Everything about Special Discrete Distributions

Mymensingh Engineering College
Department of Computer Science and Engineering
COURSE NAME: Introduction to Probability and Statistics
COURSE CODE: Assignment (Group-9)
TOPIC: Special Discrete Distributions
Submitted to:
Mahfujur Rahman Rasel
Lecturer, Mathematics
Mymensingh Engineering College
Submitted By:
Md. Shakil Anower Samrat | Roll: 307 | Reg: 2850
Asef Shahriar Shan | Roll: 312 | Reg: 2855
Sazid Hasnine | Roll: 319 | Reg: 2862
Md Sujon Ahmed | Roll: 328 | Reg: 2870
Date of submission: 21-03-2024 Signature:
MATH-3205;

write it in vast details`Special Discrete Distributions
Introduction
Explanation of discrete distributions
Importance in various fields like statistics, probability theory, and data science
Overview of special discrete distributions to be covered`
Please write in English language.
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Explanation of Discrete Distributions:
Discrete distributions are a fundamental concept in probability theory and statistics. Unlike
continuous distributions, which have outcomes that can take on any value within a range, discrete
distributions deal with outcomes that can only take on distinct, separate values. This makes them
particularly useful for modeling scenarios where outcomes are countable, such as the number of
students in a classroom, the number of defects in a product, or the number of goals scored in a
soccer match.
Importance in Various Fields:
Statistics: Discrete distributions are used extensively in statistical analysis to describe the
probabilities of different outcomes in experiments or observations. They form the basis for
understanding and making inferences from data.
Probability Theory: In probability theory, discrete distributions help us calculate the
probabilities of events occurring in a discrete sample space. This is essential for making
predictions and understanding the likelihood of various outcomes.
Data Science: With the rise of data science and its focus on analyzing large datasets, discrete
distributions play a crucial role. They are used for modeling and understanding categorical
data, counts, frequencies, and other discrete variables commonly found in real-world datasets.
Overview of Special Discrete Distributions to be Covered:
In this detailed exploration of special discrete distributions, we will cover several important
distributions, each with its unique characteristics, formulas, and applications. Here's an overview of
what will be discussed:
1. Bernoulli Distribution
Definition and parameters
Probability mass function (PMF)
Mean and variance
Applications in trials with binary outcomes, like coin flips
2. Binomial Distribution
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Introduction
Special Discrete Distributions

Mean and variance
Connection to Bernoulli trials
Applications in scenarios with a fixed number of independent Bernoulli trials
3. Geometric Distribution
Mean and variance
Applications in modeling the number of trials needed until the first success occurs
4. Negative Binomial Distribution
Mean and variance
Applications in modeling the number of trials until a specified number of failures occur
5. Poisson Distribution
Mean and variance
Applications in modeling the number of events occurring in a fixed interval of time or
space
6. Hypergeometric Distribution
Mean and variance
Applications in sampling without replacement scenarios, like drawing items from a finite
population
7. Multinomial Distribution
Mean and variance
Applications in scenarios with more than two outcomes, such as rolling multiple dice or
categorizing data into multiple classes
8. Pascal Distribution
Mean and variance
Applications in modeling the number of trials until a specified number of successes occur
9. Zero-Inflated Poisson (ZIP) Distribution

Mean and variance
Applications in modeling count data with excessive zeros
These distributions are foundational in probability theory and statistics, providing powerful tools for
analyzing and interpreting data in various fields. We will delve into each distribution, exploring its
properties, formulas, mean, variance, and practical applications. Understanding these special
discrete distributions is essential for anyone working with data, making predictions, or drawing
conclusions from experiments and observations.
You
write it in vast details `Bernoulli Distribution
Mean and variance
Applications in trials with binary outcomes, like coin flips`
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Bernoulli Distribution
Definition and Parameters:
The Bernoulli distribution is one of the simplest and most fundamental discrete probability
distributions. It models a random experiment with two possible outcomes: success (often denoted
as 1) and failure (denoted as 0). The distribution is named after Jacob Bernoulli, a Swiss
mathematician. The key parameter of the Bernoulli distribution is , which represents the
probability of success in a single trial.
Parameter : Probability of success (where )
Probability Mass Function (PMF):
The probability mass function (PMF) of a Bernoulli distribution is straightforward due to its
simplicity. It is defined as:
Where:
is the probability that the random variable takes on the value .
When (success), the probability is .
When (failure), the probability is .
p
p 0 ≤ p ≤ 1
P(X = x) = {
p
1 − p
if x = 1
if x = 0
P(X = x) X x
x = 1 p
x = 0 1 − p

Mean and Variance:
The mean (expected value) and variance of a Bernoulli distribution are important measures of its
central tendency and variability, respectively.
Mean : The mean of a Bernoulli distribution is simply the probability of success:
Variance : The variance of a Bernoulli distribution is given by:
Applications in Trials with Binary Outcomes (e.g., Coin Flips):
The Bernoulli distribution finds numerous applications, especially in scenarios where there are only
two possible outcomes in each trial. Some common examples include:
Coin Flips: A classic example is a fair coin flip, where the outcome can be heads (success) with
probability or tails (failure) with probability .
Product Quality Control: In manufacturing, a product might be classified as defective (failure)
or non-defective (success) based on certain criteria.
Medical Testing: Consider a medical test that gives a positive result (success) if a patient has a
certain disease and a negative result (failure) otherwise.
Marketing Response: Analyzing whether customers respond (success) or not (failure) to a
marketing campaign.
Sports Outcomes: Modeling whether a team wins (success) or loses (failure) a game.
Example Calculation:
Let's say we have a biased coin where the probability of getting heads (success) is . We can
calculate the mean and variance:
Mean:
Variance:
So, for this biased coin:
The mean (expected value) is .
The variance is .
Conclusion:
The Bernoulli distribution is a foundational concept in probability theory, serving as the building
block for more complex distributions like the binomial and geometric distributions. Its simplicity
and clear interpretation make it a valuable tool in modeling scenarios with binary outcomes.
μ
μ = E(X) = p
σ2
σ =
2
Var(X) = p(1 − p)
p = 0.5 1 − p = 0.5
p = 0.3
μ = p = 0.3
σ =
2
p(1 − p) = 0.3 × (1 − 0.3) = 0.3 × 0.7 = 0.21
0.3
0.21

