant Example

Flipping a coin, say, 5 times. Each flip is a trial; the outcome can be ‘heads’ (success) or ‘tails’ (failure); the number of flips is predetermined; each flip does not affect the others; and the probability of getting heads on each flip is the same. This meets the definition of a binomial experiment.

Binomial Distribution Formula

\[ P(X = x) = \binom{n}{x} \; p^{x} \; (1-p)^{n-x} \] With:

  • n = total trials,

  • x = number of desired successes,

  • p = probability of success for each trial,

  • (1−p) = probability of failure for each trial.

Note: So: a binomial experiment is the one that is conducted (the actual experiment), whereas a binomial distribution is the mathematical model for the probability of the outcomes of that experiment.

Source:

7 Binomial Distribution

The Binomial Distribution

Binomial distribution is a probability distribution that describes the likelihood of a certain number of successes occurring in n trials in a binomial experiment.

This distribution is used when:

  • Each trial has only 2 possible outcomes: success or failure.

  • The probability of success remains the same in each trial (p).

  • Each trial is independent.

  • Random variable X represents the number of successes in n trials.

Binomial Distribution Formula

\[ P(X = x) = \binom{n}{x} \; p^{x} \; (1-p)^{n-x} \]

Relevant Example

For example: Flipping a coin 5 times. The probability of getting heads is 0.5.
Calculate the chance of exactly 3 heads.

Given:

  • n = 5

  • x = 3

  • p = 0.5

\[ P(X = 3) = \binom{5}{3}(0.5)^3(0.5)^{2} \]

\[ = \frac{5!}{3! \, 2!} \times (0.5)^5 \]

\[ = 10 \times 0.03125 = 0.3125 \]

So the probability of getting exactly 3 heads out of 5 tosses is 0.3125 or 31.25%.

Source: