Discrete Random Variables
and Probability Distributions
Dr. Papia Sultana
Associate Professor
Department of Statistics
Rajshahi University
Outline



Probability distribution
Probability distribution function and mass
function
Parent distribution






Binomial distribution
Poisson distribution
Geometric distribution
Negative Binomial distribution
Hyper-geometric distribution
Multinomial distribution
09.09.2011
Dr. Papia Sultana
2
Probability distribution
Arrangement of probabilities of
outcome of an event.
 Ex. Tossing three coins simultaneously.
Outcome of interest is “number of
head” (X) in a toss.
Values of X:
0
1
2
3

(No of Head)
(TTT)
(TT H)
(T HH)
(HHH)
Probabilities: 1/8
3/8
3/8
1/8
09.09.2011
Dr. Papia Sultana
3
Probability distribution function
Let X be a random variable, then the
function
F ( x)  Pr(X  x)
is called distribution function of X.
Ex. Consider previous example.
Values of X:
0
1
2
3
(x)
F(x): 1/8
09.09.2011
4/8
7/8
Dr. Papia Sultana
1
4
Distribution mass function
Let X be a one-dimensional discrete random
variable, then the probability function P( X  x)
is called the probability mass function if it
satisfies
(a) P( x)  0
(b)
 P( x )  1
x
09.09.2011
Dr. Papia Sultana
5
Parent distribution
The distribution of a true sample. It is, actually,
distribution of population. It can be obtained limiting
the parameters to the true value of distribution of any
sample.
Binomial
Uniform
Poisson
Normal
Geometric
Gamma
Negative Binomial
Weibul
Hyper-geometric
Cauchy
Multinomial
Beta
.............so on
09.09.2011
Dr. Papia Sultana
6
Binomial distribution
Bernoulli trials:
Tossing a coin: Head or Tail
More Example: Germination of a seed,
gender of newborn baby, defectiveness of
products of a factory, ...........
09.09.2011
Dr. Papia Sultana
7
Binomial distribution
Let us consider n Bernoullian trials with
outcome “success” or “failure”.
SSFSFFFSFSSFF...........FSF
# of S=X,
# of F=n-X,
09.09.2011
P
q=1-p
Dr. Papia Sultana
8
Binomial distribution
These X success out of n trial can occur in
x n x
n 
 
ways with probability p q
.
X
 n  x n x
P( X  x)    p q ;
 x
09.09.2011
x  0,1,2,....,n
Dr. Papia Sultana
9
Binomial distribution
Properties:

mean=np, variance=npq, skewness=
1  6 pq
kurtosis= 3 
npq
q p
,
np q
mgf: M x t   q  pe 
It does not follow additive properties.
t n


09.09.2011
Dr. Papia Sultana
10
Binomial distribution

Example: The number X of seeds that
germinate in n=10 independent trials with
p=0.8 is b(10,0.8), that is,
10
x
10  x
P( X  x)   0.8 0.2 ; x  0,1,2,....,10
x 
 Distribution function
6 10
 
x
10  x
F ( X  6)  P( X  6)    0.8 0.2
x 0  x 
09.09.2011
Dr. Papia Sultana
11
Binomial distribution
09.09.2011
Dr. Papia Sultana
12
Poisson distribution




Number of customers coming at a
restaurant per hour.
Number of patients coming to take
service at a clinic per day.
Number of telephone calls at a helpline
center per minute.
.........
09.09.2011
Dr. Papia Sultana
13
Poisson distribution



Let the probability of success for each trial
is  / n .
Probability of success for each trial is
indefinitely small.
n is sufficiently large (much larger than
09.09.2011
Dr. Papia Sultana
x ).
14
Poisson distribution
pdf is

e 
P( X  x) 
; x  0,1,2,........
x!
09.09.2011
x
Dr. Papia Sultana
15
Poisson distribution

Example: Telephone calls enter a Hall
switchboard of Rajshahi University on the
average two every 3 minutes. Therefore,   2
per 3-minutes period and pdf is
2
x
e 2
P( X  x) 
; x  0,1,2,........
x!
09.09.2011
Dr. Papia Sultana
16
Poisson distribution
Properties:

mean=variance=
kurtosis= 3 



, skewness=
1

,
1

 ( et 1)
mgf: e
It follows additive properties.
09.09.2011
Dr. Papia Sultana
17
Poisson distribution
09.09.2011
Dr. Papia Sultana
18
Geometric distribution
P( X  x)  q x p; x  0,1,2,........
09.09.2011
Dr. Papia Sultana
19
Geometric distribution
Example: Bob is a high school basketball player.
He is a 70% free throw shooter. That means
his probability of making a free throw is 0.70
and the number of successes is 1 . During the
season, what is the probability that Bob makes
his first free throw on his fifth shot?
The probability of success (P) is 0.70, the
number of trials (x) before the success is 4
P( X  4)  0.7 * 0.34  0.00567
09.09.2011
Dr. Papia Sultana
20
Geometric distribution
Properties:



mean=
q
p
q
,variance= 2 ,
p
t 1
mgf: p(1  qe )
It does not follow additive properties.
09.09.2011
Dr. Papia Sultana
21
Geometric distribution
Properties:



mean=
q
p
q
,variance= 2 ,
p
t 1
mgf: p(1  qe )
It does not follow additive properties.
09.09.2011
Dr. Papia Sultana
22
Negative-Binomial distribution
When variance is larger than mean.
Example:
 Deaths of insects,
 Insect bites,
 .............
09.09.2011
Dr. Papia Sultana
23
Negative-Binomial distribution