Whether in coin flips, quality control, medical testing, or marketing analytics, understanding the
Bernoulli distribution allows for the analysis and prediction of events with two possible outcomes.
You
write it in vast details`Geometric Distribution
Mean and variance
Applications in modeling the number of trials needed until the first success occurs`
write it in vast detiles`Negative Binomial Distribution
PMF
Mean and variance
Applications in modeling the number of trials until a specified number of failures occur`
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Negative Binomial Distribution
The Negative Binomial distribution is a discrete probability distribution that models the number of
trials needed until a specified number of failures occurs, where each trial results in either a
success or a failure. It is called "negative" because it counts the number of failures before reaching a
certain number of successes.
Parameters:
: The number of failures we are interested in achieving.
: The probability of success on an individual trial.
The probability mass function of the Negative Binomial distribution describes the probability of
observing failures before achieving the desired successes. It is given by:
Where:
is the number of failures before the th success.
is the binomial coefficient, representing the number of ways to arrange failures
among trials.
r
r
p
k r
P(X = k) = p (1 −
( k
k+r−1
) r
p)k
k r
( k
k+r−1
) k
k + r − 1

is the probability of success on a single trial.
is the probability of failure on a single trial.
Mean and Variance:
The mean and variance of the Negative Binomial distribution are determined by its parameters
and .
Mean :
The mean represents the expected number of trials needed to achieve successes.
Variance :
The variance measures the spread or variability in the number of trials needed.
Applications in Modeling the Number of Trials Until a Specified Number of Failures:
The Negative Binomial distribution has several practical applications in various fields due to its
ability to model situations where we are interested in the number of trials until a certain number of
failures occur. Some examples include:
Sports Analysis: Modeling the number of losses a team might have until reaching a certain
number of wins.
Epidemiology: Studying the number of failed attempts to contract a disease before becoming
immune (after a certain number of successful immunizations).
Engineering and Quality Control: Estimating the number of defective items produced before
reaching a specified number of non-defective items.
Marketing and Sales: Predicting the number of unsuccessful sales calls before a desired
number of successful sales are achieved.
Suppose a basketball player has a free throw success rate of . We are interested in the
number of free throws ( ) the player will miss before making successful shots.
PMF for (3 misses before 5 successes):
Mean:
On average, the player is expected to take 6.25 free throws before making 5 successful
shots.
p
1 − p
r
p
μ
μ = E(X) = p
r
r
σ2
σ =
2
Var(X) = p2
r(1−p)
p = 0.8
r = 5 5
k = 3
P(X = 3) = (0.8) (0.2)
( 3
3+5−1
) 5 3
μ = =
0.8
5
6.25 trials

Variance:
The variance indicates the variability in the number of misses before reaching 5
successful shots.
Conclusion:
The Negative Binomial distribution is a versatile tool for modeling the number of trials needed until
a specified number of failures occur in a sequence of independent Bernoulli trials. Its ability to
capture the waiting time for a fixed number of successes makes it valuable in various fields, from
sports analytics to quality control and beyond. By understanding its parameters, PMF, mean, and
variance, analysts can effectively model and predict outcomes where success is achieved after a
certain number of failures.
You
write it in vast details`Poisson Distribution
PMF
Mean and variance
Applications in modeling the number of events occurring in a fixed interval of time or space`
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Poisson Distribution
The Poisson distribution is a discrete probability distribution that describes the number of events
that occur in a fixed interval of time or space when these events occur with a known constant
average rate . It is named after the French mathematician Siméon Denis Poisson.
Parameter : The average rate at which events occur in the given interval.
is typically greater than zero ( ).
The probability mass function of the Poisson distribution is given by:
Where:
is the number of events that occur in the interval.
is the base of the natural logarithm, approximately equal to 2.71828.
σ =
2
=
0.82
5×(1−0.8)
1.5625 trials2
λ
λ
λ λ > 0
P(X = k) = k!
e λ
−λ k
k
e

is the average rate of events.
is the factorial of .
Mean and Variance:
The mean and variance of the Poisson distribution are both equal to the parameter .
Mean :
The mean represents the average number of events that occur in the interval.
Variance :
The variance also equals , indicating the spread or variability in the number of events.
Applications in Modeling the Number of Events in a Fixed Interval:
The Poisson distribution is widely used in various fields to model the number of events occurring in
a fixed interval of time or space when the events are rare and independent of each other. Some
common applications include:
Traffic Flow: Modeling the number of cars passing through a toll booth in a given time period.
Call Center Analysis: Predicting the number of calls a call center receives in an hour.
Accident Analysis: Estimating the number of accidents at an intersection in a day.
Natural Events: Counting the number of earthquakes in a region over a year.
Biological Processes: Modeling the number of mutations in a DNA sequence over time.
Suppose on average customers arrive at a store every hour. We can model this using a Poisson
distribution with .
PMF for (2 customers arriving in an hour):
Mean:
On average, customers are expected to arrive per hour.
Variance:
The variance is also , indicating the variability in the number of customers arriving.
Conclusion:
λ
k! k
λ
μ
μ = E(X) = λ
σ2
σ =
2
Var(X) = λ
λ
4
λ = 4
k = 2
P(X = 2) = 2!
e 4
−4 2
μ = 4
4
σ =
2
4
4

