• Probability is the chance of an outcome of an experiment.

It is
the measure of how likely an outcome is to occur.
 Definitions of some probability terms
 Experiment: Any process of observation or measurement or
any process which generates well defined outcome.
 Probability Experiment: It is an experiment that can be
repeated any number of times under similar conditions and it is
possible to enumerate the total number of outcomes with out
predicting an individual out come.
 Some Basic Probability Terms
 Outcome :The result of a single trial of a random experiment

 Sample Space: Set of all possible outcomes of a probability

experiment.
 Event: It is a subset of sample space.

 Elementary Event: having only a single element.

 Compound Event: If an event has more than one sample points

 Equally Likely Events: Events which have the same equal

chance of occurrence.
 Mutually Exclusive Events: Two events which cannot happen
at the same time.A  B    PA  B   0
 Independent Events: Two events are independent if the
occurrence of one does not affect the probability of the other
occurrence.
 Dependent Events: Two events are dependent if the
first event affects the outcome or occurrence of
the second event in a way the probability is
changed.

Counting Rules
 In order to calculate probabilities, we have to know
 The number of elements of an event
 The number of elements of the sample space.
 That is in order to judge what is probable, we have
to know what is possible.
 Counting rules
 In order to determine the number of outcomes, one
can use several rules of counting.
 Addition rule
 The multiplication rule
 Permutation rule
 Combination rule
 The addition rule
If 1st procedure designed by 1 can be performed in n1 ways.
2nd procedure designed by 2 can be performed in n2 ways.
suppose further more that, it is not possible that both
procedures 1 and 2 are performed together then the number of
ways in which we can perform 1or 2 procedure is:
n1+n2 ways,
If we have another procedure that is designed by k with
possible way of nk we can conclude that there is:

n1+n2+…+nk possible ways.

Example: suppose we planning a trip and are deciding by
bus and train transportation. If there are 3 bus routes and 2
train routes to go from A to B. find the available routes for
the trip.
Solution:
There are 3+2 =5 routes for someone to go from A to B.
 The Multiplication rule
 If a choice consists of k steps of which the first can be made
in n1 ways, the second can be made in n2 ways…, the kth can

be made in nk ways, then the whole choice can be made in n1

* n2*…*nk ways

 Example: Distribution of Blood Types There are four blood

types, A, B, AB, and O. Blood can also be Rh+ and Rh-.
Finally, a blood donor can be classified as either male or
female. How many different ways can a donor have his or her
blood labeled?
Solution
Since there are 4 possibilities for blood type, 2 possibilities
for Rh factor, and 2 possibilities for the gender of the
donor, there are 4 2 2, or 16, different classification
categories, as shown.
 Permutation
• An arrangement of n objects in a specified order
 Permutation Rules:
1. The number of permutations of n distinct objects taken
all together is n!
n! n!
n! n * (n  1) * (n  2) * ..... * 3 * 2 *1 P  = n! . In definition 0! = 1! = 1
n  n! 0!
n n

2. The arrangement of n objects in a specified order using r

objects at a time n Pr

3. The number of permutations of n objects in which k 1 are

alike k2 are alike ---- etc is

n!
n Pk 
k1!*k 2 * ... * k n
• Example:

1. Suppose we have a letters A,B, C, D

a. How many permutations are there taking all
the four?
b. How many permutations are there two letters
at a time?

2. How many different permutations can be made

from the letters in the word “CORRECTION”?
 Combination rule
• Selection of objects without regard to order is denoted by
n n n!
n C r or     
r and is given by the formula:  r  ( n  r )!*r!
Example:
1. In how many ways 5 patients be chosen out of 9 patients?

 Out of 5 Mathematician and 7 Statistician a committee

consisting of 2 Mathematician and 3 Statistician is to be

formed. In how many ways this can be done if
a. There is no restriction
b. One particular Statistician should be included
c. Two particular Mathematicians can not be included on
the committee.
 Approaches to measuring Probability
There are four different conceptual approaches to the
study of probability theory. These are:
 The classical approach.
 The relative frequency approach.
 The axiomatic approach.
The classical approach
Short coming of the classical approach
This approach is not applicable when:
 The total number of outcomes is infinite.

