4 1 probability and discrete probability distributions

.
Probabilities
and Probability Distribution

Probability
In a given situation, we would like to be able
to calculate the probability of something. To
do this we need to make our discussion more
formal. We begin with an experiment which is
simply an activity that results in a definite
outcome, which is called sample space.

Sample Space ():
 The sample space S of statistical experiment is
the set of all possible outcomes of an experiment.
For example, consider a set of six balls numbered 1,
2, 3, 4, 5, and 6. If we put the six balls into a bag
and without looking at the balls, we choose one
ball from the bag, then, this is an experiment
which has 6 possible outcomes i.e.
S= {1,2,3,4,5,6}

Random Variables
 A random variable is a numerical value determined
by the outcome of an experiment that varies from
trial to trial.
 A statistical experiment is any process by which
several measurements are obtained.

Example
 Consider a random experiment in which a coin is tossed
three times. Let x be the number of heads. Let H
represent the outcome of a head and T the outcome of a
tail.
 The possible outcomes for such an experiment will be:
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
 Thus the possible values of x (number of heads) are
x=0: TTT
x=1: TTH, THT, HTT
x=2: THH, HTH, HHT
x=3: HHH
 From the definition of a random variable, x as defined in
this experiment, is a random variable.
P(x=0) =1/8
P(x=1) =3/8
P(x=2) =3/8
P(x=3) =1/8
If the
coin is
fair

Examples of Statistical Experiments
 Counting the number of books in the College
Library
 Counting the number of mistakes on a page
of text
 Measuring the amount of rainfall during the
month of January

Types of Random Variables
Two different classes of random variables:
1. A discrete random variable
2. A continuous random variable

Discrete Random Variable
 A discrete random variable is a quantitative random
variable that can take on only a finite number of values
or a countable number of values.
Examples:
 The number of children per family
 The number of cavities a patient has in a year.
 The number of bacteria which survive treatment with
some antibiotic.
 The number of times a person had a cold in Gaza Strip.

Continuous Random Variable
A continuous random variable is a quantitative
random variable that can take infinite number of
values within an interval
Example: the amount of rainfall in during the
month of January.

Probability distribution
 A probability distribution is the listing of all possible
outcomes of an experiment and the corresponding
probability.
 Depending on the variable, the probability
distribution can be classified into:
 Discrete probability distribution
 Continuous probability distribution

Discrete probability distribution
 A discrete probability distribution is a table, graph,
formula, or other device used to specify all possible
values of a discrete random variable along with
their respective probabilities.
Examples:
 The number of children per family
 The number of cavities a patient has in a year.
 The number of bacteria which survive treatment
with some antibiotic.
 The number of times a person had a cold in Gaza
Strip.

Probability
distribution of
number of children
per family in a
population of 50
families
x
Frequency of
occurring of x P(X=x)
0 1 1/50
1 4 4/50
2 6 6/50
3 4 4/50
4 9 9/50
5 10 10/50
6 7 7/50
7 4 4/50
8 2 2/50
9 2 2/50
10 1 1/50
50 50/50

Bar chart Graphical representation of the Probability
distribution of number of children per family for population of
50 families
0
1/50
2/50
3/50
4/50
5/50
6/50
7/50
8/50
9/50
10/50
1 2 3 4 5 6 7 8 9 10 x
Number of children/family
Probability
0
P(x)

Toss of Two Coins Roll of a Die Sex of Three-child Family
E P(E) E P(E) E P(E)
HH 1/4 1 1/6 3 boys• 0.125
HT 1/4 2 1/6 2 boys. 1 girl .375
TH 1/4 3 1/6 1 boy, 2 girls .375
TT 1/4 4 1/6 3 girls .125
1.0 5 1/6 1.000
6 1/6
1.0

Continuous probability distribution
 A continuous probability distribution can assume an
infinite number of values within a given range – for
variables that take continuous values.
 The distance students travel to class.
 The time it takes an executive to drive to work.
 The length of an afternoon nap.
 The length of time of a particular phone call.

