1. science
2. mathematics
3. algebra

Chapter 2
Probability
LEARNING OBJECTIVES
• Understand and describe sample spaces and events
• Interpret probabilities and use probabilities of outcomes
to calculate probabilities of events in discrete sample
spaces
• Calculate the probabilities of joint events such as
unions and intersections from the probabilities of
individual events
• Interpret and calculate conditional probabilities of
events
• Determine the independence of events and use
independence to calculate probabilities
• Use Bayes’ theorem to calculate conditional
Sample Space and Events
• An experiment is any
action or process that
generates observations
• The sample space of an
experiment, denoted S, is
the set of all possible
outcomes, or sample
points.
• Example: Toss a fair coin 3
times in a row
– The sample space has 8
sample points.
– S={HHH, THH, HHT, THT,
HTH, TTH, HTT, TTT}
• An event is a subset of the
sample space S
• Example: look at 3 different
events of previous example
– The event of 3 heads,
• A= {HHH}
– The event of 2 heads,
• B={HHT, HTH, THH}
– The event that the last toss is a
head,
• C={HHH, HTH, THH, TTH}
Set Relations
• Suppose S is the
universal set, with two
subsets, A and B
• A set, A, is a subset of
B if all elements of A
belong to B, A⊂B
S
B
• The union of two events
A and B, denoted by
A∪B, and read “A or
B,” is the event
consisting of all
elements that are either
in A, in B, or in both
• Or the union A∪B = { x
| x ∈ A or x ∈ B}
A
S
A
B
Set Relations-Cont.
• The intersection of two
• The complement of an
events A and B, denoted by
event A, denoted by A´,
A∩B, and read “A and B, ”
is the set of all elements
is the event consisting of
in S that are not
all elements that are in both
contained in A
• The intersection A∩B={x |
x ∈A and x ∈B} is the
S
A´
subset of S which contains
A
all elements that are in both
A&B
S
A
B
Set Relations-Cont.
• Sets A and B are
mutually exclusive or
disjoint, if and only if
A∩B = Ø, or events A
& B have no elements
in common
S
A
B
• Any number of sets,
A1, A2, A3,… are
mutually exclusive if
and only if Ai∩Aj = Ø
for i≠j.
• A1 ∩A2 = Ø A2 ∩ A3 = Ø
• A1 ∩ A3= Ø A2 ∩ A4 = Ø
• A1 ∩ A4 = Ø A3 ∩ A4 = Ø
S
A1
A2
A3
A4
Example:1
• For an experiment, let
– A = {0,1,2,3,4},
– B = {3,4,5,6}, and
– C = {1,3,5}
•
•
•
•
•
•
•
Determine:
AB
AC
AB
AC
A´
{AC}´
•
•
•
•
•
•
•
Solution:
AB={0,1,2,3,4,5,6}
AC={0,1,2,3,4,5}
AB={3,4}
AC={1,3}
A´={5,6}
{AC}´={6}
Example:2
• The rise time of a reactor is
measured in minutes (and
fractions of times). Let the
sample space positive, real
numbers. Define the events A
and B as follows:
A={x | x<72.5} and B={x |
x>52.5}
• Describe each of the
following events.
• a) A´
• b) B´
• c) AB
• d) AB
•
•
•
•
Solution:
a) A = {x | x  72.5}
b) B = {x | x  52.5}
c) A  B = {x | 52.5 < x <
72.5}
• d) A  B = {x | x > 0}
Example:3
• In an injection-molding operation,
several characteristics of each
molded part are evaluated
• Let A denote the event that a part
meets customer shrinkage
requirements, B denote the event
that a part meets customer color
requirements, and C denote the
event that a critical length meets
customer requirements
• a) Construct a Venn diagram that
includes these events and indicate
the region in the diagram in which a
part meets all customer
requirements. Shade the areas that
represent the following
• b) BC
• c) A´B
• d) AB
• Solution
Class Problem
• Disks of polycarbonate plastic
from a supplier are analyzed for
scratch resistance and shock
resistance. The results from 100
disks are summarized below.
Shock Resistance
High low
Scratch
High 70
9
Resistance
low
16
5
• Let A denote the event that a disk
has high shock resistance, and let
B denote the event that a disk has
high scratch resistance. Determine
the number of disks in A∩B, A´,
and A∪B
• Solution:
Number of samples in
A∩B=…
Number of samples in A' =
…
Number of samples in A∪B
=…
Interpreting Probabilities
• The assignment of a
• Example: Toss a fair coin
weight between 0 and 1 to
three times in a row
indicate the likelihood of
– The probability of getting
3 heads P(A)= 1/8
the occurrence of an
– The probability of getting
event.