The Poisson distribution is a powerful tool for modeling rare events that occur with a known
average rate in a fixed interval of time or space. Its simple and elegant formula allows for easy
calculation of probabilities for various numbers of events. By understanding its parameters, PMF,
mean, and variance, analysts can effectively model and predict the occurrence of events in
scenarios ranging from traffic flow and call center operations to natural phenomena and biological
processes. The Poisson distribution remains a cornerstone in probability theory and statistics for its
wide applicability and ease of use.
You
write it in vast details`Hypergeometric Distribution
PMF
Mean and variance
Applications in sampling without replacement scenarios, like drawing items from a finite
population`
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Hypergeometric Distribution
The Hypergeometric distribution is a discrete probability distribution that describes the probability
of drawing a specific number of successes from a finite population without replacement. It is
commonly used in situations where the population size is fixed, and each draw changes the
probability of subsequent draws.
Parameters:
: Total number of items in the population.
: Total number of successes in the population.
: Number of draws (samples) without replacement.
The probability mass function of the Hypergeometric distribution is given by:
Where:
is the number of successes in the sample.
is the number of ways to choose successes from items.
is the number of ways to choose failures from items.
N
K
n
P(X = k) =
(n
N
)
(k
K
)( n−k
N−K
)
k
(k
K
) k K
( n−k
N−K
) n − k N − K

is the total number of ways to choose items from a population of size .
Mean and Variance:
The mean and variance of the Hypergeometric distribution are calculated as follows:
Mean :
The mean represents the average number of successes in the sample.
Variance :
The variance measures the spread or variability in the number of successes.
Applications in Sampling Without Replacement Scenarios:
The Hypergeometric distribution is particularly useful in situations where we are sampling from a
ﬁnite population without replacement. Some common applications include:
Quality Control: Checking a batch of products for defects without replacing the items.
Election Polling: Surveying a limited population to estimate the support for a candidate.
Biological Sampling: Estimating the proportion of a species in a habitat without capturing and
releasing individuals.
Inventory Management: Selecting a random sample from a warehouse to check for damaged
items.
Lottery and Gambling: Calculating the odds of drawing speciﬁc numbers in games like poker
or bingo.
Suppose we have a deck of cards with hearts. We want to know the probability of drawing
exactly hearts when drawing cards without replacement.
PMF for :
Mean:
On average, we expect to draw approximately hearts.
Variance:
Conclusion:
(n
N
) n N
μ
μ = E(X) = N
nK
σ2
σ =
2
Var(X) = n (N
K
) ( N
N−K
) (N−1
N−n
)
52 13
2 5
k = 2
P(X = 2) =
(5
52
)
(2
13
)( 5−2
52−13
)
μ = =
52
5⋅13
≈
52
65
1.25
1.25
σ =
2
5 ⋅ (52
13
) ( 52
52−13
) (52−1
52−5
)

The Hypergeometric distribution is a valuable tool for modeling sampling without replacement
scenarios, providing probabilities for the number of successes in a sample from a finite population.
By understanding its parameters, PMF, mean, and variance, analysts can make informed decisions
in fields such as quality control, election polling, biological sampling, and more. Whether estimating
defective items in a batch or predicting the outcome of a limited survey, the Hypergeometric
distribution offers a precise way to calculate probabilities in situations where each draw changes
the likelihood of subsequent draws.
You
write it in vast details`Multinomial Distribution
PMF
Mean and variance
Applications in scenarios with more than two outcomes, such as rolling multiple dice or categorizing
data into multiple classes`
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Multinomial Distribution
The Multinomial distribution is a generalization of the binomial distribution to multiple categories. It
describes the probability of observing each possible outcome of multiple independent categorical
variables, where each variable can have more than two categories.
Parameters:
: The total number of trials or events.
: Probabilities of each of the categories, where .
The probability mass function (PMF) of the Multinomial distribution is given by:
Where:
are the counts of each category observed.
is the factorial of .
are the factorials of the counts for each category.
are the probabilities of each category.
Mean and Variance:
n
p , p , ..., p
1 2 k k p +
1 p +
2 ... + p =
k 1
P(x , x , ..., x ) =
1 2 k p p ...p
x !x !...x !
1 2 k
n!
1
x1
2
x2
k
xk
x , x , ..., x
1 2 k
n! n
x !, x !, ..., x !
1 2 k
p , p , ..., p
1 2 k