 Outcomes are not equally likely.

 The Frequentist Approach
(based on repeatability of events)
 This is based on the relative frequencies of outcomes
belonging to an event.
 The probability of an event A is the proportion of

outcomes favorable to A in the long run when the

experiment is repeated under same condition
 E.g. If records show that 60 out of 100,000 bulbs

produced are defective. What is the probability of a

newly produced bulb to be defective?
Solution: Let A be the event that the newly produced bulb
is defective N 60
P( A)  lim A
  0.0006
N  N 100,000
Axiomatic Approach:
 Conditional probability
Conditional Events: If the occurrence of one event has
an effect on the next occurrence of the other event then
the two events are conditional or dependent events.

Male Female Total

Right handed 38 42 80
Left handed 12 8 20
Total 50 50 100
1. What is the probability of left-handed given that
it is a male?
P(LH | M) = 12/50 = 0.24
2. What is the probability of female given that they
were right-handed?
P(F| RH) = 42/80 = 0.525
3. What is the probability of being left-handed?
P(LH) = 20/100 = 0.20
Conditional probability of an event
The conditional probability of an event A given that B
has already occurred, denoted by p ( A B )
p( A  B)
p( A B) = , p( B)  0
p( B)

Remark: (1) p( A' B)  1  p( A B)

'
(2) p( B A)  1  p( B A)
 Independent Events
► Often there are two events such that the occurrence or

non-occurrence of one does not in any way affect the

occurrence or non-occurrence of the other. if events A
and B are independent, P(B/A) = P(B)

Example: Sex of offspring. The chance of a male is

approximately ½. Regardless of the sexes of previous
offspring, the chance of the next child is a male is still ½.
 Summary of basic Properties of probability
 Probabilities are real numbers on the interval from 0 to 1; i.e., 0
 P(A)  1
 If an event is certain to occur, its probability is 1, and if an event
is certain not to occur, its probability is 0.
 If two events are mutually exclusive (disjoint), the probability
that one or the other will occur equals the sum of the
probabilities; P(A or B) = P(A) + Pr(B)
 If A and B are two events, not necessarily disjoint, then
P( A or B) = P (A) +P (B) – P( A and B)
 The sum of the probabilities that an event will occur and that it
will not occur is equal to 1; i.e., P(A) = 1 – P(A)
 If A and B are two independent events, then P ( A and B) = P (A)
* P (B)
 6. Probability distribution
A random variable (r.v): is a function which associates a
number to each possible outcome of an experiment,
usually denoted by capital letters.

A random variable takes a possible outcome and

assigns a number to it.
Random variables are of two types:
Discrete random variable: are variables which can
assume only a specific number of values. They have values
that can be counted.
Examples:
1.Dead/alive
2.Number of car accidents per week.
3.Number of patients.
4.Number of bacteria per two cubic centimeter of water.
Continuous random variable: are variables that can
assume all values between any two given values.
Examples:
1.weight of patients at hospital.
2. blood pressure
3.Life time of light bulbs.
 Definition: a probability distribution consists of a
value of a random variable can assume and the
corresponding probabilities of the values.
Practice Problem:
The number of patients seen in JUSH in any given hour
is a random variable represented by x. The probability
distribution for x is:
x 10 11 12 13 14
P(x) 0.4 0.2 0.2 0.1 0.1
Find the probability that in a given hour:
a. p(x=14)= .1
exactly 14 patients arrive
b. p(x12)= (.2 + .1 +.1) = .4
At least 12 patients arrive
c. At most 11 patients arrive
p(x≤11)= (.4 +.2) = .6
Properties of Probability Distribution:
1. P ( x)  0, if X is discrete.
f ( x)  0, if X is continuous.