Last trial must be a success.
There are x failures preceding the r-th
success.
 r r
x
P( X  x)    p (q) ;
x 
09.09.2011
Dr. Papia Sultana
x  0,1,2,....
24
Negative-Binomial distribution
Properties:



rq
mean= p
rq
,variance= 2 ,
p
r
mgf: (1 / p  qe / p)
It does not follow additive properties.
09.09.2011
t
Dr. Papia Sultana
25
Negative-Binomial distribution
Example:
Bob is a high school basketball player. He is a 70%
free throw shooter. That means his probability of
making a free throw is 0.70. During the season, what
is the probability that Bob makes his third free throw
on his fifth shot?
This is an example of a negative binomial experiment.
The probability of success (P) is 0.70, the number of
trials (x) is 5, and the number of successes (r) is 3.
Thus, the probability that Bob will make his third
successful free throw on his fifth shot is 0.18522.

09.09.2011
Dr. Papia Sultana
26
Hypergeometric distribution

When the population is finite and the
sampling is done without replacement,
so that the events are stochastically
dependent, although random.
09.09.2011
Dr. Papia Sultana
27
Hypergeometric distribution

Consider an urn with N balls, M of
which are white and N-M are red.
Suppose that we draw a sample of n
balls at random (without replacement)
from the urn, then the probability of
getting k white balls out of n (k<n) will
follow hypergeometric distribution.
09.09.2011
Dr. Papia Sultana
28
Hypergeometric distribution

pdf:
Hg(N,M,n)
 M  N  M 
 

k  n  k 

P( X  k ) 
;
N
 
n 
09.09.2011
k  0,1,2,....,min(n, M )
Dr. Papia Sultana
29
Hypergeometric distribution


Applications:
Industrial acceptance: An acceptance sampling
scheme is based upon drawing a random sample
of size n from a batch of N items, assumed to
contain an unknown number d of defectives. If
the number of defectives X in the sample is too
large, the batch as whole is rejected. Deciding on
the threshold to determine what “too large”
should mean is based upon the fact that
X~Hg(N,n,d). The parameter of interest is d.
09.09.2011
Dr. Papia Sultana
30
Hypergeometric distribution

Capture-recapture experiment: A population
of N individuals of a species exists in a
closed eco-system. a first random sample of
size n1 is taken, all are tagged and returned
to the wild. Later asecond random sample of
size n2 is taken. let X be the number of
tagged individuals caught in the second
sample. Then X ~ Hg( N , n1 , n2 ) and the
parameter of interest is N.
09.09.2011
Dr. Papia Sultana
31
Hypergeometric distribution

System faults: A system has an unknown
number N of faults, and two independent
inspection processes detect n1 , n2
faults respectively, of which X are
common. Then X ~ Hg( N , n1 , n2 ) and
the parameter of interest is N.
09.09.2011
Dr. Papia Sultana
32
Hypergeometric distribution

Example: In a pond there were 200
fishes. A catch of 50 fishes made and
returned them alive into the pond
marking each with a red spot. After a
reasonable period of time, another
catch of 30 fishes was made. what is
the probability that exactly 11 of the
spotted fishes were caught?
09.09.2011
Dr. Papia Sultana
33
Hypergeometric distribution
 50150 
 

k  30  k 

P( X  k ) 
;
 200


 30 
k  0,1,2,....,30
 50150
 

11 19 

P(X  11) 
 200


 30 
09.09.2011
Dr. Papia Sultana
34
Hypergeometric distribution
Properties:

nM
mean=
N
09.09.2011
nM ( N  M )(N  n)
,variance=
,
2
N ( N  1)
Dr. Papia Sultana
35
Multinomial distribution


This is an generalization of binomial
distribution.
When there are more than two
outcomes of a trial, it follows
multinomial distribution.
09.09.2011
Dr. Papia Sultana
36
Multinomial distribution
Example: In a large population of plants, there are
three possible alleles S, I and F at one locus
resulting in six genotypes SS,II, FF, SI, SF and IF.
Let 1 , 2 ,3 denote the allele frequencies of S,
I, F. The labels and frequencies of the genotypes
are
Label:
1
2
3
4
5
6
Genotype:
SS
II
FF SI
SF IF
p2 p3 p4 p5 p6
Frequency: p1
2
2
2



HWE Freq:
1
2
3 21 2 213 2 23

09.09.2011
Dr. Papia Sultana
37
Multinomial distribution

A1 , A2 ,, Ak are k mutually
Let
exclusive and exhaustive outcomes of a
trial with probabilities p1 , p2 ,, pk .
Let A1 occurs x1 times, A2 occurs x2 times,
......, and Ak occurs xk times.
n!
p( x1 , x2 ,, xk ) 
p1x1 p2x2  pkxk ;
x1! x2! xk !
09.09.2011
Dr. Papia Sultana
k
x
i 1
i
n
38
Multinomial distribution
Properties:

mgf: M x t    p1e  p2e   pk e
t1
t2

E( X i )  npi

V ( X i )  npi (1  pi )

Cov( X i , X j )  npi p j
09.09.2011
Dr. Papia Sultana

tk n
39
Multinomial distribution

Example:
In a recent three-way election for a country,
candidate A received 20% of the votes, candidate B received
30% of the votes, and candidate C received 50% of the votes.
If six voters are selected randomly, what is the probability that
there will be exactly one supporter for candidate A, two
supporters for candidate B and three supporters for candidate
C in the sample?
6!
p( x1  1, x2  2, x3  3) 
0.210.32 0.53  0.135
1!2!3!
09.09.2011
Dr. Papia Sultana
40
Thank you!
09.09.2011
Dr. Papia Sultana
41