The mean and variance of the Multinomial distribution can be calculated as follows:
Mean :
The mean for each category is the number of trials times the probability of that
category.
Variance :
The variance for each category is the number of trials times the probability of that
category times the complement of its probability.
Applications in Scenarios with More Than Two Outcomes:
The Multinomial distribution is commonly used in scenarios where there are more than two possible
outcomes, each with a known probability. Some examples include:
Rolling Multiple Dice: When rolling multiple dice, each with different numbers of sides, and
calculating the probabilities of different combinations of outcomes.
Categorizing Data: In data analysis, when data can be categorized into multiple classes (e.g.,
types of products, customer segments, survey responses).
Genetics: Modeling the distribution of different genetic traits in a population.
Election Results: Predicting the distribution of votes among multiple candidates in an
election.
Market Research: Analyzing preferences among multiple product options in a survey.
Suppose we have a six-sided die and we roll it times. We want to calculate the probability of
getting each outcome (1 through 6) for each roll.
PMF for rolling once, twice, three times, once, twice, and once:
Mean:
Expected number of ones:
Expected number of twos:
Expected number of threes:
Expected number of fours:
Expected number of fives:
Expected number of sixes:
Variance:
Variance of ones:
Variance of twos:
μ
μ =
i npi
i
σ2
σ =
i
2
np (1 −
i p )
i
i
10
1 2 3 4 5 6
P(1, 2, 3, 4, 5, 6) = 1!2!3!1!2!1!
10!
(6
1
)
1
(6
1
)
2
(6
1
)
3
(6
1
)
1
(6
1
)
2
(6
1
)
1
10 × =
6
1
6
10
10 × ×
6
1
2 = 6
20
10 × ×
6
1
3 = 6
30
10 × =
6
1
6
10
10 × ×
6
1
2 = 6
20
10 × =
6
1
6
10
10 × ×
6
1
=
6
5
36
50
10 × ×
6
1
=
6
5
36
50

Variance of threes:
Variance of fours:
Variance of fives:
Variance of sixes:
Conclusion:
The Multinomial distribution is a valuable tool for modeling scenarios with multiple outcomes, each
with a known probability. By understanding its parameters, PMF, mean, and variance, analysts can
make informed decisions in various fields such as dice games, market research, genetics, and
election polling. Whether calculating probabilities for rolling different combinations of dice or
predicting the distribution of product preferences among consumers, the Multinomial distribution
provides a powerful framework for handling multiple categorical outcomes.
You
write it in vast details`Pascal Distribution
PMF
Mean and variance
Applications in modeling the number of trials until a specified number of successes occur`
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Pascal Distribution
The Pascal distribution, also known as the Negative Binomial distribution with a positive integer
value of the parameter , models the number of trials needed to achieve a specified number of
successes ( ) in a sequence of independent Bernoulli trials. It is sometimes referred to as the
"waiting time" distribution, as it represents the number of trials needed until a certain number of
successes occur.
Parameters:
: The number of successes we are interested in achieving.
: The probability of success on an individual trial.
The probability mass function (PMF) of the Pascal distribution is given by:
Where:
10 × ×
6
1
=
6
5
36
50
10 × ×
6
1
=
6
5
36
50
10 × ×
6
1
=
6
5
36
50
10 × ×
6
1
=
6
5
36
50
r
r
r
p
P(X = k) = p (1 −
(r−1
k−1
) r
p)k−r

is the number of trials needed to achieve the successes.
is the binomial coefficient, representing the number of ways to achieve successes in
the first trials.
is the probability of success on an individual trial.
is the probability of failure on an individual trial.
Mean and Variance:
The mean and variance of the Pascal distribution are:
Mean :
The mean represents the expected number of trials needed to achieve successes.
Variance :
The variance measures the spread or variability in the number of trials needed.
Applications in Modeling the Number of Trials Until a Specified Number of Successes:
The Pascal distribution is widely used in scenarios where we are interested in the number of trials
required to achieve a certain number of successes. Some common applications include:
Product Development: Modeling the number of tests needed to reach a specific number of
successful prototypes.
Insurance Claims: Estimating the number of claims processed until a certain number of high-
value claims occur.
Sports Statistics: Predicting the number of games needed for a team to win a certain number
of matches.
Clinical Trials: Calculating the number of patients needed in a trial to observe a certain
number of successful treatments.
Marketing Campaigns: Estimating the number of promotions needed to achieve a specific
number of conversions.
Suppose a basketball player makes of their free throws. We are interested in the number of
free throws they need to make successful shots.
PMF for (8 free throws until 5 successes):
Mean:
k r
(r−1
k−1
) r − 1
k − 1
p
1 − p
μ
μ = E(X) = p
r
r
σ2
σ =
2
Var(X) = p2
r(1−p)
70%
5
k = 8
P(X = 8) = (0.7) (1 −
(5−1
8−1
) 5
0.7)8−5
μ = ≈
0.7
5
7.14

On average, the player is expected to make approximately free throws to achieve 5
successful shots.
Variance:
The variance is approximately , indicating the variability in the number of free throws
needed.
Conclusion:
The Pascal distribution is a useful probability distribution for modeling the number of trials needed
to achieve a specified number of successes in a sequence of independent trials. By understanding
its parameters, PMF, mean, and variance, analysts can make informed decisions in various fields
such as product development, insurance claims processing, and clinical trials. Whether estimating
the number of tests to reach a goal or predicting the duration of a marketing campaign, the Pascal
distribution provides a framework for understanding waiting times until a specified number of
successes occur.
You
write it in vast details`Zero-Inflated Poisson (ZIP) Distribution
PMF
Mean and variance
Applications in modeling count data with excessive zeros`
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Zero-Inflated Poisson (ZIP) Distribution
The Zero-Inflated Poisson (ZIP) distribution is a mixture distribution that combines aspects of the
Poisson distribution and the point mass at zero. It is particularly useful for modeling count data with
an excessive number of zeros, where some of the zeros are due to a separate process that leads to
an excess of zeros beyond what the Poisson distribution would predict.
Parameters:
: The probability of excess zeros (a point mass at zero).
: The rate parameter for the Poisson distribution component.
The probability mass function (PMF) of the ZIP distribution is given by:
7.14
σ =
2
≈
0.72
5×(1−0.7)
2.04
2.04
p
λ

