Download chapter 1 eqt 272

EQT 272
PROBABILITY
AND STATISTICS
ROHANA BINTI ABDUL HAMID
INSTITUT E FOR ENGINEERING MATHEMATICS (IMK)
UNIVERSITI MALAYSIA PERLIS
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Page 1
CHAPTER 1
PROBABILITY
1.1 Introduction
1.2 Sample space and algebra of sets
1.3 Properties of probability
1.4 Tree diagrams and counting techniques
1.5 Conditional probability
1.6 Bayes’s theorem
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1.7 Independence
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WHY DO COMPUTER ENGINEERS NEED
TO STUDY PROBABILITY???????
1.
2.
3.
4.
5.
Signal processing
Computer memories
Optical communication systems
Wireless communication systems
Computer network traffic
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Probability and statistics are
related in an important way.
Probability is used as a tool; it
allows you to evaluate the reliability
of your conclusions about the
population when you have only
sample information.
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Probability
• Probability is a measure of the likelihood
of an event A occurring in one experiment
or trial and it is denoted by P (A).
number of ways thattheevent A can occur ( A)
P( A) 
totalnumber of outcomes( S )
n( A)

n( S )
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Experiment
• An experiment is any process of
making an observation leading to
outcomes for a sample space.
Example:
-Toss a die and observe the number that
appears on the upper face.
-A medical technician records a person’s blood
type.
-Recording a test grade.
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The mathematical basis of probability is the
theory of sets.
• Sets
A set is a collection of elements or components
• Sample Spaces, S
A sample space consists of points that
correspond to all possible outcomes.
• Events
An event is a set of outcomes of an experiment
and a subset of the sample space.
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• Experiment:
Tossing a die
• Sample space:
S ={1, 2, 3, 4, 5, 6}
• Events:
A: Observe an odd number
B: Observe a number less than 4
C: Observe a number which could
divide by 3
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Basic Operations
S
B
A
Figure 1.1: Venn diagram representation of
events
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1. The union of events A and B, which is denoted as A  B ,
- is the set of all elements that belong to A or B or both.
- Two or more events are called collective exhaustive events if the
unions of these events result in the sample space.
2. The intersection of events A and B, which is denoted by A  B,
- is the set of all elements that belong to both A and B.
- When A and B have no outcomes in common, they are said to
be mutually exclusive or disjoint sets.
3. The event that contains all of the elements that do not belong to
Freecomplement
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an event A is called the
of A and is denoted byPage
A 10
Example 1.2
• Given the following sets;
A= {2, 4, 6, 8, 10}
B= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C= {1, 3, 5, 11,….}, the set of odd numbers
Find A  B , A  B and C
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Answer
• A  B = {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
• A  B = {2, 4, 6, 8, 10}
• C = {2, 4, 6, 8,…}, the set of even
numbers
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Example 1.3
• A survey finds that 56% of people are
married. They ask the same group of
people, and 67% have at least one child.
There are 41% that are married and have
at least one child. Describe this results
with a Venn diagram.
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Exercise 1
A group of 100 factory workers were questioned by
a popular health magazine and 48% were found to
take regular exercise. When asked about their
eating habits, 67% replied that they always have
breakfast. Not only that, 32% always have
breakfast and exercise regularly. Describe this
results with a Venn diagram.
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1) 0  P ( A)  1
2) P ( A)  P ( A)  1
3) P ( A  B)  P ( A)  P ( A  B )
4) P ( A  B )  P ( B )  P ( A  B )
5) P ( A  B)  1  P ( A  B )
6) P (( A  B ))  P ( A  B )
7) P (( A  B ))  P ( A  B)
8) P ( A  ( A  B ))  P ( A  B )
9)
S
B
A
A B
A  B
A  B
P ( B )  P[( A  B )  ( A  B )]
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Theorem 1.1 : Laws of
Probability
a) P( A)  1 – P  A 
b) P( A  B)  P  A   P  B  – P( A  B)
c) P( A  B  C )  P  A   P  B   P  C  – P( A  B) – P( A  C ) – P( B  C )  P( A  B  C )
d) If A and B are mutually exclusive events, then P( A  B)  0
e) If A1 and A2 are the subset of S where A1  A2 , then P  A1   P  A2 
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Two fair dice are thrown. Determine
a) the sample space of the experiment
b) the elements of event A if the outcomes of both
dice thrown are showing the same digit.
c) the elements of event B if the first thrown giving
a greater digit than the second thrown.
d) probability of event A, P(A) and event B, P(B)
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Solutions 1.4
a) Sample space, S
1
2
3
4
5
6
1
(1, 1)
(1, 2)
(1, 3)
(1, 2)
(1, 5)
(1, 6)
2
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
3
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
4
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
5
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
6
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
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Solutions 1.4
b) A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
c) B = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3),
(5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n( A) 6 1
d) P  A 