2.  PX  x   1 ,
x
if X is discrete.

 f ( x)dx
x
 1 , if is continuous.

Note:
•If X is a continuous random variable then
b
P ( a  X  b)  a
f ( x ) dx
•Probability of a fixed value of a continuous random
variable is zero.
 P ( a  X  b)  P ( a  X  b)  P ( a  X  b)  P ( a  X  b)

•If X is discrete random variable the

b 1
P ( a  X  b)   P ( x )
x  a 1
b 1
P ( a  X  b)   p ( x )
xa
b
P ( a  X  b)   P ( x )
x  a 1
b
P ( a  X  b)   P ( x )
xa

•Probability means area for continuous random variable.

 Introduction to expectation
Definition:
•Let a discrete random variable X assume the values X1,

X2, …, Xn with the probabilities P(X1), P(X2), ….,P(Xn)

respectively. Then the expected value of X ,denoted as

E(X) is defined as:
E ( X )  X 1P( X 1 )  X 2 P( X 2 )  ....  X n P( X n )
n
  X i P( X i )
i 1
•Let X be a continuous random variable assuming the
b

values in the interval (a, b) such that  f ( x)dx  1

b a

,then E ( X )   x f ( x)dx
a

•The variance of X is given by:

Variance of X  var( X )  E ( X )  [ E ( X )]
2 2

Where:
n
E ( X )   xi P( X  xi ) , if X is discrete
2 2

i 1

  x f ( x)dx ,
2
if X is continuous.
x
 There are some general rules for mathematical expectation.
Let X and Y are random variables and k is a constant.

RULE 1 E (k )  k
RULE 2 Var (k )  0
RULE 3 E (kX )  kE ( X )

RULE 4 Var (kX )  k 2Var ( X )

RULE 5 E ( X  Y )  E ( X )  E (Y )
Common Discrete probability distributions

•Binomial Distribution
•Poisson Distribution

Common Continuous Probability Distributions

•Normal Distribution
 A) The Binomial distribution
Assumptions of a binomial distribution.
The experiment consists of n identical trials.
Each trial has only one of the two possible mutually
exclusive outcomes, success or a failure.
The probability of each outcome does not change from
trial to trial, and
The trials are independent, thus we must sample with
replacement.
• The probability distribution of the binomial
random variable X, the number of successes in
n independent trials is:
 n  X n X
f (x )  P (X  x )    p q , x  0,1,2,...., n
x 
 
n 
 
• Where  x 
is the number of combinations of n
distinct objects taken x of them at a time.
n  n!
  
x  x !( n  x )!
 
x !  x (x  1)(x  2)....(1)
* Note: 0! =1
 The parameters of the binomial distribution
are n and p
  E (X )  np
 2  var(X )  np (1  p )
 Exercise
Suppose that in a certain malarias area past experience

indicates that the probability of a person with a high fever will

be positive for malaria is 0.7. Consider 3 randomly selected

patients (with high fever) in that same area.

a) What is the probability that no patient will be positive for
malaria?
b) What is the probability that exactly one patient will be
positive for malaria?
c) What is the probability that exactly two of the patients will
be positive for malaria?
d) What is the probability that all patients will be positive for
malaria?
 B) The Normal distribution
♣ The Normal Distribution is by far the most important
probability distribution in statistics.

♣ The normal distribution is a theoretical, continuous probability

distribution whose equation is:

1  x   2
-  
e 2  
1
f(x)  for - < x < +
2 
 Characteristics of the Normal Distribution
♣ It is a probability distribution of a continuous variable. It
extends from minus infinity( -) to plus infinity (+).
♣ It is unimodal, bell-shaped and symmetrical about x =.
♣ The mean, the median and mode are all equal
♣ The total area under the curve above the x-axis is one square
unit.

♣ The curve never touches the x-axis.

