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STAT 111
Chapter Two
Probability
Probability
1.
2.
3.
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
advance,
Any performance of the experiment results in an outcome
that is not known in advance,
The experiment can be repeated under identical 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 .
1.
2.
S
Discrete
Finite
Continuous
(infinite)
Infinite
(Countable)
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.

A coin is tossed two times.
H
H
‫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
(1,5)
(3,5)
(5,5)
A= {sum of the two numbers is 5 }
5
(1,3)
(3,3)
(5,3)
B= {sum of the numbers is at least 6}
3
A=Φ
(1,1)
(3,1)
(5,1)
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,
1)
Construct a sample space for this experiment. Let 3 denotes 3
appears , let A denote 3 does not appear
S = { 3 , A 3 , AA3 , …} .
2)
List the elements in E , the event that 3 appears before the fifth
roll
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.
1)
The union of two events A and B denoted by A  B 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  An
n 1
is defined to be that event which consists of all points that
are in An for at least one value of n = 1,2, ...
Some Relations from Set Theory
The intersection of two events A and B, denoted by A  B and 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 by n 1 , is
defined to be that event consisting of all points that are in all of
An, n= 1 , 2 , …
2)
3)
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
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:
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

 
P  An    An
 n 1  n 1
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 the
same probability.
Thus if S contains n elements wi, i.e. if S = { w1, …,wn}
then
1


P wi 
n
with this assignment
number of elements in A n A 
PA  

number of elements in S n S
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.
1)
P( Φ ) = 0
2)
Let A1, A2, ..., An a collection of pairwise disjoint events in S, then
3)
4)
5)
6)
7)
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 )
Example 1
Example 1:
If two dice are rolled what is the probability
that
1)
The sum of upturned faces will equal
7?
Psum  7  
2)
6 1

36 6
The sum of upturned faces will equal
2 or 12?
1
1 2
Psum  2  Psum  12 


36 36 6
3)
6
(1,6)
(2,6)
(3, 6)
(4, 6)
(5,6)
(6, 6)
5
(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
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 
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
1)
P ( A U B ) = 0.59 + 0.30 – 0.21 = 0.68
2)
P (A ∩ Bc) = P ( A ) – P ( A ∩ B ) = 0.59 - 0.21 = 0.38
3)
P ( Ac U Bc ) = 1- P ( A ∩ B ) = 1 – 0.21 = 0.79 or
P (AcUBc)=P(Ac)+P(Bc)–P(Ac∩Bc)=0.41 +0.7 -0.32=0.79
4)
P ( Ac ∩ Bc) = 1- P (A U B ) = 1-0.68 =0.32
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
1.
P(A U B U C)?
A
S
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
2.
P(Ac ∩ (B U C))?
C
A
B
P(Ac ∩ (B U C))=P(B) + P(C) = 0.3 + 0.2 = 0.5
S
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?
n(S)=4x4=16
n(two digits are the same)=4x1=4
P(two digits are the same)=4/16=1/4
1.
The number is odd?
n(number is odd)=4x2=8
P(number is odd)=8/16=1/2
2.
Remember
n A
P  A 
nS 
Remember
P  A 
Example 6
n A
nS 
If 10 people, including A and B, are randomly arranged in a
line,
1.
What is the probability that A and B are next to each
other?
2!(10  1)!
10!
2.
What would probability be if the people were randomly
arranged in a circle?
2!(10  2)!
10!
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}
Remember
P(A1)=4/8=1/2
(A  B )c  Ac U Bc (De Morgan' s laws)
P(A2)=4/8=1/2
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