n( S ) 36 6
n( B) 15 5
P  B 


n( S ) 36 12
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Consider randomly selecting a UniMAP Master Degree
international student, and let A denote the event that the
selected individual has a Visa Card and B has a
Master Card. Suppose that P(A) = 0.5 and P(B) = 0.4
and P( A  B) = 0.25.
a) Compute the probability that the selected individual
has at least one of the two types of cards ?
b) What is the probability that the selected individual
has neither type of card?
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Solutions 1.5
a) P( A  B)  P  A   P  B  – P( A  B)
= 0.5  0.4 – 0.25  0.65
b) 1  P( A  B)  1 – 0.65  0.35
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1.4.1 Tree diagrams
• Some experiments can be generated in
stages, and the sample space can be
displayed in a tree diagram.
• Each successive level of branching on the
tree corresponds to a step required to
generate the final outcome.
• A tree diagram helps to find simple
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events.
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• A box contains one yellow and two red
balls. Two balls are randomly selected and
their colors recorded. Construct a tree
diagram for this experiment and state the
simple events.
Y1
R1
R2
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First ball
Second ball
R1
Y1
R2
Y1
R1
R2
RESULTS
Y1R1
Y1R2
R1Y1
R1R2
Y1
R2Y1
R1
R2R1
R2
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Exercise 2
• 3 people are randomly selected from voter
registration and driving records to report
for jury duty. The gender of each person is
noted by the county clerk. List the simple
events by creating a tree diagram.
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1.4.2 Counting technique
• We can use counting techniques or counting
rules to
# find the number of ways to accomplish the
experiment
# find the number of simple events.
# find the number of outcomes
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Permutations
Counting
rules
Combinations
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• This counting rule count the
number of outcomes when the
experiment involves selecting r
objects from a set of n objects
when the order of selection is
important.
n
n!
Pr 
( n  r )!
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• The number of ways to arrange
an entire set of n distinct items is
n
Pn  n!
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• "The password of the safe was
472".
• We do care about the order.
"724" would not work, nor would
"247". It has to be exactly 4-7-2.
• To help you to remember, think
"Permutation ... Position"
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• There are basically two types of
permutation:
Repetition is Allowed: such as a lock.
It could be "333".
No Repetition: for example the first
three people in a running race. You
can't be first and second.
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Repetition is Allowed
• When you have n things to choose from
... you have n choices each time!
• When choosing r of them, the
permutations are:
• n × n × ... (r times)
• (In other words, there are n possibilities for the
first choice, THEN there are n possibilites for
the second choice, and so on, multplying each
time.)
• Which is easier to write down using an
exponent of r: nFree
× nPowerpoint
× ... (rTemplates
times) = nr
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• Example:
• In a lock , there are 10 numbers to
choose from (0,1,..9) and you
choose 3 of them:
• 10 × 10 × ... (3 times)
= 103 = 1,000 permutations
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No Repetition
• In this case, you have to reduce
the number of available choices
each time.
• For example, what order could 16
pool balls be in?
• After choosing a ball, you can't
choose it again.
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• So, your first choice would have 16
possibilites, and your next choice would
then have 15 possibilities, then 14, 13,
etc. And the total permutations would
be:
• 16 × 15 × 14 × 13 × ... =
20,922,789,888,000
• But maybe you don't want to choose
them all, just 3 of them, so that would be
only:
• 16 × 15 × 14 = 3,360
• In other words, there are 3,360 different
ways that 3 pool balls could be selected
out of 16 balls. Free Powerpoint Templates
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• Suppose you have 3 books, A, B and C
but you have room for only two on your
bookshelf. In how many ways can you
select and arrange the two books when
the order is important.
A
B
C
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A
B
A
C
A
B
C
B
A
2.AC
C
A
3.BC
C
A
C
B
1.AB
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4.BA
5.CA
6.CB
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n
3
n!
Pr 
( n  r )!
3!
P2 
( 3 2 )!
6
There are 6 ways to select and
arrange the books
in order.
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Exercise 3
Three lottery tickets are drawn from a
total of 50. If the tickets will be distributed
to each of the employees in the order in
which they are drawn, the order will be
important. How many simple events are
associated with the experiment?
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• This counting rule count the
number of outcomes when the
experiment involves selecting r
objects from a set of n objects
when the order of selection is not
important.
n!
nC   n  
r r 
  r ! n  r  !
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• "My fruit salad is a combination
of apples, grapes and bananas"
We don't care what order the fruits
are in, they could also be
"bananas, grapes and apples" or
"grapes, apples and bananas", its
the same fruit salad.
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• Suppose you have 3 books, A, B and C
but you have room for only two on your
bookshelf. In how many ways can you
select and arrange the two books when
the order is not important.
A
B
C
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A
B
1.AB
A
C
2.AC
A
B
C
3.BC
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n!
nC   n  
r r 
  r ! n  r  !
3
3!
C2 
2!( 3 2 )!
3
There are 3 ways to select and arrange
the books when the order is not
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important
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Exercise 4
Suppose that in the taste test, each
participant samples 8 products and is
asked the 3 best products, but not in any
particular order. Calculate the number of
possible answer test.
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TRY!
• A company produces 10 microchips
during a night staff. 6 of these turn
out to be defective. Suppose 3 of
the chips were sent to a customer.
a)What is the probability of the
customer receiving 2 defective
chips?
b)What is the probability no defective
chips?
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• Definition:
For any two events A and B with P(B) > 0,
the conditional probability of A given
that B has occurred is defined by
P( A  B)
P( A | B) 
P( B)
GIVENFree Powerpoint Templates
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A study of 90 students was done by UniMAP first year
students. The results are given in the table :
Area/Gender
Male (C)
Female (D)
Total
Urban (A)
35
10
45
Rural (B)
25
20
45
Total
60
30
90
If a student is selected at random and have been told
that the individual is a male student, what is the
probability of he is from
urban
area?
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Answer;
P( A | C )
P( A  C )