Features of a Discrete Distribution
 The main features of a discrete probability distribution
are:
 The probability of a particular outcome, P(Xi), is
between 0 and 1.00.
 The sum of the probabilities of the various outcomes
is 1.00. That is,
P(X1) + … + P(XN) = 1
 The outcomes are mutually exclusive. That is,
P(X1and X2) = 0 and
P(X1or X2) = P(X1)+ P(X2)

Mutually exclusive events, Ei,
 Two events A and B are mutually exclusive if
the event (A and B) contain no elements.
BA

For mutually exclusive events A and B
P(A or B) = P(A) + P(B)
BA

Probability Distributions
Important Probability Distributions:
• The Binomial Distribution
• The Poisson Distribution
• The Hypergeometric Distribution
• The Normal Distribution
• The t Distribution
• The χ2
Distribution
• The F Distribution

Discrete Probability Distribution
Models
Discrete Probability Distribution
To be studied
Binomial Poisson

A binomial experiment has the following
conditions:
1. There are n repeated trials
Binomial Distribution

 2. Each trial has only two possible outcomes-
success or failure, girl or boy, sick or well, dead
or alive, at risk or not at risk, infected–not
infected, white–nonwhite, or simply positive–
negative etc
1 1 0 1 0 0 1 0 0 0 0 0 1 0 0 0 00 0 0
1 1 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 0 1 0

3. The probabilities of the two outcomes remains
constant from trial to trial.
The probability of success denoted by p.
The probability of a failure, 1- p, is denoted by q.
Since each trial results in success or failure,
p + q = 1 and q = 1 – p.

4. The outcome of each trial is
independent of the outcomes of any other
trial; that is, the outcome of one trial has
no effect on the outcome of any other
trial.

5. When the conditions of the binomial experiment are
satisfied, Our interest is in the number of successes r
occurring in the n trials.
Example
If a certain drug is known to cause a side effect 10% of the
time and if five patients are given this drug, what is the
probability that four or more experience the side effect?
More Examples
 # Defective items in a batch of 5 items
 # Correct on a 33 question exam
 # Customers who purchase out of 100 customers who
enter store

where
 n = the number of trials in an experiment
 r = the number of successes
 n - r = the number of failures
 p = the probability of success
 q = 1 - p, the probability of failure
qp
rnr
n
successrp rnr −










=
−
..
)!(!
!
)(
The Binomial probability
formula
 The probability of obtaining r successes in n
trials with a probability p of success in each trial
can be calculated using the formula;

"!" represents the factorial function.
n! = (n)(n - 1) (n - 2)(n - 3) . . . (1). For example,
4! = (4)(3)(2)(1) = 24
By definition, 0! = 1
qp
rnr
n
successrp rnr −










=
−
..
)!(!
!
)(

Example
 What is the probability of having two girls and one boy, 3
boys, and at least one boy in a three-child family if the
probability of having a boy is 0.5?
 Solution:
From the calculations in equation
It can be seen that
qp
rnr
n
successrp rnr −










=
−
..
)!(!
!
)(

For two girls and one boy
P(2G, 1B)= = 3(.125) = .375
For 3 boys
a. considering n = 3, r = 3, i.e. boy is the success
P(3B)= = 1 X (.125)X 1 = .125
b. considering n = 3, r = 0, i.e. girl is the success or boy is the failure
P(3B)= = 1 X 1 X (.125) = .125
For at least 1 boy
P (3B) ) + P(2B, 1G) + P(1B, 2G) = 0.125 + 0.375 +0.375 = 0.875
n = 3, r = 2, p = 1/2, q = 1/2
qp
rnr
n
successrp rnr −










=
−
..
)!(!
!
)(

Example:
Ten individuals are treated surgically. For each individual
there is a 70% chance of successful surgery. Among these
10 people, the number of successful surgeries follows a
binomial distribution with n =10, and p =0.7.
What is the probability of exactly 5 successful
surgeries?
qp
rnr
n
successrp rnr −










=
−
..
)!(!
!
)(
3.07.0
)!510(!5
!10
)5( 510.5. −










=
−
=rp
= (252)(0.75
)(0.35
) = 0.1029
n = 10, r = 5, p = 0.7, q = 1 – p = 0.3

Mean & Variance of the Binomial Distribution
 The mean is found by:
 The variance is found by:
np=µ
)1(2
pnp −=σ

EXAMPLE
 p =.2 and n=14.
 Hence, the mean is:
µ= n p = 14(.2) = 2.8.
 The variance is:
σ2
= n p (1- p ) = (14)(.2)(.8) =2.24.