♣ It is determined by two quantities: its mean (  ) and SD (  )

♣ An observation from a normal distribution can be related to a

standard normal distribution (SND) which has a published
table.
 Properties of Normal Distribution
• The normal Distribution is a family of
Bell-shaped and symmetric distributions as the allocation is
symmetric: one-half (.50 or 50%) lies on either side of the mean.

Each is characterized by a different pair of mean, , and

variance, . That is: [X~N()].
 Standard normal distribution
♣ Since the values of  and  will depend on the particular
problem in hand and tables of the normal distribution cannot
be published for all values of  and , calculations are
made by referring to the standard normal distribution which
has  = 0 and  = 1.

♣ Thus an observation x from a normal distribution with mean

 and standard deviation  can be related to a Standard
normal distribution by calculating :
SND = Z = (x -  )

Properties of the Standard Normal Distribution:
 Same as a normal distribution, but also...
 Mean is zero
 Variance is one
 Standard Deviation is one
 Areas under the standard normal distribution curve
have been tabulated in various ways.
 The most common ones are the areas between Z= 0
and a positive value of Z
The Standard Normal Distribution
The standard normal random variable, Z, is the normal random
variable with mean = 0 and standard deviation = 1: Z~N(0,12).

Standard Normal Distribution

0 .4

0 .3

{
=1
f( z)

0 .2

0 .1

0 .0

-5 -4 -3 -2 -1 0 1 2 3 4 5

= 0
Z
 Given a normal distributed random variable X
with Mean μ and standard deviation σ
a X  b
P ( a  X  b)  P (   )
  

a b
P ( a  X  b)  P ( Z )
 
Examples
1. Find the area under the standard normal distribution

which lies
a. Between Z  0 and Z  0.96

b. Between
Z  1.45 and Z  0

c. To the right of Z  0.35

d. Between Z  0.67 and Z  0.75

 Solutions
 a. Area  P(0  Z  0.96)  0.3315
 b. Area  P(1.45  Z  0)
 P (0  Z  1.45)
 0.4265
Area  P( Z  0.35)
 c.
 P(0.35  Z  0)  P( Z  0)
 P(0  Z  0.35)  P( Z  0)
 d.  0.1368  0.50  0.6368
Area  P(0.67  Z  0.75)
 P(0.67  Z  0)  P(0  Z  0.75)
 P(0  Z  0.67)  P(0  Z  0.75)
 0.2486  0.2734  0.5220
 Finding Probabilities of the Standard Normal
Distribution: P(0 < Z < 1.56)
Standard Normal Probabilities
Standard Normal Distribution zz .00
.00 .01
.01 .02
.02 .03
.03 .04
.04 .05
.05 .06
.06 .07
.07 .08
.08 .09
.09
0.0
0.0 0.0000
0.0000 0.0040
0.0040 0.0080
0.0080 0.0120
0.0120 0.0160
0.0160 0.0199
0.0199 0.0239
0.0239 0.0279
0.0279 0.0319
0.0319 0.0359
0.0359
0.4 0.1
0.1 0.0398
0.0398 0.0438
0.0438 0.0478
0.0478 0.0517
0.0517 0.0557
0.0557 0.0596
0.0596 0.0636
0.0636 0.0675
0.0675 0.0714
0.0714 0.0753
0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.3 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6
0.6 0.2257
0.2257 0.2291
0.2291 0.2324
0.2324 0.2357
0.2357 0.2389
0.2389 0.2422
0.2422 0.2454
0.2454 0.2486
0.2486 0.2517
0.2517 0.2549
0.2549
f(z)

0.2 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
0.1 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.56 1.0
1.0 0.3413
0.3413 0.3438
0.3438 0.3461
0.3461 0.3485
0.3485 0.3508
0.3508 0.3531
0.3531 0.3554
0.3554 0.3577
0.3577 0.3599
0.3599 0.3621
0.3621
{

