Fityanandra Athar Adyaksa (52250059) November 30,
2025
ESSENTIALS OF PROBABILITY
You can think of probability as the voice of reason in statistics. It
offers a coherent framework to navigate uncertainty so that our
decisions are guided by logic, not just guesswork.
Instead of leaning on intuition or relying on conjecture, probability
transforms data analysis from a guessing game into an exact science by
giving us the critical tools to quantify likelihoods, interpret
patterns, and analyze real-world phenomena. That’s why, for anyone
serious about research or evidence-based practice, having a strong
command of probability isn’t optional—it’s the foundation of everything
you do.
1 Basic Probability Concepts
1.1 Definition of Probability
Probability is defined as the likelihood of an event occurring,
calculated using the formula of the number of favorable outcomes divided
by the total number of possible outcomes. For example, in a single coin
flip, the probability of getting heads is \(\frac{1}{2}\) or 50%, since there is one
favorable outcome (heads) out of two possibilities (heads or tails). The
general formula is
\[
P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of
possible outcomes}}
\]
1.2 Sample Space
The Sample Space (\(S\)) encompasses
the entire collection of possible outcomes from an experiment. For two
coin flips, the sample space consists of four possibilities: HH, HT, TH,
and TT (where H denotes heads and T denotes tails).The Sample Space is
best represented with a Tree Diagram, which shows the branches of the
first flip (H or T) followed by the second flip (H or T), enabling the
systematic identification of all outcomes.
Since these are independent events (one flip doesn’t affect the
next), each outcome has an equal probability of:\[0.5 \times 0.5 = 0.25\]or 25%.
1.3 Basic Probability Rules
The core of probability is governed by two fundamental rules:
1.3.1 The Bounds Rule
The probability of any event \(A\) must
always range between 0 and 1, where 0 means impossible and 1 means
certain.
\[
0 \le P(A) \le 1
\]
1.3.2 The Sum Rule
The sum of probabilities of all outcomes in the sample space always
equals 1; for instance, with two coin flips:
1.3.3 The Multiplication Rule (Independent Events)
For independent events \(A\) and \(B\), the probability of both occurring is
the product of their individual probabilities:
\[
P(A \cap B) = P(A) \times P(B)
\]
such as the probability of two heads: \(P(H) \times P(H) = 0.5 \times 0.5 =
0.25\).
1.4 Complement Rule
The Complement Rule states that the probability of an event not
occurring is 1 minus the probability that it will occur. It is
highly efficient for calculating “at least one” or “not all” scenarios.
\[
P(A^c) = 1 - P(A)
\]
For example, the probability of not getting two tails (\(TT\)) in two coin flips is:
\[
1 - P(TT) = 1 - 0.25 = \textbf{0.75}
\]
This is equivalent to summing the probabilities of the three other
outcomes (\(HH + HT + TH\)).
1.5 Application Example
To calculate the probability of at least one tail in two flips, we
identify outcomes HT, TH, and TT (total probability 0.75).
However, this is calculated most efficiently using the Complement
Rule:
\[
P(\text{at least one T}) = 1 - P(\text{No Tails}) = 1 - P(HH)
\]
\[
P(\text{at least one T}) = 1 - 0.25 = \textbf{0.75}
\]
This practice reinforces understanding of sample spaces and the
Complement Rule in solving basic probability problems.
2 Independent and Dependent Events
2.1 Independent Events
Independent events occur when the outcome of one event does not affect
the probability of another event. For example, rolling a die and
flipping a coin are independent because the die roll does not influence
the coin flip result. The probability of both events \(A\) and \(B\) occurring together (the Intersection)
is calculated using the simple multiplication rule:
\[P(A \cap B) = P(A) \times
P(B)\]
Calculating Probability of Independent Events To find the probability
of rolling a 5 on a 6-sided die and getting heads on a coin flip, first
determine individual probabilities: \(P(5) =
\frac{1}{6}\) and \(P(\text{heads}) =
\frac{1}{2}\). Multiply them: \(P(5
\cap \text{heads}) = \frac{1}{6} \times \frac{1}{2} =
\frac{1}{12}\) (or approximately \(0.0833\)).
2.2 Dependent Events
Dependent events occur when the outcome of one event affects the
probability of the next event. This typically happens in scenarios
without replacement, such as drawing items from a box. For instance,
drawing marbles from a box with 7 green and 3 blue (total 10) without
replacement changes the sample space after each draw. The formal
calculation for dependent events involves Conditional Probability—the
probability of \(B\) given that \(A\) has already occurred (\(P(B|A)\)):
\[P(A \cap B) = P(A) \times
P(B|A)\]
Probability of Dependent Events (Green then Blue) The probability of
drawing a green marble first (\(P(G_1) =
\frac{7}{10}\)) followed by a blue marble (\(P(B_2|G_1) = \frac{3}{9}\)) without
replacement is: \[P(\text{green then blue}) =
\frac{7}{10} \times \frac{3}{9} = \frac{21}{90} = \frac{7}{30} \approx
0.233\] Note how the second probability adjusts due to the
reduced total (9 marbles left).
