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Random Variable & Discrete Distribution
Random Variable
Theoretical Probability Distribution
• Random Variable
• Discrete Probability Distributions
A variable that assumes a numerical
description for the outcome of a
random experiment (by chance).
Usually is denoted by a capital letter.
X, Y, Z, ...
Very useful for mathematically
modeling the distribution of variables.
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Discrete Random Variable
Continuous Random Variable
Discrete Random Variable:
Continuous Random Variable:
A random variable assumes
discrete values by chance.
A random variable that can
take on any value within a
specified interval by chance.
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Discrete Random Variable
Discrete Probability Distribution
Example: (Toss a balanced coin)
If a balanced coin is tossed, Head and Tail are
equally likely to occur,
X = 1, if Head occurs, and
X = 0, if Tail occurs.
P(X=1) = .5 = 1/2 and P(X=0) = .5 = 1/2
P(Head) = P(X=1) = P(1) = .5
P(Tail) = P(X=0) = P(0) = .5
Probability mass function:
P(X=x) =.5 , if x = 0, 1,
and P(X=x) = 0 , elsewhere.
1/2
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P(all possible outcomes) = P(X=1 or 0)
= P(X=1) + P(X=0)
= 1/2 + 1/2
= 1.0
Total probability is 1.
Total probability is 1.
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Random Variable - 1
Random Variable & Discrete Distribution
Discrete Uniform Probability
Distribution
Discrete Random Variable
Example: What is probability of getting a
number less than 3 when roll a balanced
die?
Probability mass function:
P(X=x) =1/6, if x=1,2,3,4,5,or 6,
and P(X=x) = 0 elsewhere.
Probability Mass Function for
Discrete Uniform Distribution:
P(x) = c, c is a constant
1/6
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2 3 4 5 6
P( X < 3 ) = P( X ≤ 2) = P(X = 1) + P(X = 2)
= 1/6 + 1/6 = 2/6
• Balanced Coin: P(x) = 1/2, for x = 0,1
• Balanced Die:
P(x) = 1/6, for x = 1,2,3,4,5,6
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Discrete Random Variable
Balanced Die:
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Other Discrete Distributions
P(x) = 1/6, for x = 1,2,3,4,5,6
P(X=3) = 1/6
P(X=5) = 1/6
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Bernoulli
Binomial
Poisson
Geometric
…
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Bernoulli Trial
Bernoulli Distribution Model
(Bernoulli Probability Distribution)
Definition: Bernoulli trial is a random
experiment whose outcomes are classified as
one of the two categories. (S , F) or (Success,
Failure) or (1, 0)
Example:
Tossing a coin, observing Head or Tail
Observing patient’s status Died or Survived.
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Random Variable - 2
Random Variable & Discrete Distribution
Bernoulli Probability Distribution
Example: In a random experiment of
casting a balanced die, we are only
interested in observing 6 turns up or
not. It is a Bernoulli trail.
Example: (Tossing a balanced coin)
P(S) = P(X=1) = p = .5
P(F) = P(X=0) = 1 − p = .5
P(6) = P(X=1) = p = 1/6
Bernoulli Distribution
P(6’) = P(X=0) =1 − 1/6 = 5/6
.5
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Bernoulli Probability Distribution
Bernoulli Distribution
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0
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Binomial Experiment
Binomial Distribution Model
(Binomial Probability Distribution)
A random experiment involving a
sequence of independent and
identical Bernoulli trials.
Example:
Toss a coin ten times, and observing Head
turns up.
Roll a die 3 times, and observing a 6 turns up
or not.
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Binomial Probability Model
In a random sample of 5 from a large
population, and observing subjects’ disease
status. (Almost binomial)
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Binomial Probability Model
In a binomial experiment involving n
independent and identical Bernoulli trials each
with probability of success p, the probability of
having x successes can be calculated with the
binomial probability mass function, and it
is, for x = 0, 1, …, n,
A model to find the probability of
having x number successes in a
sequence of n independent and
identical Bernoulli trials.
P( X = x) =
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n!
⋅ p x ⋅ (1 − p ) n − x
x!⋅(n − x)!
⎛ n⎞
= ⎜⎜ ⎟⎟ ⋅ p x ⋅ (1 − p ) n − x
⎝ x⎠
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Random Variable - 3
Random Variable & Discrete Distribution
(6, 6’, 6’)
Factorial
Binomial Probability
0! = 1
3! = 1·2·3 = 6
⎛ 5⎞
⎝ 2⎠
Example: ⎜ ⎟ =
Identify n = 3, p = 1/6, x = 2
5!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
=
= 10
(5 − 2)!2! 3 ⋅ 2 ⋅1 ⋅ 2 ⋅1
3!