write it in vast details according to your answers`Conclusion
Summary of special discrete distributions covered
Importance of understanding these distributions in various fields
Resources for further study`
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Conclusion
In this detailed exploration, we have covered several special discrete distributions, each with its
unique characteristics, applications, and significance in various fields. Let's summarize the
distributions we discussed, highlight their importance, and provide resources for further study.
Special Discrete Distributions Covered:
1. Bernoulli Distribution:
Definition: Describes a single trial with two possible outcomes (success or failure).
PMF:
Mean and Variance:
Applications: Coin flips, binary outcomes in experiments.
2. Geometric Distribution:
Definition: Models the number of trials needed until the first success occurs.
PMF:
Mean and Variance:
Applications: Modeling waiting times, success probabilities.
3. Negative Binomial Distribution:
Definition: Models the number of trials until a specified number of failures occur.
PMF:
Mean and Variance:
Applications: Sports analytics, epidemiology, quality control.
4. Poisson Distribution:
Definition: Describes the number of events occurring in a fixed interval with a known
average rate.
PMF:
Mean and Variance:
Applications: Traffic flow, call center analysis, accident prediction.
5. Hypergeometric Distribution:
Definition: Models the number of successes in a sample drawn without replacement.
P(X = x) = p (1 −
x
p)1−x
μ = p, σ =
2
p(1 − p)
P(X = k) = (1 − p) p
k−1
μ = , σ =
p
1 2
p2
1−p
P(X = k) = p (1 −
( k
k+r−1
) r
p)k
μ = , σ =
p
r 2
p2
r(1−p)
P(X = k) = k!
e λ
−λ k
μ = λ, σ =
2
λ

PMF:
Mean and Variance:
Applications: Quality control, election polling, biological sampling.
6. Multinomial Distribution:
Definition: Generalizes the binomial distribution to multiple categories.
PMF:
Mean and Variance:
Applications: Rolling multiple dice, categorizing data into classes.
7. Pascal Distribution:
Definition: Models the number of trials needed until a specified number of successes
occur.
PMF:
Mean and Variance:
Applications: Product development, insurance claims, clinical trials.
8. Zero-Inflated Poisson (ZIP) Distribution:
Definition: Combines aspects of the Poisson distribution with a point mass at zero.
PMF:
Mean and Variance:
Applications: Healthcare data, ecology, insurance claims, epidemiology.
Importance of Understanding These Distributions:
Statistical Modeling: Special discrete distributions provide essential tools for modeling
various real-world phenomena with discrete outcomes.
Data Analysis: Understanding these distributions helps in analyzing and interpreting data
correctly, especially in fields like healthcare, finance, and social sciences.
Predictive Analytics: They enable us to make predictions and forecasts based on historical
data and probabilistic models.
Decision Making: These distributions assist decision-makers in risk assessment, resource
allocation, and strategy development.
Scientific Research: Researchers rely on these distributions to model and analyze
experimental results and observational data.
Resources for Further Study:
Books:
"Introduction to Probability" by Joseph K. Blitzstein and Jessica Hwang
P(X = k) =
(n
N
)
(k
K
)( n−k
N−K
)
μ = , σ =
N
nK 2
N (N−1)
2
nK(N−K)(N−n)
P(X = x , x , ..., x ) =
1 2 k p p ...p
x !x !...x !
1 2 k
n!
1
x1
2
x2
k
xk
μ =
i np , σ =
i i
2
np (1 −
i p )
i
P(X = k) = p (1 −
(r−1
k−1
) r
p)k−r
μ = , σ =
p
r 2
p2
r(1−p)
P(X = x) = {
(1 − p) + p ⋅ e−λ
p ⋅ x!
e λ
−λ x
if x = 0
if x = 1, 2, 3, ...
μ = p ⋅ λ, σ =
2
p ⋅ λ 1 + λ − p ⋅ λ
( )

now write all the answer after topic and questions
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Bernoulli Distribution
Questions:
1. What is the definition of the Bernoulli distribution?
Answers:
1. Definition: The Bernoulli distribution is a discrete probability distribution that models a single
trial with two possible outcomes: success (usually denoted as 1) and failure (usually denoted as
0).
p & text{if } x = 1
1-p & text{if } x = 0
end{cases} ]
where ( p ) is the probability of success and ( 1-p ) is the probability of failure.
Mean and Variance:
Mean:
Variance:
Applications:
Modeling coin flips (success = heads, failure = tails).
Binary outcomes in experiments (e.g., success = event occurs, failure = event does not
occur).
Geometric Distribution
Questions:
1. What is the Geometric distribution used to model?
2. Explain the probability mass function (PMF) of the Geometric distribution.
3. What are the mean and variance of a Geometric random variable?
4. Provide an example of a situation where the Geometric distribution is applicable.
μ = p
σ =
2
p(1 − p)
2. What are the mean and variance of a Bernoulli random variable?
3. Give an example of an application of the Bernoulli distribution in real life.
2.
3.
Simple Questions

1. Usage: The Geometric distribution models the number of trials needed until the first success
occurs in a sequence of independent Bernoulli trials.
2. PMF: The probability mass function of the Geometric distribution is:
where is the probability of success on an individual trial.
3. Mean and Variance:
Mean:
Variance:
4. Example:
The number of attempts until a basketball player makes their first free throw (with
probability ).
Negative Binomial Distribution
Questions:
1. Define the Negative Binomial distribution.
2. What is the probability mass function (PMF) of the Negative Binomial distribution?
3. Provide formulas for the mean and variance of a Negative Binomial random variable.
4. Give an example where the Negative Binomial distribution is appropriate.
Answers:
1. Definition: The Negative Binomial distribution models the number of trials needed to achieve
a specified number of failures ( ) in a sequence of independent Bernoulli trials.
2. PMF: The probability mass function of the Negative Binomial distribution is:
Mean:
Variance:
4. Example:
Modeling the number of attempts until a basketball player makes 5 free throws (with
probability ).
Poisson Distribution
Questions:
1. What is the Poisson distribution used to model?
P(X = k) = (1 − p) p
k−1
p
μ = p
1
σ =
2
p2
1−p
p
r
P(X = k) = p (1 −
( k
k+r−1
) r
p)k
μ = p
r
σ =
2
p2
r(1−p)
p
Answer:

2. Describe the probability mass function (PMF) of the Poisson distribution.
3. Provide the formulas for the mean and variance of a Poisson random variable.
4. Give an example of a real-life situation where the Poisson distribution is applicable.
Answers:
1. Usage: The Poisson distribution models the number of events occurring in a ﬁxed interval of
time or space, when events occur independently at a constant rate.
2. PMF: The probability mass function of the Poisson distribution is:
where is the average rate of occurrence.
Mean:
Variance:
4. Example:
Number of customer arrivals at a store in an hour, with an average of arrivals per hour.
Hypergeometric Distribution
Questions:
1. What does the Hypergeometric distribution model?
2. Explain the probability mass function (PMF) of the Hypergeometric distribution.
3. Provide formulas for the mean and variance of a Hypergeometric random variable.
4. Give an example where the Hypergeometric distribution is used.
Answers:
1. Usage: The Hypergeometric distribution models the number of successes in a sample drawn
without replacement from a ﬁnite population.
2. PMF: The probability mass function of the Hypergeometric distribution is:
Mean:
Variance:
4. Example:
Quality control: Sampling without replacement from a batch of products to check for
defective items.
P(X = k) = k!
e λ
−λ k
λ
μ = λ
σ =
2
λ
λ
P(X = k) =
(n
N
)
(k
K
)( n−k
N−K
)
μ = N
nK
σ =
2
N (N−1)
2
nK(N−K)(N−n)

Multinomial Distribution
Questions:
1. What is the Multinomial distribution used for?
2. Describe the probability mass function (PMF) of the Multinomial distribution.
3. What are the mean and variance of a Multinomial random variable?
4. Provide an example where the Multinomial distribution is applicable.
Answers:
1. Usage: The Multinomial distribution is used to model scenarios with more than two outcomes,
where each outcome has a known probability.
2. PMF: The probability mass function of the Multinomial distribution is:
Mean: for each category
Variance: for each category
4. Example:
Rolling multiple dice with different numbers of sides and calculating probabilities of
different combinations.
Pascal Distribution
Questions:
1. Define the Pascal
You
write all the solutions after topic and questions`Certainly! Here are some math questions related to
special discrete distributions:
### Bernoulli Distribution:
1. **Question 1:**
- In a series of Bernoulli trials with success probability ( p = 0.3 ), what is the probability of getting
exactly 2 successes in 5 trials?
2. **Question 2:**
- A biased coin has a probability of ( p = 0.6 ) of landing heads. What is the probability of getting
at least 3 heads in 5 flips?
### Geometric Distribution:
3. **Question 3:**
- In a game, a player has a ( frac{1}{4} ) chance of winning each round. What is the probability
P(X = x , x , ..., x ) =
1 2 k p p ...p
x !x !...x !
1 2 k
n!
1
x1
2
x2
k
xk
μ =
i npi i
σ =
i
2
np (1 −
i p )
i i

These questions cover various aspects of special discrete distributions and their applications. They
involve calculating probabilities, expected values, and understanding the distributions' properties.
Feel free to try solving these questions or let me know if you'd like me to provide solutions!`
ChatGPT
Certainly! Here are the solutions to the math questions related to special discrete distributions:
The probability mass function (PMF) for the Bernoulli distribution is:
Where:
(number of trials)
(number of successes)
(probability of success)
So, the probability of getting exactly 2 successes in 5 trials is approximately .
Question 2:
A biased coin has a probability of of landing heads. What is the probability of getting at
least 3 heads in 5 flips?
Solution 2:
To find the probability of getting at least 3 heads in 5 flips, we need to calculate the probabilities of
getting 3, 4, or 5 heads and add them up.
P(X = k) = p (1 −
(k
n
) k
p)n−k
n = 5
k = 2
p = 0.3
P(X = 2) = (0.3) (1 −
(2
5
) 2
0.3)5−2
P(X = 2) = (0.3) (0.7)
(2
5
) 2 3
P(X = 2) = (0.3) (0.7)
2!(5−2)!
5! 2 3
P(X = 2) = (0.3) (0.7)
2×1
5×4 2 3
P(X = 2) = 10 × 0.09 × 0.343
P(X = 2) = 0.30942
0.30942
p = 0.6
, what is the probability of getting
p = 0.3
In a series of Bernoulli trials with success probability
exactly 2 successes in 5 trials?
Q 1:
Solution 1:
Bernoulli Distribution:
Math Examples:

So, the probability of getting at least 3 heads in 5 ﬂips is approximately .
Geometric Distribution:
Question 3:
In a game, a player has a chance of winning each round. What is the probability that the
player wins on the 4th round?
Solution 3:
The probability mass function (PMF) for the Geometric distribution is:
Where:
(probability of success on each round)
(number of trials until success)
So, the probability that the player wins on the 4th round is .
Question 4:
Suppose a light bulb has a probability of burning out each day. What is the expected
number of days until it burns out?
Solution 4:
The mean of a Geometric distribution with probability is . In this case, .
So, the expected number of days until the light bulb burns out is 20 days.
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
P(X ≥ 3) = (0.6) (1 −
(3
5
) 3
0.6) +
5−3
(0.6) (1 −
(4
5
) 4
0.6) +
5−4
(0.6) (1 −
(5
5
) 5
0.6)5−5
P(X ≥ 3) = (0.6) (0.4) +
(3
5
) 3 2
(0.6) (0.4) +
(4
5
) 4 1
(0.6) (0.4)
(5
5
) 5 0
P(X ≥ 3) = (0.6) (0.4) +
3!(5−3)!
5! 3 2
(0.6) (0.4) +
4!(5−4)!
5! 4 1
(0.6) (0.4)
5!(5−5)!
5! 5 0
P(X ≥ 3) = (0.6) (0.4) +
2×1
5×4 3 2
(0.6) (0.4) +
1
5 4 1
1 × (0.6)5
P(X ≥ 3) = 10 × 0.216 × 0.16 + 5 × 0.1296 × 0.4 + 0.07776
P(X ≥ 3) = 0.3456 + 0.2592 + 0.07776
P(X ≥ 3) = 0.68256
0.68256
4
1
P(X = k) = (1 − p) p
k−1
p = 4
1
k = 4
P(X = 4) = (1 − )
4
1 4−1
4
1
P(X = 4) = ( ) ×
4
3 3
4
1
P(X = 4) = ×
64
27
4
1
P(X = 4) = 256
27
256
27
20
1
p p
1
p = 20
1
Mean(μ) = =
20
1
1
20

Negative Binomial Distribution:
Question 5:
A basketball player has a 70% free-throw success rate. What is the probability that she makes
her 8th successful free throw on her 10th attempt?
Solution 5:
The probability mass function (PMF) for the Negative Binomial distribution is:
Where:
(number of successes)
(probability of success)
(total attempts)
You
Continue writing please
ChatGPT
So, the probability that she makes her 8th successful free throw on her 10th attempt is
approximately .
Question 6:
A company receives customer complaints at a rate of 3 per day. What is the probability that
they will receive exactly 5 complaints in the next 2 days?
Solution 6:
This scenario can be modeled using the Negative Binomial distribution with (number of
successes) and (probability of success in a day). Since we are interested in the complaints over
2 days, the total number of trials will be 2.
P(X = k) = p (1 −
( k
k+r−1
) r
p)k
r = 8
p = 0.7
k = 10
P(X = 10) = (0.7) (1 −
( 10
10+8−1
) 8
0.7)10
P(X = 10) = (0.7) (0.3)
(10
17
) 8 10
P(X = 10) = (0.7) (0.3)
10!(17−10)!
17! 8 10
P(X = 10) = (0.7) (0.3)
10×9×8×7×6×5×4×3×2×1
17×16×15×14×13×12×11 8 10
P(X = 10) = (0.7) (0.3)
262144
6188 8 10
P(X = 10) = 0.02303 × 0.0576
P(X = 10) = 0.0013276
0.0013276
r = 5
p = 3
1
k
2
Now

The probability that the company will receive exactly 5 complaints in the next 2 days is
approximately .
Poisson Distribution:
Question 7:
A call center receives an average of 10 calls per hour. What is the probability that they will
receive 15 calls in a given hour?
Solution 7:
The Poisson distribution with parameter (average rate of calls per hour) is used to model this
scenario.
So, the probability that they will receive 15 calls in a given hour is approximately .
Question 8:
The number of typographical errors in a book follows a Poisson distribution with a mean of 2
errors per page. What is the probability of ﬁnding no errors on a randomly chosen page?
Solution 8:
The Poisson distribution with (mean number of errors per page) is used here.
So, the probability of ﬁnding no errors on a randomly chosen page is approximately .
P(X = 5) = ( 5
5+2−1
) (3
1
)
5
(3
2
)
2
P(X = 5) = (5
6
) (3
1
)
5
(3
2
)
2
P(X = 5) = 5!(6−5)!
6!
(3
1
)
5
(3
2
)
2
P(X = 5) = 5!
6
(3
1
)
5
(3
2
)
2
P(X = 5) = ×
5
1
×
243
1
9
4
P(X = 5) = 1215
4
P(X = 5) ≈ 0.00329218
0.00329218
λ = 10
P(X = 15) = 15!
e ×10
−10 15
P(X = 15) = 15×14×13×12×11×10!
e ×10
−10 15
P(X = 15) = 15×14×13×12×11×10×9×8×7×6×5×4×3×2×1
e ×10
−10 15
P(X = 15) = 15!
e ×10
−10 15
P(X = 15) ≈ 0.03468
0.03468
λ = 2
P(X = 0) = 0!
e ×2
−2 0
P(X = 0) = e−2
P(X = 0) ≈ 0.1353
0.1353

Hypergeometric Distribution:
Question 9:
A deck of cards has 52 cards, 13 of which are hearts. If you draw 5 cards, what is the
probability of getting exactly 2 hearts?
Solution 9:
The Hypergeometric distribution with (total cards), (hearts), and (cards drawn)
is used here.
You
ChatGPT
So, the probability of getting exactly 2 hearts when drawing 5 cards from a deck is approximately
.
Question 10:
A batch of 100 products contains 10 defective items. If 5 items are randomly selected, what is
the probability that exactly 2 of them are defective?
Solution 10:
This situation can be modeled using the Hypergeometric distribution with (total products),
(defective items), and (items selected).
N = 52 K = 13 n = 5
P(X = 2) =
(5
52
)
×
(2
13
) (3
39
)
P(X = 2) =
5!(52−5)!
52!
×
2!(13−2)!
13!
3!(39−3)!
39!
P(X = 2) =
5!47!
52!
×
2!11!
13!
3!36!
39!
P(X = 2) =
5×4×3×2×1
52×51×50×49×48
×
2×1
13×12
3×2×1
39×38×37
P(X = 2) = 311875200
78×14454
P(X = 2) = 311875200
1129392
P(X = 2) = 311875200
1129392
P(X = 2) ≈ 0.00363
0.00363
N = 100
K = 10 n = 5