Example: Two Green Marbles
For drawing two green marbles without replacement:
First green: \(P(G_1) =
\frac{7}{10}\)
Second green: \(P(G_2|G_1) =
\frac{6}{9}\) (Since one green marble is gone) Thus, \(P(\text{two greens}) = \frac{7}{10} \times
\frac{6}{9} = \frac{42}{90} = \frac{7}{15} \approx
0.4667\).
Summary
Independent events use simple multiplication \(P(A) \times P(B)\).
Dependent events require sequential adjustment using the
Conditional Probability formula \(P(A) \times
P(B|A)\).
Recognize “without replacement” as the key indicator for
dependence in probability calculations.
3 Union of Events
3.1 Understanding the Union of Events
The union of two events, denoted as \(A
\cup B\), represents the probability that either event A occurs,
or event B occurs, or both occur simultaneously. This concept is crucial
when probability questions use the word “or” – indicating we want the
chance of at least one of the events happening. The key challenge is
avoiding double-counting outcomes that belong to both events.
3.2 The Union Formula
The probability of the union is calculated using the
inclusion-exclusion formula:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Here, \(P(A \cap B)\) is the
intersection – the probability that both events happen together.
Subtracting this term prevents counting overlapping outcomes twice when
adding \(P(A)\) and \(P(B)\).
3.2.1 Special Case: Mutually Exclusive Events
If events A and B are mutually exclusive (they cannot occur
together), then \(P(A \cap B) = 0\).
The formula simplifies to:
\[
P(A \cup B) = P(A) + P(B)
\]
This makes calculation straightforward since there’s no overlap.
3.2.2 Detailed Example: Rolling Two Dice
Consider rolling two six-sided dice (36 total outcomes in the sample
space). Calculate the probability of rolling two even numbers OR at
least one 2:
Event A (two even numbers): 9 favorable outcomes → \(P(A) = \frac{9}{36} = 0.25\)
Event B (at least one 2): 11 favorable outcomes → \(P(B) = \frac{11}{36} \approx 0.3056\)
Intersection \(A \cap B\) (two
evens AND at least one 2): 5 overlapping outcomes → \(P(A \cap B) = \frac{5}{36} \approx
0.1389\)
This result correctly accounts for all unique outcomes satisfying
either condition.
4 Exclusive and Complete Events
4.1 Mutually Exclusive and Exhaustive Events
4.1.1 What Are Mutually Exclusive Events?
Mutually exclusive events are events that cannot happen
simultaneously – if one occurs, the other definitely cannot. Think of
flipping a coin: you can get heads or tails, but never both in a single
flip. Mathematically, their intersection has zero probability:
\[
P(A \cap B) = 0
\]
This means no outcome belongs to both events. A real-world example:
rolling a die and getting a 1 versus getting an
even number (2, 4, 6) – these overlap, so they are
not mutually exclusive.
4.1.1.1 Probability Rule for Mutually Exclusive Events
For mutually exclusive events \(A\)
and \(B\), the probability of
either occurring (union) is simply the sum of their
individual probabilities:
\[
P(A \cup B) = P(A) + P(B)
\]
Example: Probability of rolling a 1
or a 6 on a fair die:
Exhaustive events are a collection of events that
together cover every possible outcome in the sample
space. At least one must occur; their union equals the entire sample
space. For a die roll, events {1}, {2}, {3}, {4}, {5}, {6} are
exhaustive because one number always appears.
The total probability sums to certainty:
\[
\sum_{i=1}^{n} P(E_i) = 1
\]
Key Insight: Exhaustive events don’t need to be
mutually exclusive, but when combined with mutual exclusivity, they form
a complete partition of the sample space.
4.2.1.1 Union of Exhaustive Events
When events are both mutually exclusive and
exhaustive, their union is guaranteed:
\[
P(E_1 \cup E_2 \cup \dots \cup E_n) = 1
\]
Example: Coin flip outcomes {Heads, Tails} –
mutually exclusive (no overlap) and exhaustive (covers all
possibilities), so \(P(\text{Heads} \cup
\text{Tails}) = 1\).
4.2.1.2 Overlapping vs. Non-Overlapping Exhaustive Events
Non-overlapping exhaustive (ideal partition):
Like die faces – no gaps, no overlaps.
Overl apping exhaustive: Events cover all
outcomes but share some (e.g., “even” or “multiple of 3” on a die covers
everything but overlaps at 6).
Venn diagrams visualize this: mutually exclusive = separate circles;
exhaustive = circles filling the sample space rectangle.
Mistake to Avoid: Adding probabilities of
non-mutually exclusive events without subtracting overlap leads to
overestimation.