Using the Binomial
Probability Table

 The formula for calculating the binomial
distribution can be quite cumbersome and
leaves plenty of room for human error in
calculation
 An alternative and much simpler way to
determine the probability of r is to use
binomial tables

 Find the section labeled with your value of n.
 Find the entry in the column headed with your value of p
and row labeled with the r value of interest.
Using the Binomial Probability Table

Using the Binomial Probability Table
n = 8, p = 0.7, find P(r =6):

Example
 The probability that a patient recovers from a
rare blood disease is 0.4. If 10 people are
known to have contracted this disease, what
is the probability that
(a) exactly 3 survive,
(b) at least 8 survive, and
(c) from 2 to 5 survive?

Let x be the number of people that survive. Consider the
binomial probability table with n = 10, and P = 0.4.
We find that
a. exactly 3 survive
P (r = 3) = 0.2150
b. at least 8 survive
P ( ) = P (r = 8) + P (r = 9) + P (r = 10)
= 0.0106 + 0.0016+ 0.0001
= 0.0123
c. from 2 to 5 survive
P( ) = P (r = 2) + P (r = 3) + P (r = 4) + P (r =
5)
= 0.1209 + 0.2150 + 0.2508 + 0.2007
8≥r
52 ≤≤ r
Solution

Example
 A biologist is studying a new hybrid tomato. It
is known that the seeds of this hybrid tomato
have probability 0.70 of germinating. The
biologist plants 10 seeds.

 a. What is the probability that exactly 8 seeds will
germinate?
 Solution
This is a binomial experiment with n = 10 trials. Each
seed planted represents an independent trial. We’ll
say germination is success, so the probability of
success on each trial is 0.70.
n = 10 p = 0.70 q = 0.30 r = 8
We wish to find P(8), the probability of exactly eight
successes.
In binomial probability Table, find the section with n =
10. Then find the entry in the column headed by P =
0.70 and the row headed by the r value 8. This entry
is 0.233.

 b. What is the probability that at least 8 seeds will
germinate?
 Solution
In this case, we are interested in the probability of 8 or
more seeds germinating. This means we are to compute
P(r 8). Since the events are mutually exclusive, we can
use the addition rule.
P(r 8) = P(r = 8 or r = 9 or r = 10) = P (8) + P(9) +
P(10)
We already know the value of P(8) and P(9) and P(10).
Use the same part of the Table but find the entries in the
row headed by the r value 9 and then the r value 10. Be
sure to use the column headed by the value P, 0.70.
P(9) = 0.121 and P(10) = 0.028
Now we have all the parts necessary to compute P(r 8)
P(r 8) = P(8) + P(9) + P(10)
= 0.233 + 0.121 + 0.028
≥

Binomial Probability Distribution
 To construct a binomial distribution, let
 n be the number of trials
 r be the number of observed successes
 P be the probability of success on each trial
 The formula for the binomial probability distribution
is:
qp
rnr
n
rp rnr −










=
−
..
)!(!
!
)(

r P(r)
0 0.031
1 0.156
2 0.312
3 0.312
4 0.156
5 0.031
P=0.5, n=5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 1 2 3 4 5
r
p(r)

These distributions can be easily constructed by using
the binomial probability distribution Table to find the
P(r) values for the specified n and P
r P(r)
0 0.031
1 0.156
2 0.312
3 0.312
4 0.156
5 0.031
P=0.5, n=5
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 1 2 3 4 5
r
p(r)
P=0.5, n=5

p =0.3 p =0.5
p =0.7
Notice that when p < 0.5 the
distribution is skewed right, and
when p >0.5 the distribution is
skewed left. When p =0.5, the
distribution is symmetric.
Characteristics of Binomial
Distribution

Common English expression and corresponding inequalities
(consider a binomial experiment with n trials and r success)
≥
≤
Expression Inequalities
Four or more successes r 4
At least four successes That is, r = 4, 5, 6, ……., n
No fewer than four successes
Not less than four successes
Four or fewer successes r 4
At most four successes That is, r = 0, 1, 2, 3, or 4
No more than four successes
The number of successes does not exceed four
More than four successes r > 4
The number of successes exceeds four That is r = 5, 6, 7, ………., n
Fewer than four successes r < 4
The number of successes is not as large as four That is, r = 0, 1, 2, 3

Poisson Probability Distribution

Poisson Probability Distribution
The Poisson probability distribution is often used to
model a discrete variable which is applied to
experiments with random and independent
occurrences of an event.
The occurrences are considered with respect to (per
unit)
a time interval,
a length interval,
a fixed area
or a particular volume.

