2.1 Simple Probability

Interpretation:
Probability can be defined as the chance of an event occurring by calculating the number of desired outcomes with the total number of possibilities on a scale of 0 to 1, where a value of 1 means that the probability is very accurate and vice versa.

Formula:
\[P = \frac{\text{Total Number Of Favourable Outcomes}}{\text{Total Number Of Possible Outcomes}}\]

Example: Probability of drawing a Queen from a deck of cards.

# Definisikan Ruang Sampel (Total Kartu)
Total_Cards <- 52

# Definisikan Kejadian A: Mendapatkan kartu Queen (ada 4 Queen)
Favourable_Outcomes_A <- 4

# Hitung Probabilitas P(A)
P_A_Queen <- Favourable_Outcomes_A / Total_Cards

# Tampilkan Hasil
cat("Total Cards:", Total_Cards, "\n")

## Total Cards: 52

cat("Favourable Outcomes (Queen):", Favourable_Outcomes_A, "\n")

## Favourable Outcomes (Queen): 4

cat("Probability P(Queen):", round(P_A_Queen, 4), "\n")

## Probability P(Queen): 0.0769

2.2 Sample Space

Interpretation:
The Sample Space is the set of all possible outcomes of a random experiment. It defines the entire universe of possible results.

(Insert sample space Venn diagram image here if needed)

Formula (Conceptual):

Sample Space:
\[S = \{h_1, h_2, ..., h_n\}\]

Total Outcomes:
\[N = |S|\]

Note:
\(h_n\) is the n-th individual outcome, and \(|S|\) represents the total number of possible outcomes (Total Outcomes), denoted by N.

Example: Sample Space of rolling two six-sided dice.

# Ruang Sampel (S) saat melempar dua dadu (dadu 1, dadu 2)
# Setiap hasil adalah pasangan terurut (d1, d2).
# Total kemungkinan hasil: 6 * 6 = 36

Total_Outcomes_N_Dice <- 6 * 6

# Contoh beberapa hasil dalam ruang sampel S
Example_Outcomes <- c("(1, 1)", "(1, 2)", "(2, 1)", "(6, 6)")

# Tampilkan hasil
cat("Example Outcomes in S:", Example_Outcomes, "\n")

## Example Outcomes in S: (1, 1) (1, 2) (2, 1) (6, 6)

cat("Total Outcomes (N) for Two Dice:", Total_Outcomes_N_Dice, "\n")

## Total Outcomes (N) for Two Dice: 36

2.3 Complement Rule

Interpretation:
The Complement Rule states that the probability of an event not happening is equal to1 minus the probability of the event happening.
The complement of event A is written as A′ (A prime) or Aᶜ.

This rule is useful when calculating probabilities of events that are easier to evaluate by first determining what does not occur.

Formula:
\[ P(A') = 1 - P(A) \]

Example: Probability of not drawing a Queen from a standard deck of 52 cards.

# Probabilitas menggunakan Aturan Komplemen (Complement Rule)

# Probabilitas mendapatkan Queen (ada 4 Queen dari 52 kartu)
P_Queen <- 4 / 52  

# Aturan komplemen: probabilitas TIDAK mendapatkan Queen
P_Not_Queen <- 1 - P_Queen  

# Tampilkan hasil
cat("P(Queen):", round(P_Queen, 4), "\n")             # Menampilkan peluang mendapatkan Queen

## P(Queen): 0.0769

cat("P(Not Queen):", round(P_Not_Queen, 4), "\n")     # Menampilkan peluang tidak mendapatkan Queen

## P(Not Queen): 0.9231

Conclusion Fundamental probability concepts help us understand how likely events are to occur. By knowing the sample space, event types, and basic rules—such as addition, multiplication, complements, independent, and dependent events—we can analyze uncertain situations more accurately and make better predictions.

3.1 Independent Events

Interpretation:
Independent events occur when the outcome of one event does not affect the probability of the other event.
Examples include flipping a coin and rolling a dice—neither action influences the other.

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

Example:
A coin is flipped and a dice is rolled.
Find the probability that the coin shows= Tails= and the dice shows an even number.

Probability of Tails: \[ P(\text{Tails}) = \frac{1}{2} \]

Probability of even number (2, 4, 6): \[ P(\text{Even}) = \frac{3}{6} = \frac{1}{2} \]

Thus: \[ P(\text{Tails and Even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

# Contoh kejadian independent
# Kejadian A = mendapatkan Tails
# Kejadian B = mendapatkan angka genap pada dadu

P_Tails <- 1/2        # peluang Tails
P_Even  <- 3/6        # angka genap (2,4,6) dari 6 angka

# Hitung peluang gabungan karena independent
P_Combined_Ind <- P_Tails * P_Even

cat("P(Tails):", P_Tails, "\n")

## P(Tails): 0.5

cat("P(Even Number):", P_Even, "\n")

## P(Even Number): 0.5

cat("P(Tails and Even):", P_Combined_Ind, "\n")

## P(Tails and Even): 0.25

3.2 Dependent Events

Interpretation:
Dependent events occur when the outcome of the first event affects the probability of the second event.
A common example is drawing objects without replacement, where the total number of items changes after the first draw.

Formula:
\[ P(A \cap B) = P(A) \times P(B|A) \]

Example:
A box contains 4 blue balls and 3 red balls (7 total).
Two balls are drawn without replacement.
Find the probability that the first ball is blue and the second ball is red.