The probability that exactly 2 of the 5 selected items are defective is approximately .
Multinomial Distribution:
Question 11:
A six-sided die is rolled 10 times. What is the probability of getting each outcome (1 through 6)
exactly once?
Solution 11:
The Multinomial distribution with (number of trials) and for (probability
of each outcome) is used here.
The probability of getting each outcome exactly once is calculated as:
So, the probability of getting each outcome (1 through 6) exactly once when rolling a six-sided die
10 times is approximately .
Pascal Distribution:
Question 12:
A machine produces light bulbs with a 5% defect rate. What is the probability that the 4th
defective bulb will be produced on the 8th attempt?
Solution 12:
P(X = 2) =
( 5
100
)
×
(2
10
) (3
90
)
P(X = 2) =
5!(100−5)!
100!
×
2!(10−2)!
10!
3!(90−3)!
90!
P(X = 2) =
5!95!
100!
×
2!8!
10!
3!87!
90!
P(X = 2) =
5×4×3×2×1
100×99×98×97×96
×
2×1
10×9
3×2×1
90×89×88
P(X = 2) = 75287520
45×71220
P(X = 2) ≈ 0.042
0.042
n = 10 p =
i 6
1
i = 1, 2, ..., 6
P(Each outcome once) = 1!1!1!1!1!1!
10!
(6
1
)
1
(6
1
)
1
(6
1
)
1
(6
1
)
1
(6
1
)
1
(6
1
)
1
10!
(6
1
)
6
10!
(6
1
)
6
P(Each outcome once) = 1!6
10!
(6
1
)
6
P(Each outcome once) = 720
10!
(6
1
)
6
P(Each outcome once) = ×
720
10!
46656
1
P(Each outcome once) = ×
720
3628800
46656
1
P(Each outcome once) ≈ 0.000776
0.000776

The Pascal distribution with (number of successes) and (probability of success) is
used here.
The probability that the 4th defective bulb will be produced on the 8th attempt is approximately
.
Zero-Inﬂated Poisson (ZIP) Distribution:
Question 13:
In a certain area, the number of accidents per day follows a ZIP distribution with and
. What is the probability of no accidents in a day?
Solution 13:
The ZIP distribution has a probability mass function (PMF) given by:
Where:
So, the probability of no accidents in a day is .
Question 14:
Suppose in a ZIP-distributed dataset, and . Calculate the expected number of
non-zero counts.
You
r = 4 p = 0.05
P(X = 8) = (0.05) (1 −
(4−1
8−1
) 4
0.05)8−4
P(X = 8) = (0.05) (0.95)
(3
7
) 4 4
P(X = 8) = (0.05) (0.95)
3!(7−3)!
7! 4 4
P(X = 8) = (0.05) (0.95)
3×2×1
7×6×5 4 4
P(X = 8) = 35 × 0.0000125 × 0.81450625
P(X = 8) = 0.00377668
0.00377668
p = 0.2
λ = 3
P(X = 0) = p + (1 − p)e−λ
p = 0.2
λ = 3
P(X = 0) = 0.2 + (1 − 0.2)e−3
P(X = 0) = 0.2 + 0.8e−3
P(X = 0) = 0.2 + 0.8(0.0498)
P(X = 0) = 0.2 + 0.03984
P(X = 0) = 0.23984
0.23984
p = 0.3 λ = 2

ChatGPT
Solution 14:
The expected number of non-zero counts in a ZIP distribution is given by:
Where:
So, the expected number of non-zero counts in this ZIP-distributed dataset is 1.4.
Conclusion
Summary of special discrete distributions covered:
Bernoulli Distribution: Models a single trial with two outcomes.
Geometric Distribution: Models the number of trials until the first success.
Negative Binomial Distribution: Models the number of trials until a specified number of
successes.
Poisson Distribution: Models the number of events occurring in a fixed interval.
Hypergeometric Distribution: Models sampling without replacement.
Multinomial Distribution: Models scenarios with more than two outcomes.
Pascal Distribution: Models the number of trials until a specified number of successes.
Zero-Inflated Poisson (ZIP) Distribution: Models count data with excessive zeros.
Importance of understanding these distributions:
These distributions are fundamental in probability and statistics, providing tools to model and
analyze various real-world phenomena.
Understanding these distributions helps in making informed decisions, predicting outcomes,
and analyzing data in fields like finance, biology, engineering, and more.
They serve as the building blocks for more complex statistical models and analyses.
Resources for further study:
Books:
"Introduction to Probability" by Joseph K. Blitzstein and Jessica Hwang
"Statistics for Engineers and Scientists" by William Navidi
"Probability and Statistics for Engineers and Scientists" by Ronald E. Walpole
Online Courses:
E(X) = λ(1 − p)
p = 0.3
λ = 2
E(X) = 2(1 − 0.3)
E(X) = 2 × 0.7
E(X) = 1.4
This Project was Completed by:
Md Shakil Anower Samrat
Sazid Hasnine
Md Sujon Ahmed
Asef Shahriar Shan

Everything about Special Discrete Distributions

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