1.1
1.1 0.3643
0.3643 0.3665
0.3665 0.3686
0.3686 0.3708
0.3708 0.3729
0.3729 0.3749
0.3749 0.3770
0.3770 0.3790
0.3790 0.3810
0.3810 0.3830
0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
0.0 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
-5 -4 -3 -2 -1 0 1 2 3 4 5 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
Z 1.5
1.5 0.4332
0.4332 0.4345
0.4345 0.4357
0.4357 0.4370
0.4370 0.4382
0.4382 0.4394
0.4394 0.4406
0.4406 0.4418
0.4418 0.4429
0.4429 0.4441
0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
Look in row labeled 1.5 1.9
2.0
2.0
2.1
0.4713
0.4772
0.4772
0.4821
0.4719
0.4778
0.4778
0.4826
0.4726
0.4783
0.4783
0.4830
0.4732
0.4788
0.4788
0.4834
0.4738
0.4793
0.4793
0.4838
0.4744
0.4798
0.4798
0.4842
0.4750
0.4803
0.4803
0.4846
0.4756
0.4808
0.4808
0.4850
0.4761
0.4812
0.4812
0.4854
0.4767
0.4817
0.4817
0.4857
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
and column labeled .06 to 2.2
2.2
2.3
2.3
0.4861
0.4861
0.4893
0.4893
0.4864
0.4864
0.4896
0.4896
0.4868
0.4868
0.4898
0.4898
0.4871
0.4871
0.4901
0.4901
0.4875
0.4875
0.4904
0.4904
0.4878
0.4878
0.4906
0.4906
0.4881
0.4881
0.4909
0.4909
0.4884
0.4884
0.4911
0.4911
0.4887
0.4887
0.4913
0.4913
0.4890
0.4890
0.4916
0.4916

find P(0 z 1.56) =

2.4
2.4 0.4918
0.4918 0.4920
0.4920 0.4922
0.4922 0.4925
0.4925 0.4927
0.4927 0.4929
0.4929 0.4931
0.4931 0.4932
0.4932 0.4934
0.4934 0.4936
0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.6
2.6 0.4953
0.4953 0.4955
0.4955 0.4956
0.4956 0.4957
0.4957 0.4959
0.4959 0.4960
0.4960 0.4961
0.4961 0.4962
0.4962 0.4963
0.4963 0.4964
0.4964
0.4406 2.7
2.7
2.8
2.8
0.4965
0.4965
0.4974
0.4974
0.4966
0.4966
0.4975
0.4975
0.4967
0.4967
0.4976
0.4976
0.4968
0.4968
0.4977
0.4977
0.4969
0.4969
0.4977
0.4977
0.4970
0.4970
0.4978
0.4978
0.4971
0.4971
0.4979
0.4979
0.4972
0.4972
0.4979
0.4979
0.4973
0.4973
0.4980
0.4980
0.4974
0.4974
0.4981
0.4981
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
3.0
3.0 0.4987
0.4987 0.4987
0.4987 0.4987
0.4987 0.4988
0.4988 0.4988
0.4988 0.4989
0.4989 0.4989
0.4989 0.4989
0.4989 0.4990
0.4990 0.4990
0.4990
Example: A random variable X has a normal distribution
with mean 80 and standard deviation 4.8. What is the
probability that it will take a value
•Less than 87.2
•Greater than 76.4
•Between 81.2 and 86.0
Soluti X is normal with mean,   80, s tan dard deviation ,   4.8
on: X   87.2  
a P ( X  87.2)  P (



)

87.2  80
 P( Z  )
4.8
 P( Z  1.5)
 P( Z  0)  P(0  Z  1.5)
 0.50  0.4332  0.9332
Exercise: Diskin et al. studied common breath metabolites such as

ammonia, acetone, isoprene, ethanol and acetaldehyde in five

subjects over a period of 30 days. Each day, breath samples
were taken and analyzed in the early morning on arrival at the
laboratory. For subject A, a 27-year-old female, the ammonia
concentration in parts per billion (ppb) followed a normal
distribution over 30 days with mean 491 and standard
deviation 119. What is the probability that on a random day,
the subject‘s ammonia concentration is between 292 and 649
ppb?