2 heads P(B)=3/8
• The probability of an
– The probability that the
event is defined in terms
last toss is a head P(C) =
of an experiment and a
4/8 =1/2
sample space.
Axioms of Probability
1.
For any event A, P (A) 0
•
2.
P(S)=1
•
3.
The chance of occurring
should be at least 0
The maximum possible
probability is assigned to S
Let A1, A2, A3,…, An,… be
a finite or infinite sequence
of mutually exclusive
events. Then
P(A1∪A2∪A3…) = P (A1) +
P(A2) + P(A3)+…=∑ P(Ai)
• Example
• If an experiment has the three
possible and mutually
exclusive outcomes A, B, and
C, check in each case whether
the assignment of
probabilities is permissible:
• P(A) = 1/3, P(B)= 1/3, and
P(C) = 1/3
• P(A) = 0.64, P(B)= 0.38, and
P(C) = -.02
• P(A) =0.35, P(B)=0.52, and
P(C) =0.26
• P(A) =0.57, P(B)=0.24, and
P(C) =0.19
Class Problem
• The sample space of a random
experiment is {a, b, c, d, e}
with probabilities 0.1, 0.1, 0.2,
0.4, and 0.2, respectively. Let
A denote the event {a, b, c},
and let B denote the event {c,
d, e}. Determine the following
• a) P(A)
• b) P(B)
• c) P(A')
• d) P(A∪B)
• e) P(A∩B)
•
•
•
•
•
•
Solution:
a) P(A) =
b) P(B) =
c) P(A') =
d) P(AB) =
e) P(AB) =
Rules of Probability
Complement Rule
• The probability of
impossible events is 0:
P(Ø) =0
• Complement rule:
P(A´)= 1- P(A)
S
A´
A
• Proof
• From axioms 3 for finite
case, let k=2, A1=A and
A2=A'
• By definition, A∪A' =S
while A and A' are
mutually exclusive
• 1=P(S)=P(A∪A')=P(A)+
P(A' )
• P(A´)= 1- P(A)
Addition Rule
• For any two events A
and B
P(A1∪A2)=P(A1)+P(A2) –
P(A1∩A2)
• By Venn diagram
• The probability of a
union of more than two
events
P(A1∪A2∪A3)=
P(A1)+ P(A2)+ P(A3)
+P(A1∩A2∩A3)
-P(A1∩A2)
-P(A2∩A3)
-P(A1∩A3)
Example: 1
• If P(A) =0.3, P(B)=0.2,
and P(AB) =0.1,
determine the following
probabilities
•
•
•
•
•
•
a) P(A´)
b) P(AB)
c) P(A´B)
d) P(AB´)
e) P[(AB)´]
f) P(A´B)
• Solution
• a) P(A') = 1- P(A) = 0.7
• b) P(AB) = P(A) + P(B) P(AB) = 0.3+0.2 - 0.1 = 0.4
• c) P(A´B)+ P(AB) = P(B).
Therefore, P(A´B)= 0.2 0.1 = 0.1
• d) P(A) = P(AB) + P(AB´)
Therefore, P(AB´) = 0.3 0.1 = 0.2
• e) P((AB) ') =1 - P(AB)=
1 - 0.4 = 0.6
• f) P(A´B)= P(A') + P(B) P(A´  B)= 0.7 + 0.2 - 0.1 =
0.8 from part c
Example:2
• Denote the six events
1,2,3,4,5, and 6 associated
with tossing a six-sided die
once by E1, E2, E3, E4, E5,
and E6
• Suppose the die is constructed
so that any of the three even
outcomes is twice as likely to
occur as any of the three odd
outcomes
• Determine P(A) where the
event A is even
• Determine P(B) where the
event B is less than or equal to
3
•
•
•
•
Solution:
P(E1)=P(E3)=P(E5)=…
P(E2)=P(E4)=P(E6)=…
Define A={outcome is
even}= E2  E4  E6
• P(A)= P(E2)+P(E4)+P(E6)=…
• B={outcome ≤3}= E1E2E3
• P(B)= P(E1)+P(E2)+P(E3)=…
Class Problem
•
Disks of polycarbonate plastic from a
supplier are analyzed for scratch
resistance and shock resistance. The
results from 100 disks are summarized
below.
