Probabillity is a foundational pillar of statistical reasoning,
offering a systematic and coherent framework for understanding
uncertainty and guiding informed decision-making. Rather than relying on
intuition or conjecture, probability enables us to quantify the
likelihood of various outcomes, interpret patterns within data, and
analyze phenomena that arise from natural or experimental processes. A
strong command of probability concepts is essential for effective data
analysis, scientific research, and evidence-based practice.
2 Fundamental
Concept
2.1 Definition of
Probability
Probability is defined as the likelihood that an event will occur. It
is calculated using the formula: Total number of favorable outcomes /
Total number of possible outcomes.
Example: The probability of getting “heads” when flipping a coin is
1/2, or 50% (0.5), as there is only one favorable outcome (heads) out of
two possible outcomes (heads or tails).
2.2 Probability of
Independent Events
To find the probability of two independent events occurring together,
multiply the probability of each event.
Example: The probability of getting “heads” twice in a row when
flipping a coin twice is 0.5 = 0.25 or 25%.
2.3 Sample Space
The sample space refers to the entire set of possible outcomes. When
flipping a coin twice, a sample space diagram can be constructed to
determine all possible outcomes: HH, HT, TH, and TT, for a total of four
possible outcomes. The probability of each individual outcome is 0.5 =
0.25.
Example Problem: The probability of getting at least one “tails” in
two coin flips is the sum of the probabilities for the outcomes HT, TH,
and TT: 0.25 + 0.25 + 0.25 = 0.75.
2.4 Rules of
Probability
All probability problems must always satisfy two conditions: The
probability of an event occurring must always have a value between 0 and
1, inclusive of 0 and 1.
A probability of 0 means the event will never happen.
A probability of 1 means the event will always happen.
A probability of 0.5 means the event is expected to occur 50% of
the time.
(The probabilities of all possible outcomes must always sum up to
1).
2.5 The Complement
Rule
This rule states that the probability of an event not occurring is
equal to 1 minus the probability of the event occurring.
The formula is: P(A^c) = 1 - P(A), where P(A^c) is the probability of
event A not occurring (the complement of A).
Example: What is the probability of not getting two “tails” when
flipping two coins? P() = 1 - P(). P() = 1 - 0.25 = 0.75. This result is
the same as summing the probabilities of all other outcomes (HH + HT +
TH): 0.25 + 0.25 + 0.25 = 0.75. The video concludes by noting that there
are many ways to solve probability problems, and users should use the
method that works best for them.
3 Independent and
Dependent
3.1 Independent
Events
Definition: The occurrence of one event does not affect the
probability of the other event occurring.
Example: Rolling a die and flipping a coin. The result of the die
(e.g., getting a 6) does not change the probability of the coin landing
on Heads or Tails.Formula for Joint Probability (A and B):\[P(A \text{ and } B) = P(A) \times
P(B)\]Example Problem: Calculating the probability of getting a 5
on a die and Heads on a coin is \(1/6 \times
1/2 = 1/12\).
3.2 Dependent Events
Definition: The occurrence of one event affects the probability of
the subsequent event. The probability of the second event relies on the
outcome of the first event.
Example: Drawing two marbles sequentially without replacement from a
bag containing green and blue marbles.Reason for Dependency: Because the
first marble is not put back, the total number of marbles and the number
of specific-colored marbles for the second draw change, thus altering
the probability.Formula for Joint Probability (A and B):\[P(A \text{ and } B) = P(A) \times P(B \mid
A)\](Where \(P(B \mid A)\) is
the probability of B occurring after A has occurred.)
Example Problem: Calculating the probability of drawing a green
marble, then a blue marble (without replacement). The first probability
is \(7/10\), but the second probability
becomes \(3/9\) because only 9 marbles
are left. The result is \(7/10 \times 3/9 =
7/30\).
4 Union of Events
4.1 Basic Probability
Concepts
Sample Space: The entire set of possible outcomes in a statistical
experiment.
Example: Rolling two 6-sided dice results in 36 possible outcomes.
Simple Probability: The total number of favorable outcomes divided by
the total number of possible outcomes in the sample space.
4.2 Probability of the
Union of Events
Definition: The probability of the union calculates the chance of at
least one of two events occurring (Event A OR Event B). The keyword “OR”
in a probability question typically indicates the need to use the union
of events formula.Formula for Union Probability (A or B):\[P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and
} B)\]Importance of Subtraction: The term \(P(A \text{ and } B)\) must be subtracted
because when \(P(A)\) and \(P(B)\) are added together, the outcomes
common to both events (the intersection) are counted twice
(duplication). This subtraction removes the duplicate outcomes and.
4.3 Example Problem
(Rolling Two Dice)
Event A: Rolling two even numbers.\(P(A) =
9/36\)Event B: Rolling at least one 2.\(P(B) = 11/36\)Intersection (A and B):
Rolling two even numbers and at least one 2.The result is found by
observing the overlap (intersection) between the two events, which is 5
outcomes.\(P(A \text{ and } B) =
5/36\)Union (A or B): Rolling two even numbers or at least one
2.Using the formula: \(P(A \text{ or } B) =
P(A) + P(B) - P(A \text{ and } B)\)\(P(A \text{ or } B) = 9/36 + 11/36 - 5/36 =
15/36\)
5 Exclusive and
Exhaustive
5.1 Mutually Exclusive
Events
Interpretation: This refers to two or more events that cannot happen
at the same time during a single experiment.