Event Type
Intersection
Union Formula
Example
Mutually Exclusive
\(P(A \cap B) = 0\)
\(P(A) + P(B)\)
Heads or Tails
Exhaustive
Covers sample space
Sum = 1
All die outcomes
Both
Partition
Sum = 1, no overlap
{1,2,3,4,5,6}
Mastering these concepts builds foundation for advanced topics like
Bayes’ theorem and conditional probability.
5 Binomial Experiment
5.1 What is a Binomial Experiment?
A binomial experiment models scenarios with repeated
independent trials, each having exactly two possible
outcomes: success (with probability \(p\)) or failure (with
probability \(q = 1 - p\)). The “bi”
prefix highlights these two outcomes, similar to bicycle (two wheels) or
binoculars (two lenses). Real-world examples include coin flips (heads =
success), quality control inspections (defective = failure), or medical
trials (recovery = success).
5.1.1 The Four Essential Conditions
For an experiment to qualify as binomial, it must satisfy all
four conditions:
Fixed number of trials (\(n\)): The experiment repeats a
specific number of times (e.g., flip coin 3 times).
Two outcomes per trial: Success or failure only (no
other possibilities).
Constant success probability (\(p\)): \(p\) remains the same for every trial.
Independent trials: One trial’s outcome doesn’t
affect others.
If any condition fails, it’s not binomial (e.g., drawing cards
without replacement violates independence).
Example 1: Coin Flips (Simple Binomial)
Question: Flip a fair coin 3 times. Probability of
exactly 1 head?
# Memuat library yang diperlukan (pastikan Anda sudah menginstal ggplot2: install.packages("ggplot2"))
library(ggplot2)
# --- Parameter Binomial ---
n_trials <- 5 # Jumlah percobaan (n)
p_success <- 0.2 # Probabilitas sukses (p)
# Membuat vektor untuk jumlah keberhasilan yang mungkin (x)
x_values <- 0:n_trials
# Menghitung probabilitas binomial untuk setiap x
# dbinom(x, size=n, prob=p) adalah fungsi R untuk P(X=x)
probabilities <- dbinom(x_values, size = n_trials, prob = p_success)
# Membuat data frame untuk plotting
binomial_data <- data.frame(
Successes = factor(x_values),
Probability = probabilities
)
# Membuat plot menggunakan ggplot2
ggplot(binomial_data, aes(x = Successes, y = Probability)) +
geom_bar(stat = "identity", fill = "skyblue", color = "darkblue") +
geom_text(aes(label = round(Probability, 3)), vjust = -0.5, size = 3.5) +
labs(
title = paste("Binomial Probability Distribution (n =", n_trials, ", p =", p_success, ")"),
x = "Number of Successes (k)",
y = "Probability P(X=k)"
) +
theme_minimal() +
scale_y_continuous(limits = c(0, max(probabilities) * 1.1))
# Menambahkan Insight Kunci
cat("The graph shows a probability distribution where most outcomes are expected to occur around the Mean mu = n*p =", n_trials * p_success)
## The graph shows a probability distribution where most outcomes are expected to occur around the Mean mu = n*p = 1
Binomial distribution forms the basis for more advanced models like
Poisson or normal approximations.
6 Binomial Distribution
6.1 Understanding the Binomial Distribution
The binomial distribution shows the probabilities of
getting exactly \(k\) successes in
\(n\) independent trials, each with
success probability \(p\). It builds
directly from the binomial formula reviewed earlier:
\[
P(k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
# Load the required library
library(ggplot2)
n_simple <- 2
p_simple <- 0.5
x_simple <- 0:n_simple
prob_simple <- dbinom(x_simple, size = n_simple, prob = p_simple)
df_simple <- data.frame(Successes = factor(x_simple), Probability = prob_simple)
ggplot(df_simple, aes(x = Successes, y = Probability)) +
# Create a bar chart (column chart)
geom_bar(stat = "identity", fill = "lightgreen", color = "darkgreen") +
# Add probability labels on top of the bars
geom_text(aes(label = round(Probability, 2)), vjust = -0.3) +
# Set titles and labels
labs(
title = paste("Simple Binomial Distribution (n=2, p=0.5)"),
x = "Number of Successes (k)",
y = "P(X=k)"
) +
# Use a clean theme
theme_minimal()
Visualized as a bar chart, the x-axis shows possible successes (0 to
\(n\)), and y-axis shows their
probabilities. For coin flips (\(n=2\),
\(p=0.5\)): P(0 heads)=0.25, P(1
head)=0.50, P(2 heads)=0.25 – symmetric and bell-shaped.
6.1.1 Key Parameters of Binomial Distribution
For a binomial random variable \(X\):
Mean (expected successes): \(\mu = np\)
Variance: \(\sigma^2 =
np(1-p)\)
Standard deviation: \(\sigma = \sqrt{np(1-p)}\)
These center the distribution. For \(n=10\), \(p=0.5\): \(\mu=5\), \(\sigma
\approx 1.58\), peaking around 5 successes.
6.2 Effect of Changing \(p\) (Shape
Control)
The parameter \(p\) determines
skewness while keeping \(n\) fixed:
