STAT 111
Chapter Two
Probability
Probability
Many statistical principles and procedures are
based on the important concept of probability. The
purpose of this chapter is to define such concept and
discuss some of its properties. A random experiment
is an experiment in which
All possible outcomes of the experiment are known in
.1
advance,
Any performance of the experiment results in an
.2
outcome that is not known in advance,
The experiment can be repeated under identical
.3
conditions
Examples
1)
2)
3)
4)
Tossing a coin once or several times.
Obtaining blood types from a group of
individuals.
Determining the sex of a newborn.
Tossing two dice.
Definition
Definition: The sample space of an experiment,
denoted by S, is the set of all possible outcomes of that
experiment.

Sample spaces are either
Discrete : which contains a finite number of elements ,
or and infinite but countable number of elements .
Continuous : which contains an infinite number of
sample points constituting a continuum , such as all
points on a line segment or all the points in a plane .

S
Discrete
Finite
Continuous
(infinite)
Infinite
(Countable)
.1
.2
Example
Describe an appropriate sample space for the
experiments below. Determine the number of elements
and state whether the sample space is discrete or not.
H
H
A coin is tossed two times.
‫طرق حساب فراغ العينة إذا كان محدود‬
‫أوالً باستخدام الشجرة البيانية‬
T
S = {HH, HT, TH, TT} n(S) = 4, discrete
H
T
T



Example (continued)
A coin is tossed and a dye is rolled
‫ثانيا ً باسخدام الجداء الديكارتي‬
S = (H,T)x(1,2, ...,6)
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)},
n(s) = 2x6=12, discrete


Two numbers are selected from the set {1, 2, 3} without repetition of
digits.
‫ثالثا ً باستخدام الشبكة التربيعية‬
S = {(1,2), (1,3), (2,1), (2,3), (3,1), (3,3) }
n(S) = 3x2=6, discrete
3
(1,3)
2
(1,2)
(2,3)
(3,2)
(2,1)
1
1
2
(3,1)
3
Example (continued)
A coin is tossed until the first head appears.
S = {H, TH, TTH, ...,∞), and so the coin is tossed an infinite
number of times, here ∞, refers to the case when a head
never appears.

A light bulb is observed so that the length of its useful
life might be recorded.
S = {(0, ∞)}, since one could not say with certainty that
the bulb would have burned out by any given time => S
is continuous.

Events
Definition An event is any subset of the sample space S. An
event A occurs, if the outcome of the experiment is in A.
Example Two cards are drawn, randomly with replacement
from three cards carrying the number 1,3, 5. Describe the
sample space, find the events
(3,5)
(5,5)
A= {sum of the5two(1,5)
numbers
is 5 }
(1,3) is(3,3)
(5,3) 6}
B= {sum of the numbers
at least
3
(1,1)
(3,1)
(5,1)
A=Φ
1
B= {(1,5), (3, 3), (3, 5), (5,1), (5, 3), (5, 5)}
1
3
5
Example
Example:
An experiment consists of rolling a die until a 3 appears,
Construct a sample space for this experiment. Let 3 denotes 3
(1
appears , let A denote 3 does not appear
S = { 3 , A 3 , AA3 , …} .
List the elements in E , the event that 3 appears before the fifth roll
(2
E = { 3 , A3 , AA3 , AAA3 }
Some Relations from Set Theory
An event is nothing but a set, so that
relationships and results from elementary set
theory can be used to study events. The following
concepts from set theory will be used to construct
new events from given events.
The union of two events A and Bdenoted
by A  B
(1
A
and read 'A or B' is the event consisting of all
outcomes which are either in A or in B or in both
events, that is
A U B= {x ϵS :x ϵA or xϵ B}
If A1, A2, ... are events, the union of these events, denoted,
by
is defined to be that event which consists of all points that
are in An for at least one value of n = 1,2, ...

n
n 1
Some Relations from Set Theory
The intersection of two events A and B, denoted by A  B and
(2
read 'A and B' is the event consisting of all outcomes which are
in both A and B, that is,

A ∩ B={x ϵ S :x ϵ A and x ϵ B}
An

If A1, A2, ... are events, the intersection of these events, denoted
n 1 by
, is defined to be that event consisting of all points that are in
all of
An, n= 1 , 2 , …
The complement of an event A , denoted by A ( or Ac ) is the
set of all outcomes in S which are not contained in A . That is
Ac = { x ϵ S : x
(3
ϵ
A}
Some Relations from Set Theory
Definition
When A and B have no outcomes in common (the
intersection of A and B contains no outcomes,
i.e. A ∩ B= Φ), they are said to be mutually exclusive of
disjoint events.
Definition A nonempty collection of subsets F of a set
S is called a σ-filed of subsets of S provided the
following properties hold.
1. S  F
2. If A  F then A c  F


n 1
n 1
3. If A n  F, n  1,2,..., then  A n and  A n are both in F
Probability
Definition
A probability measure P (or simply probability) is a real-valued function
having domain F satisfying the following properties.
Axioms of Probability:

 
P  An    An
 n 1  n 1
A
1- 0≤ P(A) ≤ 1
ϵ F
2- P(S)=1
3- If An, n = 1,2, ..., are mutually disjoint sets in S, then
A probability space , denoted by ( S , F , P ) is a σ– field of subsets F and a
probability measure P defined on S .
Probability
Definition
An assignment of probability is said to be equally likely
(or uniform) if each elementary event in S is assigned
1
the same probability.
Pw  
n
with this assignment
Thus if S contains n elements
wi, i.e. if S = { w1, …,wn}
number of elements in A n A 
PA  

then
number of elements in S n S
i
Properties of probability
In the following, some additional properties of probability measure P will
be presented. These properties follow from the definition of a
probability
measure.
P( Φ ) = 0
(1
Let A1, A2, ..., An a collection of pairwise disjoint events in S, then
(2
If A is an event in S, then P(A)=1-P(A)
Let A and B be events in S , then P (A ) = P (A ∩ B ) + P ( A ∩ Bc)
Let A and B be event in S, then P(A U B ) = P(A)+ P(B) - P(A ∩ B)
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) +
P(A ∩ B∩C)
If A  B , then P ( A ) ≤ ( B )
(3
(4
(5
(6
(7
Example 1
Example 1:6
If two dice are rolled what is the probability that
5
The sum of upturned faces will equal 7?
(1
Psum  7  
6 1

The sum of upturned faces will
36equal6 2 or
12?
(1,6)
(2,6)
(3, 6)
(4, 6)
(5,6)
(6, 6)
(1,5)
(2,5)
(3, 5)
(4,5)
(5, 5)
(6,5)
4
(1,4)
(2,4)
(3, 4)
(4, 4)
(5, 4)
(6, 4)
3
(1,3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
(6, 3)
2
(1,2)
(2,2)
(3, 2)
(4, 2)
(5, 2)
(6, 2)
1
(1,1)
(2, 1)
(3,1)
(4,1)
(5, 1)
(6,1)
1
2
3
4
5
6
(2
1
1 2
Psum  2   Psum  12  


36 36 6
The sum of upturned faces will be an even
number or a number less than 6? Let A =
{sum is even), B = {number less than 6}
PA  B  PA   PB  PA  B 
(3
18 10 4 24
 