Poisson Random Variable
 1. Our interest is in the Number of events that occur in
Time interval, length, area, volume
Examples
 The number of serious injuries in a particular factory in a year.
 The number of times a three-old child has an ear infection in a
year.
 Number of blood cells in unit area of haemocytometer.
 The number of live insects per square after spraying a large area
of land with an insecticide and counting the number of living
insects in a randomly selected squares.
 # Customers arriving in 20 minutes
 the number of colonies growing in 1 ml of culture medium.

Formula of Poisson Probability
Distribution
The Poisson distribution can be described
mathematically using the formula:
where
λ (Greek letter lambda) is the mean number of occurrence
of an event in a particular interval of time, volume, area,
and so forth.
e is the the base of natural logs = 2.71828.
r is the number of occurrence of event (r = 0, 1, 2, 3, …) in
corresponding interval of time, volume, area, and so
forth.
!
)(
r
e
rP
r λ
λ −
=

λ
A count was made of the red blood corpuscles in each of the 64
compartments of a haemocytometer gave the following results:
In this case, the given area is the hematocytometer compartment.
An event is the number of RBCs.
The total number of RBCs has been
2 ×1 + 3 ×5 + 4 ×4 + 5 ×9 + 6 ×10 + 7 ×10 + 8 ×8 + 9 ×6 +10 ×4 +
11 ×3 + 12 ×2 + 13 ×1 + 14 ×1 = 450
The number of compartments has been 1 + 5 + 4 + 9+ 10+ 10+ 8+
6+ 4+ 3+ 2+ 1+ 1 = 64
So the mean number of RBCs per compartment is
= 7.03 ≈ 7
Since λ = 7, the probabilities of any number of corpuscles in a randomly
selected compartment can be easily determined
450
64

The shape of Poisson Distributions
 Probability distribution always skewed to the
right.
 Becomes symmetrical when λ gets large.

EXAMPLE
 An urgent Care facility specializes in caring
for minor injuries, colds, and flu. For the
evening hours of 6-10 PM the mean number
of arrivals is 4.0 per hour. What is the
probability of 2 arrivals in an hour?
2 4
4
( ) .1465
! 2!
r
e e
P r
r
λ
λ − −
= = =

Example
 Suppose we are interested in the number of snake
bite cases in a particular hospital in a year. Assume
that the number of snake bite cases at the hospital
has a Poisson (6) (i.e., λ = 6) distribution.
a. What is the probability that in a randomly chosen
year, the number of snake bite cases will be 7?
b. What is the probability that the number of such
cases will be less than 2?
c. What is the expected number of snake bite cases in
a year?

Example: Poisson Probabilities
r = number of snakebite cases at this hospital
in a year
λ = 6
Find the probability that 7 bites will occur in a
randomly selected year.
-6 7
e (6) 693.89197
P(r=7)= = 0.1378
7! 5040
=
!
)(
r
e
rP
r λ
λ −
=

Example: Poisson Probabilities
r = number of snakebite cases at this hospital
in a year
λ = 6
The second question asks us to find p(r<2).
0 -6 1 -6
(6) .e (6) .e
p(r 2) ( 0) ( 1)
0! 1!
p r p r< = = + = = +
= e-6
+ e-6
(6) =0.01735
The expected number of snake bite cases in
a year at this hospital is µ = λ = 6 cases
!
)(
r
e
rP
r λ
λ −
=

Example 3
In a study of the spatial distribution of cacti in the genus Acuna,
researchers placed a 5 m by 5 m grid over a 1002
m study area.
Suppose that Acuna cacti are distributed randomly over the
study area with on average 2.2 cacti per grid cell. Also,
suppose the number of cacti in a cell is independent of the
number in any other cell. What is the probability that there are
exactly 4 Acuna cacti in a given cell?
 Solution:
In this case, r = 4 cacti and cacti, so2.2l =
10815.
!4
)2.2(
!
)4(
2.24
====
−−
e
r
e
rP
r λ
λ