Conceptual Example: If you roll a single die, the outcomes of getting
a “2” and getting a “5” in one roll are mutually exclusive events.Key
Characteristic: The intersection of the two event sets is null (\(\text{P(A and B)} = 0\)).
5.2 Exhaustive
Events
Interpretation: This refers to a set of events that, when combined,
include all possible outcomes in the sample space (all possible results
of an experiment).
Conceptual Example: When rolling a die, the events of getting an
“even number” (2, 4, 6) and an “odd number” (1, 3, 5) are exhaustive
events because their union covers all possible results (1, 2, 3, 4, 5,
6).Key Characteristic: The union of all event sets equals the entire
sample space (\(\text{P(A or B)} =
1\)).
6 Binomial
Experiment
6.1 Core Concept of the
Binomial Distribution
The binomial probability distribution refers to the probability of a
success or a failure in an experiment that is repeated multiple times.
The prefix bi (two) refers to the two possible outcomes: success or
failure.
6.2 The Four Conditions
for a Binomial Experiment
An experiment can only be called a Binomial Experiment if it
satisfies all four of the following conditions:
Fixed Number of Trials (\(n\)):
The number of repetitions must be definite and set.
Two Possible Outcomes: Each trial must only have two results:
success or failure.
Constant Probability of Success (\(p\)): The probability of success (\(p\)) must remain the same for every
trial.
Independent Trials: The outcome of one trial does not influence
the outcome of another trial.
6.3 Application
Examples
Example 1: Flipping a Coin 3 TimesQuestion: Calculate the probability
of getting exactly one head from 3 coin flips.Analysis: This experiment
meets the four binomial conditions (\(n=3,
p=0.5\)).Manual Result: There are 3 ways to get 1 head (HTT, THT,
TTH). The total probability is \(\mathbf{0.375}\).
Example 2: Drawing Marbles with ReplacementCrucial Condition: The
drawing must be done with replacement to ensure that each trial remains
independent and the probability of success is constant.Question:
Calculate the probability of drawing exactly 2 green marbles out of 5
draws.Probabilities: \(p\)
(success/green) \(= 2/10 = 0.2\). \(1-p\) (failure/not green) \(= 8/10 = 0.8\).Manual Result: There are 10
ways to get 2 successes and 3 failures. The total probability is \(\mathbf{0.2048}\).
6.4 The Binomial
Formula
To simplify lengthy calculations, the Binomial Formula is used:\[P(k) = \binom{n}{k} p^k
(1-p)^{n-k}\]Formula Components:\(P(k)\): The probability of getting exactly
\(k\) successes.\(\binom{n}{k}\): The Combination formula (n
choose k) which calculates the number of ways to achieve \(k\) successes in \(n\) trials.\(p^k\): The probability of success (\(p\)) raised to the power of the number of
successes (\(k\)).\((1-p)^{n-k}\): The probability of failure
(\(1-p\)) raised to the power of the
number of failures (\(n-k\)).Formula
Application: Applying the formula to the Marble Example (\(n=5, k=2, p=0.2\)) yields the same answer:
\(\mathbf{0.2048}\).
7 Binomial
Distribution
7.1 Visualizing the
Binomial Distribution
The binomial distribution is visualized using a bar chart or
histogram, where:The X-axis shows the possible values of \(k\) (the number of successes).The Y-axis
shows the probability \(P(k)\) for each
value of \(k\).
7.2 Parameters of the
Binomial Distribution
If a variable \(X\) follows a
binomial distribution, we can calculate its parameters (center and
dispersion) using \(n\) (the number of
trials) and \(p\) (the probability of
success):Mean (\(\mu\)): \(\mu = n \times p\)Variance: \(np(1-p)\)Standard Deviation (\(\sigma\)): \(\sigma = \sqrt{np(1-p)}\)
7.3 Influence of \(n\) and \(p\) on the Distribution Shape
A. Influence of \(n\) (Number of
Trials)As \(n\) increases (e.g., from 2
to 10), the shape of the binomial distribution begins to resemble the
normal distribution (bell curve).
B. Influence of \(p\) (Probability
of Success)The value of \(p\) controls
the shape’s skewness:\(p = 0.5\): The
distribution will be symmetrical (not skewed), with the mean (\(\mu\)) located exactly in the center.\(p < 0.5\): The distribution will be
skewed to the right. This means there is a higher probability of getting
a small number of successes (\(k\))
(closer to 0).\(p > 0.5\): The
distribution will be skewed to the left. This means there is a higher
probability of getting a large number of successes (\(k\)) (closer to \(n\)).In general, the data will always
cluster around the Mean (\(\mu =
np\)).
7.4 Normal Approximation
Guideline
As \(n\) increases, even skewed
binomial distributions will approach a normal shape. For calculation
purposes, we can assume a normal approximation to the binomial
distribution if both of the following conditions are met:\(n \times p \ge 10\)\(n \times (1-p) \ge 10\)(Note: Some
guidelines use the value 5, but 10 is a common general guideline).