36 36 36 36
Example 2
A
Ac
Total
B
0.21
0.09
0.3
Bc
0.38
0.32
0.7
Total
0.59
0.41
1
Given P(A) = 0.59, P(B) = 0.30, P (A ∩ B) = 0.21,
find
P ( A U B ) = 0.59 + 0.30 – 0.21 = 0.68
(1
P (A ∩ Bc) = P ( A ) – P ( A ∩ B ) = 0.59 - 0.21 = 0.38 (2
P ( Ac U Bc ) = 1- P ( A ∩ B ) = 1 – 0.21 = 0.79 or
(3
P (AcUBc)=P(Ac)+P(Bc)–P(Ac∩Bc)=0.41 +0.7 -0.32=0.79
P ( Ac ∩ Bc) = 1- P (A U B ) = 1-0.68 =0.32
(4
Example 3
If A and B are two
events such that A  B.
What is P(A  B), what
is P(A ∩ B) and what is
P(A ∩ Bc)?
P(A  B)=P(B)
P(A ∩ B)=P(A)
P(A ∩ Bc)=P(A)- P(A ∩ B)
= P(A)- P(A)=0
S
B
A
Example 4
If A, B and C are mutually exclusive events and P(A) = 0.2,
P(B) = 0.3, and P(C) = 0.2, find
C S
A
P(A U B U C)?
.1
0.2
0.2
B
0.3
0.3
P(A U B U C)= P(A) + P(B) + P(C) = 0.2 + 0.3 + 0.2 = 0.7
P(Ac ∩ (B U C))?
C
A
S
.2
B
P(Ac ∩ (B U C))=P(B) + P(C) = 0.3 + 0.2 = 0.5
Example 5
A two-digit number is formed by randomly selecting, with replacement,
digits from the set {6, 7, 8, 9}. Find the probability;
The two digits are the same?
.1
n(S)=4x4=16
n(two digits are the same)=4x1=4
Remember
P(two digits are the same)=4/16=1/4
n A
P  A 
nS  is odd?
The number
.2
n(number is odd)=4x2=8
P(number is odd)=8/16=1/2
Example 6
Remember
P  A 
n A
nS 
If 10 people, including A and B, are randomly arranged in a line,
What is the probability that A and B are next to each
.1
other?
2!(10  1)!
10!be if the people were randomly
What would probability
arranged in a circle?
2!(10  2)!
10!
.2
Independent Events
Suppose that two events A and B occur independently of one another in the
sense that the occurrence or nonoccurrence of either of them has no
relation to the occurrence or nonoccurrence of the other, that is the
probability that A and B will occur is equal to the product of their individual
probabilities.
Definition
Two events A and B independent if and only if
P (A ∩ B ) = P (A ) P ( B )
More generally , for n ≥ 3 , events A1 , A2 , … , An are independent if
P ( A1 ∩ A2 ∩ … ∩An) = P (A1) P(A2)P(A3) …P ( An)
And if any subcollection containing at least two but fewer than n
events are independent .
Events that are not independent are said to be dependent. Note that 
independence of events is not to be confused with disjoint or mutually
exclusive events.
Result Assuming that A and B are independent events, then the events Ac 
and B, A and Bc, Ac and Bc are also independent.
Example 1
Let two fair coins be tossed, let A = {head on the second
throw}, B = {head on the first throw}. Show that A and B
are independent.
S = {HH, HT, TH, TT}
A = {HH, TH}
B = {HH, HT}
A∩B={HH}
P(A∩B)=1/4
P(A)P(B)=½ x ½=1/4= P(AB)
A and B are independent
Example 2
Two students A and B are both registered for a certain course.
If student A attends class 80 percent of the time and student B
attends class 60 percent of the time, and if the absence of the two
students are independent, what is the probability that at least one
of the two students will be in class on a given day?
P(A) = 0.8, P(B) = 0.6
Ac and Bc are independent
A and B are also independent
P ( A U B ) = P ( A ) + P (B) – P (A ) P (B ) = 0.8 + 0.6 – 0.6x0.8=0.92
Example 3
A coin is biased so that a head is twice as likely to occur as a tail, If the coin is tossed 3
times, what is the probability of getting 2 tails and 1 head?
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT},
let P(H) = 2W, P(T) = 1W
2W + 1W =1
W=1/3
P(H) = 2/3 and P(T) = 1/3
A = {TTH,THT,HTT},
P(TTH) = P(T) P(T) P(H)= 1/3 x 1/3 x 2/3 = 2/27 by independence
P(THT) = P(T) P(H) P(T)= 1/3 x 2/3 x 1/3 = 2/27 by independence
P(HTT) = P(H) P(T) P(T)= 2/3 x 1/3 x 1/3 = 2/27 by independence
P(A) = P(TTH) + P(THT) + P(HTT)= 2/27 + 2/27 + 2/27=6/27=2/9
Example 4
Suppose that three balanced coins are tossed. Find the probability that at least one of
them lands head.
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT},
A1=first coin lands head= {HHH, HHT, HTH, HTT}
A2=second coin lands head={HHH,HHT,THH,THT} and
A3=third coin lands head={HHH, HTH, THH, TTH}
P(A1)=4/8=1/2
Remember
c P(Ac 2)=4/8=1/2
(A  B )c  A U B (De Morgan' s laws)
P(A3)=4/8=1/2
P ( at least one lands head ) = P (A1 U A2 U A3)

P A1  A2  A3   P A  A  A
c
1
   
c
2

c c
3

 1  P A1c  A2c  A3c
1 1 1
1 7
 1  P A1c P A2c P A3c  1     1  
2 2 2
8 8



P Ac  1  P A
Example 5
A system containing five components is arranged in the manner shown in
Figure - 1, where the probabilities given indicate the chance that the
component will work. If we assume that whether a component works or
not is independent of whether any other component is working or not.
what is the probability that the system will work?
P(B)=0.90
P(D)=0.93
P(A)=0.98
P(C)=0.95
P(E)=0.97
P(system work) =P(A ∩ (B U C) ∩ (D U E))
= P(A)P(B U C)P(D U E) since independent
P(B U C)=1- P(Bc ∩ Cc)= 1- P(Bc)P( Cc)=0.995 since independent
Or ( B U C ) = P ( B ) + P ( C ) – P (B ) P (C) =0.995 since independent
Similarly, P(D U E)= 1- P(Dc ∩ Ec)= 1- P(Dc)P( Ec)=0.9979
P(system work) = P(A)P(B U C)P(D U E)=0.98x0.995x0.9979=0.973