Probability the first ball is blue: \[ P(\text{Blue First}) = \frac{4}{7} \]

After drawing a blue ball, 6 balls remain (3 red, 3 blue).
Probability the second ball is red: \[ P(\text{Red Second | Blue First}) = \frac{3}{6} \]

Thus: \[ P(\text{Blue then Red}) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7} \]

P_Blue_First <- 4/7       
P_Red_After_Blue <- 3/6   

P_Combined_Dep <- P_Blue_First * P_Red_After_Blue

cat("P(Blue First):", P_Blue_First, "\n")

## P(Blue First): 0.5714286

cat("P(Red Second | Blue First):", P_Red_After_Blue, "\n")

## P(Red Second | Blue First): 0.5

cat("P(Blue then Red):", P_Combined_Dep, "\n")

## P(Blue then Red): 0.2857143

Conclusion: The key difference between independent and dependent events lies in whether the probability of the second event changes after the first event occurs. If the probability stays the same → the events are independent. If the probability changes → the events are dependent.

4.1 Review of Basic Probability Definitions

Interpretation: The video begins with a review of basic probability concepts.
To understand the union of events, it is important to first understand:

• Sample space: the set of all possible outcomes of an experiment
  
• Probability of an event:
  \[ P = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \]

These definitions are the foundation for understanding how probabilities combine when two events occur together or overlap.

4.2 Union of Events

Interpretation: The video introduces the concept of Union of Events, written as \(A \cup B\), which indicates that:

• Event A occurs,
  
• OR event B occurs,
  
• OR both occur.

Using a Venn diagram helps visualize how the events overlap.

The formula for calculating the probability of a union is:

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

The subtraction of \(P(A \cap B)\) prevents double-counting outcomes that belong to both A and B.

Example: When rolling two dice, the sample space contains 36 possible outcomes.
Using the union formula allows us to correctly calculate the probability of events involving combined results from both dice.

5.1 Mutually Exclusive Events

Interpretation:
Two events are called mutually exclusive if they cannot occur at the same time.
If one event happens, the other one is guaranteed not to happen.

Formula:
Because they do not overlap: \[ P(A \cap B) = 0 \]

Thus, the union becomes: \[ P(A \cup B) = P(A) + P(B) \]

Example:
Rolling a single die:
- A = outcome is 2
- B = outcome is 5

These events cannot occur simultaneously, so: \[ P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]

# Mutually Exclusive Example
# Event A = roll a 2
# Event B = roll a 5

P_A <- 1/6
P_B <- 1/6

# Because A and B cannot happen together
P_union <- P_A + P_B

cat("P(A):", P_A, "\n")

## P(A): 0.1666667

cat("P(B):", P_B, "\n")

## P(B): 0.1666667

cat("P(A ∪ B):", P_union, "\n")

## P(A ∪ B): 0.3333333

5.2 Exhaustive Events

Interpretation:
Events are called exhaustive if together they cover the entire sample space.
This means no possible outcome is left out.

Formula:
If A and B together include all outcomes: \[ P(A \cup B) = 1 \]

Example:
Rolling a die:
A = outcome is even {2, 4, 6}
B = outcome is odd {1, 3, 5}

Together, they cover all 6 outcomes:
\[ P(A \cup B) = 1 \]

# Exhaustive Events Example
# A = even numbers
# B = odd numbers

P_even <- 3/6
P_odd  <- 3/6

# Because even + odd cover all possible outcomes
P_union_exhaustive <- P_even + P_odd

cat("P(Even):", P_even, "\n")

## P(Even): 0.5

cat("P(Odd):", P_odd, "\n")

## P(Odd): 0.5

cat("P(A ∪ B):", P_union_exhaustive, "\n")

## P(A ∪ B): 1

Conclusion Mutually exclusive events cannot occur at the same time, so their intersection is zero.
Exhaustive events together include all possible outcomes, so their union equals one.
These concepts help determine how events relate within the sample space and how their probabilities should be combined.

6.1 Binomial Experiment

Interpretation:
A process is called a binomial experiment if it meets the following four conditions:

1. It consists of n repeated trials.
   
2. Each trial has only two possible outcomes (success or failure).
   
3. The probability of success p remains constant.
   
4. Each trial is independent.

Example:
In the video:
- n = 3 trials
- p = 0.70 probability of success
We calculate the probability for X = 0, 1, 2, 3 successes.

6.2 Binomial Formula

Interpretation:
The binomial formula gives the probability of getting exactly k successes in n trials.

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

Example:
Find the probability of exactly 2 successes out of 3 trials
with p = 0.70.

# Binomial Experiment (Video Example)

n <- 3       # number of trials
p <- 0.70    # probability of success
x <- 0:n     # possible number of successes

# Binomial probabilities for X = 0, 1, 2, 3
prob_table <- data.frame(
  Successes = x,
  Probability = dbinom(x, n, p)
)

print("Binomial Experiment Probabilities:")

## [1] "Binomial Experiment Probabilities:"

print(prob_table)

##   Successes Probability
## 1         0       0.027
## 2         1       0.189
## 3         2       0.441
## 4         3       0.343

# Binomial Formula Example
k <- 2  # exactly 2 successes

P_exact_k <- choose(n, k) * p^k * (1 - p)^(n - k)

cat("\nProbability of exactly 2 successes (Binomial Formula):\n")

## 
## Probability of exactly 2 successes (Binomial Formula):

cat("P(X = 2) =", P_exact_k, "\n")

## P(X = 2) = 0.441

Conclusion A binomial experiment deals with repeated trials that have only success or failure. The binomial formula allows us to compute the probability of exactly k successes. Both concepts work together to describe discrete probability events.