Shock Resistance
High
low
Scratch
High
70
9
Resistance
low
16
5
a) If a disk is selected at random, what is
the probability that its scratch resistance
is high and its shock resistance is high?
b) b) If a disk is selected at random, what
is the probability that its scratch
resistance is high or its shock resistance
is high?
c) Consider the event that a disk has high
scratch resistance and the event that a
disk has high shock resistance. Are
these two events mutually exclusive?
• Solution
• Let A denote the event
that a sample has high
shock resistance and let B
denote the event that a
sample has high scratch
resistance.
a) P(AB) = …
b) P(AB) = P(A) + P(B)
- P(AB) = …
c) Because (AB) does
not equal Ø , A and B…
Equally Likely Outcomes
• In an experiment
consisting of N outcomes,
it is reasonable to assign
equal probabilities to all
N sample events
1  i 1 P( Ei )  i 1 p  p.N
N
• So, p=1/N
N
• Example
• When two dice are
rolled separately, there
are N=36 outcomes,
which are equally likely
or P(Ei)=1/36
• Let A={sum of two
numbers=7}
• P(A)=…
Counting Techniques
• Ability to count number of
elements in the sample space
without listing actually each
element
• The Product Rule for
Ordered Pairs
• If the first object of an ordered
pair can be selected in n1 ways,
and for each of these n1 ways
the second object of the pair
can be selected in n2 ways,
then the number of pairs is n1
n2
• Example
– A homeowner requires
two types of contractors,
plumbing and electrical
– 3 plumbing contractors
– 3 electrical contractors
– How many possible ways
of choosing the two types
of contractors?
• N= n1 n2 =3*3=9
Tree Diagrams
• Used to represent pictorially
all the possibilities
• Starting on the left side of the
diagram, for each possible
first element of a pair a
straight-line segment
emanates rightward
• Construct another line
segment emanating from the
tip of the branch for each
possible choice of a second
element of the pair
• A more general Product Rule
• N= n1 n2 n3 … nk
• Example
P2
E1
E2
E3
E1
E2
P3
E3
P1
E1
E2
E3
Permutations
• Any ordered sequence of k objects • Example
taken from a set of n distinct objects
– Consider the set {A,B,C,D,E}
consisting of 5 elements
is called a permutation of size k of
– Number of permutations of
the objects
size 3?
• The number of permutations of size
– By taking 5 letters three at a
k that can be formed from the n
time
objects is denoted by Pk,n
– P5,3 = 5!/(5-3)! = 60
• Obtained from the general product • Class Problem
rule
– Three awards will be given
• Pk,n=n(n-1)(n-2)…(n-k+2)(n-k+1)
for a class of 25 graduate
students
• Using factorial notation
– If each student can receive at
most one award
n(n  1)...( n  k  1)( n  k )( n  k  1)...( 2)(1)
Pk ,n 
– How many possible
(n  k )( n  k  1)...( 2)(1)
selections?
n!
• P25,3=…
P 
k ,n
(n  k )!
Combinations
• Any unordered sequence of k
objects taken from a set of n
distinct objects is called
combination of size k of the
objects
• The number of combinations
of size k that can be formed
from n distinct objects will be
n
denoted by
k
• Smaller than the number of
permutations because the
order is disregarded
(
)
• Example:
– Consider the set {A,B,C,D,E}
consisting of 5 elements
– Number of combinations of size
3?
– There are six permutations of
size 3 consisting of the elements
A, B, and C (3!=6)
– These six permutations are
equivalent to the single
combination {A,B,C}
– So, 60/3! =10
• In general
P
n!
( )

k! k!(n  k )!
n
k
k ,n
Class Problem
• Car dealer service 10 foreign
cars and 15 domestic cars on
a particular day
• There are only 6 mechanics
• If 6 cars are chosen at
random, what is the
probability that exactly 3 of
the cars are domestic and the
other cars are foreign?
• What is the probability that at
least 3 of the cars are
domestic and the other cars
are foreign?
• Solution
• Let D3={exactly 3 of the 6
cars chosen are domestic}
• P(D3)=…
• P(D3D4 D5 D6) =…
Conditional Probability
• Interested at the probability • Simple example concerns the
of an event occurring
situation in which two events
conditional on the
are mutually exclusive
knowledge that another
• P(B|A)= P(A∩B)/P(A)=0/P(B)=0
event has occurred
• Let A and B be events with
P(A) 0
A
B
• The conditional
probability of B given A is
P(B|A)= P(A∩B)/P(A)
• Note: P(B|A) is undefined
if P(A) =0.