Use of Table
 A table of the Poisson probability distribution for
selected values of λ and the number of successes
r is often available in the Appendix of statistics
books.
 To find the value of p (r = 2) when λ = 0.3, look in
the column labeled by 0.3 and the row labeled by
2. λ
r 0.1 0.2 0.3 0.4 0.5
0 .9048 .8187 .7408 .6703 .6065
1 .0905 .1637 .2222 .2681 .3033
2 .0045 .0164
3 .0002 .0011 .0033 .0072 .0126
4 .0000 .0001 .0003 .0007 .0016
.0333

Example
 In the study of certain aquatic organism, a large number
of samples were taken from a pond, and the number of
organisms in each sample was counted. The average
number of organisms per sample was found to be two.
Assuming that the number of organisms follows a
Poisson distribution, find the probability that:
a. The next sample will contain one or fewer organisms.
Solution:
By using the table we see that when λ = 2,
P(r 1) = p(r = 1) + p(r = 0) = 0.2707 + 0.1353
= 0.406
≤

Example
 In the study of certain aquatic organism, a large number
of samples were taken from a pond, and the number of
organisms in each sample was counted. The average
number of organisms per sample was found to be two.
Assuming that the number of organisms follows a
Poisson distribution, find the probability that:
a. The next sample will contain more than five organisms.
Solution:
By using the table we see that when λ = 2,
P(r > 5) = 1 - p(r 5) = 1 – [p (r = 5) + p(r = 4) + p(r = 3) +
p(r = 2) + p(r = 1) + p(r = 0)]
=1 – [0.0361 + 0.0902 + 0.1804 +0.2707 + 0.2707 + 0.1353
= 1 – 0.9834 = 0.0166
≤

The sum of two or more
Poisson distributions
 If two independent distributions are both
Poisson with means λ1 and λ2 then the sum of
the distributions is Poisson with
mean λ = λ1 + λ2 .

Example
A certain blood disease is caused by the presence of two
types of deformed corpuscles, A and B. It is known that if
the total number of deformed corpuscles exceeds 6 per
0.001cm3
of blood then the patient will contract the blood
disease.
For a particular family the number of type A per 0.001cm3
of
blood can be modelled by a Poisson distribution with mean
1.3 and the number of type B per 0.001 cm3
of blood can be
modelled by a Poisson distribution with mean 1.6.
 Find the probability that one person in this family will
contract the disease.

 SOLUTION
In this example we are interested in the total
number of deformed corpuscles per 0.001cm3
.
This has mean 1.3 + 1.6 = 2.9.
So in this case the distribution of the total
number of deformed corpuscles per 0.001cm3
is Poisson with parameter 2.9.
From the table, the probability that one person
in this family will contract the disease is
p(r = 1) = 0.1596.

Mean and Variance of a Poisson
Random Variable
If x is a Poisson random variable with parameter µ, then
Standard Deviation λσσ =2
=
λµ =Mean
λσ =2
Variance
An interesting feature of the Poisson distribution is the
fact that the mean and variance are equal.

Poisson approximation of the
 The complexity of the formula for calculating
binomial probabilities makes them difficult to
calculate when a binomial experiment has a
quite low p and high n.
 Thus, the Poisson distribution can be used
instead to approximate the probabilities of a
discrete random variable x which has a
Binomial (n, p) distribution when

n is large ( n ≥ 100),
 and p is small (p ≤ 0.01) or close to zero

and np ≤ 10

Example
 Suppose that the probability that a person suffers from certain rare
disease is 1 in 2500, that is 1/2500 = 0.0004. If we can assume that
one person’s suffering from the disease is independent from another’s
and we sampled 1000 persons, what is the probability that there are at
least one person have contracting this disease? What is the expected
number to be contracting this disease?
 Since n = 1000 is large and p = 0.0004 is small, binomial probabilities
are difficult to calculate so we use the Poisson approximation.
 We are asked to find p(r ≥ 1)
 p(r ≥ 1) = 1 – p (x = 0) = 1 -
 The expected number of suffering from that disease in 1000
is λ = np = 0.4
32968.01
!0
)4.0( 4.0
04.0
=−= −
−
e
e

4 1 probability and discrete probability distributions

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