·(1/6)2·(5/6)3-2
2! 1!
= 3·(1/6)2·(5/6)1
= .069
n!
P (X = x ) =
P(X=2) =
x!⋅(n − x )!
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Binomial Probability
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Example: In the previous problem, what is the
probability that 4 or more people have the
disease?
(Assume binomial experiment.)
Identify n = 5, x = 4, p = .10
Identify n = 5, x = 4, p = .10
P(X≥4) = P(X=4) + P(X=5)
P (X = x ) =
⋅ p x ⋅ (1 − p ) n − x
Binomial Probability
Example: If there are 10% of the population in
a community have a certain disease, what is the
probability that 4 people in a random sample of
5 people from this community has the disease?
5!
P(X=4) =
·(.10)4·(1 − .10)5-4
4! 1!
= 5·(.10) 4(.90)1
= .0004
(6’, 6, 6’)
Example: A balanced die is rolled three times
(or three balanced dice are rolled), what is the
probability to see two 6’s?
n! = 1·2·3·... ·n
Example:
(6’, 6’, 6)
5!
5! 0!
= .00045 +
n!
⋅ p x ⋅ (1 − p ) n − x
x!⋅(n − x )!
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Parameters of Binomial
Distribution
·(.10)5·(1 −.10)5-5
= .00045 + .00001 = .00046
(What is this number telling us?)
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Binomial Distribution
P(0) = .5905
n = 5, p = .10
Parameters of the distribution:
Mean of the distribution, µ = n·p
Variance of the distribution, σ2 = n·p·(1 − p)
Standard deviation, σ , is the square root of
variance.
µ = 5 x .10 = .5
P(1) = .3281
σ2 = 5 x .1 x (1−.1) = .45
P(2) = .0729
P(3) = .0081
0.590
P(4) = .0004
P(5) = .00001
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Random Variable - 4
Random Variable & Discrete Distribution
Discrete Probability Models
Poisson Distribution
Bernoulli : Two categories of outcomes.
Binomial : Number of successes in a
binomial experiment.
Poisson : Number of successes in a
given time period or in a given unit
space.
Let X represents the number of occurrences
of some event of interest over a given
interval from a Poisson process, and the λ is
the mean of the distribution, then the
probability of observing x occurrences is,
for x = 0, 1, 2, …,
−λ x
P( X = x) = e λ
x!
It can be used to approximate Binomial prob., for large n.
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Poisson Process
e = 2.71828…
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Examples of Poisson Process
The probability that a single event
occurs within an interval is proportional
to the length of the interval.
Within a single interval, an infinite
number of occurrences is possible.
The events occur independently both
within the same interval and between
consecutive non-overlapping intervals.
Number of people visiting to the
emergency room for treatment per hour.
Number of customers coming to the
Arby’s to buy sandwich per ten minutes.
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Poisson Probability
Poisson Probability
If on average there are 4 people catch flu in a
given week in a community during a certain
season, what is the probability of observing 2
people catch flu in this community in a given
week period during the season? (Assume the
number of people catching flu in a given period
of time follow a Poisson Process.)
λ=4
x=2
e −4 42
P(X=2) =
= .1465
2!
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−λ x
P ( X = x) = e λ
x!
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If on average there are 4 people catch flu in a
given week in a community during a certain
season, what is the probability of observing 2
people catch flu in this community in a given
two weeks period during the season? (Assume
the number of people catching flu in a given
period of time follow a Poisson Process.)
λ = 4x2 = 8
x=2
e −8 82
P(X=2) =
= .011
2!
−λ x
P ( X = x) = e λ
x!
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Random Variable - 5
Random Variable & Discrete Distribution
Poisson Probability
Discrete Probability Models
If on average there are 4 people catch flu in a
given week in a community during a certain
season, what is the probability of observing less
than 2 people catch flu in this community in a
given week period during the season? (Assume
the number of people catching flu in a given
period of time follow a Poisson Process.)
λ=4
x = 0 and 1
P(X<2) = P(X=0) + P(X=1)
= (e-4·40)/0! + (e-4·41)/1! = .0916
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Binomial : Number of successes in a
binomial experiment. (There is a sample
taken.)
Poisson : Number of successes in a
given time period or in a given unit
space. (No sample taken.)
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Random Variable - 6