A
B
Examples
• If the probability that a
research project will be
planned is 0.80 and the
probability that it will be
planned and well executed is
0.72, what is the probability
that a research project, which
is well planned, will also be
well executed?
• Solution
• P(A/B) =
P(A∩B)/P(B)=0.72/0.80=0.90
• If the probability that a
communication system will
have high fidelity is 0.81, and
the probability that it will
have high fidelity and high
selectivity is 0.18, what is the
probability that a system with
high fidelity will also have
high selectivity?
– A: A communication system
which has high selectivity
– B: A communication system
which has high fidelity
• Data: P(B)=0.81,
P(A∩B)=0.18
• P(A/B)=P(A∩B)/P(B)=
0.18/0.81=2/9
Class Problem
• A magazine publishes three columns entitled A, B, and C
• Reading habits of a selected readers with respect to these
columns
Read regularlyA
B
C
A∩B A∩C B∩C A∩B∩C
Probability
0.14 0.23 0.37 0.08 0.09 0.13 0.05
• P(A|B)=…
• P(A|B  C)=…
• P(A| reads at least one) = P(A|ABC)=…
• P(AB |C)=…
Class Problem
•
Disks of polycarbonate plastic from a
supplier are analyzed for scratch
resistance and shock resistance. The
results from 100 disks are summarized
below.
Shock Resistance
High
low
Scratch
High
70
9
Resistance
low
16
5
• Let A denote the event that a disk has
high shock resistance, and let B denote
the event that a disk has high scratch
resistance. Determine the following
probabilities.
a) P(A)
b) P(B)
c) P(A/B)
d) P(B/A)
•
a)
b)
c)
d)
Solution
P(A) =…
P(B) = …
P(A/B)=…
P(B/A)=…
Multiplication Rule
• From the definition of
conditional probability, we
can write
P(A∩B)= P(A) P(B/A)
• Example: The supervisor of
a construction group (20
workers) wants to get the
opinion of 2 of them about
new safety regulations. If 12
of them favor the new
regulations and the other 8
are against it, what is the
probability that both of them
chosen by the supervisor will
be against the new safety
regulation?
• Solution
• P(A)=the first worker
selected will be against the
new safety regulation=8/20
• P(B/A)=the second worker
selected will be against given
that the first one is
against=7/19
• P(A∩B)= P(A) P(B/A)=
– (8/20)(7/19)=14/95
Independent Events
• Two events A and B are
independent if and only if
P(B/A)=P(B) if P(A)0 and
P(A/B)=P(A) if P(B)0
• P(B/A)=P(B) means the
probability of event B
remains the same whether or
not event A is conditional
upon
• One event does not affect the
probability of the another
event
• P(A∩B)=P(A)P(B/A)=
P(A)P(B)
• Example 1:
– What is the probability of
getting two heads in two
flips of a balanced coin?
• Solution
– (1/2)(1/2)=1/4
• Example 2:
– If P(C)=0.65, P(D)=0.4, and
P(C∩D)=0.24, are the
events C and D
independent?
• Solution
– P(C)P(D)=(0.65)(0.4)=0.26
0.24
Class Problem
•
Disks of polycarbonate plastic from
a supplier are analyzed for scratch
resistance and shock resistance. The
results from 100 disks are
summarized below.
Shock Resistance
High
low
Scratch
High
70
9
Resistance
low
16
5
• Let A denote the event that a disk
has high shock resistance, and let B
denote the event that a disk has
scratch resistance. Are events A and
B independent?
• Solution:
– P(A∩B)= = …
– P(A) = …
– P(B) = 79/100
Total Probability Rule
• Let A1, A2, A3,…, An be a
collection of mutually
exclusive events with known
probabilities which partition S
• Consider an event B with
known conditional
probabilities
A1
A2
An
B
• How to use P(Ai) and P(B/Ai)
to calculate P(B)
• Easily achieved by
B=(A1∩B)∪… ∪(An∩B)
• Events Ai∩B are mutually
exclusive
P(B)=P(A1∩B)+… P(An∩B)
=P(A1)P(B/A1)+…+P(An)P(B/An)
• Known as the law of total
probability
• Example
– Suppose that P(A/B)=0.2,
P(A/B')=0.3, and P(B)=0.8. What
is P(A)?
• Solution
P ( A)  P ( A  B )  P ( A  B )
 P ( A B ) P ( B )  P ( A B ) P ( B )
 (0.2)(0.8)  (0.3)(0.2)
 0.16  0.06  0.22
Example
• In a certain assembly plant, • Let
three machines, B1, B2, and
– A: the product is defective
B3, make 30%, 45%, and
– B1=the product is made by machine B1
25%, respectively, of the
– B2=the product is made by machine B2
products
– B3=the product is made by machine B3
• It is known from past
• P(A)=P(B1)P(A/B1)+…+P(Bn)P(A/Bn)
experience that 2%, 3%, and
– P(B1)P(A/B1)=(0.3)(0.02)=0.006
2% of the products made by
– P(B2)P(A/B2)=(0.45)(0.03)=0.0135
each machine, respectively,
– P(B3)P(A/B3)=(0.25)(0.02)=0.005
are defective
• P(A)=0.006+0.0135+0.005=0.0245
• Select a finished product
randomly
• What is the probability that it
is defective?
Posterior Probabilities
• How to use the probabilities
P(Ai) and P(B/Ai) to calculate
the probabilities P(Ai/B)
• This is the revised
probabilities of the events Ai
conditional on the event B
• Assume P(A1), …, P(An) are
the prior probabilities of
events A1, …, An
• Observation of the event B
provides some additional
information to revise these
prior probabilities
• Called posterior probabilities
and conditional on the event B
• From the law of total
probability
P ( Ai B ) 
P ( Ai  B )
P( B )
P ( Ai ) P ( B Ai )

P( B)


P ( Ai ) P ( B Ai )
n
j 1
P( Aj ) P( B Aj )
• Known as Bayes’ Theorem
Example
• In a certain assembly plant,
three machines, B1, B2, and
B3, make 30%, 45%, and
25%, respectively, of the
products
• It is known from past
experience that 2%, 3%, and
2% of the products made by
each machine, respectively,
are defective
• Suppose a product is selected
randomly and it is defective
• What is the probability made
by machine Bi (say B3)?
• Let
–
–
–
–
A: the product is defective
B1=the product is made by machine B1
B2=the product is made by machine B2
B3=the product is made by machine B3
• Instead of asking P(A), we want
to find P(Bi/A)
• Using the Bayes’ formula
P( A ) P( B A )
P( A B) 
 P( A ) P( B A )
i
i
i
n
j 1
j
j
P( B ) P( A B )
P ( B A) 
 P( B ) P( A B )
3
3
3
n
j 1
j
• P(B3/A)=0.005/0.0245 = 10/49
j
Prior and Posterior Probabilities
• Prior probabilities
•
– It is known from past experience
that 2%, 3%, and 2% of the
products made by each machine,
respectively, are defective
Past data
– B1=the product is made by machine B1
– B2=the product is made by machine B2
– B3=the product is made by machine B3
• P(B1)=0.02
• P(B2)=0.03
• P(B3)=0.02
• Posterior probabilities
– Revision of the prior
probabilities conditional on the
event A
– A product is selected randomly
and it is defective, what is the
probability made by machine
Bi?
• P(B1/A)=0.006/0.0245 =
• P(B2/A)=0.0135/0.0245 =
• P(B3/A)=0.005/0.0245 = 10/49
Class Problem
•
•
A TV store sells 3 different brands of TVs
Of its TV sales, 50% are brand 1, 30% are brand 2, and 20% are
brand 3
• Known that 25% of brand 1 requires warranty repair work,
whereas brand 2 and 3 are 20% and 10%, respectively
1. What is the probability that a selected buyer has a TV that will
need repair while under warranty?
2. If a customer returns a TV brand, what is the probability that is a
brand 1? A brand 2? A brand 3?
Solution
Tree Diagram
P(B/A1) P(A1)=P(BA1)=0.125
P(B/A2) P(A2)=P(BA2)=0.06
Brand 2
P(B/A3) P(A3)=P(BA3)=0.02
P(B)=P[(brand1 and repair) or (brand 2 and repair) or (brand 3 and repair)]
P(B)= P(BA1)+ P(BA2)+ P(BA3)=0.125+0.06+0.02
P(A1/B)=…
P(A2/B)=…
P(A3/B)=…
=0.205
=0.205
Class Problem
• An assembly plant receives
its voltage regulators from
three different suppliers, 60%
from supplier B1, 30% from
supplier B2, and 10% from
supplier B3. If 95% of the
voltage regulators from B1,
80% of those from B2, and
65% of those from B3
perform according to
specifications, what is the
probability that a particular
voltage regulator which is
known to perform according
to specifications came from
supplier B3?
• If A denotes the event a
voltage regulator received by
the plant performs according
to specifications and B1, B2,
B3 are the events that it comes
from the respective suppliers,
then