P(C )
Probability of male
students from urban
area
35 / 90

60 / 90
 0.5833
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In 2006, Edaran Automobil Negara (EON) will
produce a multipurpose national car (MPV)
equipped with either manual or automatic
transmission and the car is available in one of
four metallic colours. Relevant probabilities
for various combinations of transmission type
and colour are given in the accompanying
table:
Transmission
Black
Grey (C)
Blue
Automatic, (A)
0.15
0.10
0.10
0.10
Manual
0.15
0.05
0.15
0.20
type/Colour
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(B)
Red
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• Let,
A = automatic transmission
B = black
C = grey
Calculate;
a) P ( A), P ( B ) and P ( A  B )
b) P ( A | B ) and P ( B | A)
c) P ( A | C ) and P ( A | C  )
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Answer;
a) P(A) = probability of MPV with
automatic transmission
P(A) = 0.15+0.10+0.10+0.10 = 0.45
P(B) = probability of black MPV
P(B) = 0.10+0.15 = 0.25
P(A∩B) = probability of black MPV with
automatic transmission
P(A∩B) = 0.10
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P(A|B) = probability of auto MPV given
that the MPV is black
0 .1
P( A  B)

P( A | B) 
 0.4
0.25
P( B)
P(B|A) = probability of black MPV given
that the MPV has automatic transmission
P( A  B)
P( B | A) 
P( A)
0 .1

0.45
 0.222
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P(A|C) = probability of auto MPV given
that the MPV is grey
0.15
P( A  C )

P( A | C ) 
 0.5
0 .3
P(C )
P(A|C’) = probability of auto MPV given
that the MPV is not grey
P( A  C )
P( A | C ) 
P(C )
0.3

0.7
 0.429
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-Used to revise previously calculated
probabilities based on new information.
-Extension of conditional probability
If A1 , A2 ,..., An is a partition of a sample space, then the posterior
probabilities of events Ai conditional on an event B can be obtained
from the probabilities P  Ai  and P  B | Ai  using the formula,
P  Ai  B  P  Ai  P  B | Ai 
P  Ai | B  


P  B
P B
P  Ai  P  B | Ai 
 P A  PB | A 
n
j 1
j
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j
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• Suppose someone told you they had a
nice conversation with someone on the
train. Not knowing anything else about
this conversation, the probability that
they were speaking to a woman is
50%.
• Now suppose they also told you that this
person had long hair. It is now more
likely they were speaking to a woman,
since women are more likely to have
long hair than men.
• Bayes' theorem can be used to
calculate the probability that the
person is a woman
.
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• Suppose it is also known that 75% of
women have long hair. Likewise,
suppose it is known that 15% of men
have long hair.
• Our goal is to calculate the
probability that the conversation
was held with a woman, given the
fact that the person had long hair.
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• Probability that the conversation was
held with a woman, given the fact
that the person had long hair
P ( women | long )
P(long | women) P( women)

P(long | women) P( women)  P(long | man) P(man)
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P( women)  0.5, P(man)  0.5
P(long | women)  0.75,
P(long | man)  0.15
P(long | women) P( women)

P(long | women) P( women)  P(long | man) P(man)
0.75(0.5)

0.75(0.5)  0.15(0.5)
 0.83
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A drilling company has estimated a 40% chance of
striking oil for their new well. A detailed test has
been scheduled for more information. Historically,
60% of successful wells have had detailed tests, and
20% of unsuccessful wells have detailed tests. Given
that this well has been scheduled for a detailed test,
what is the probability that the well will be
successful?
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• Answer:
P(successful | detailed )
P(detailed | success ) P(success )

P(detailed | success ) P(success )  P(detailed | failure ) P(failure )
P( success )  0.4, P( failure )  0.6
P(detail | success )  0.6,
P(detail | failure )  0.2
0.6(0.4)

0.6(0.4)  0.2(0.6)
 0.6667
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TRY!!!
• You have a database of 100 emails.
• 60 of those 100 emails are spam
– 48 of those 60 emails that are spam have the
word "buy"
– 12 of those 60 emails that are spam don't
have the word "buy"
• 40 of those 100 emails aren't spam
– 4 of those 40 emails that aren't spam have
the word "buy"
– 36 of those 40 emails that aren't spam don't
have the word "buy"
• What is the probability that an email is spam if it
has the word "buy"? Free Powerpoint Templates
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• Definition :
 Two events A and B are said to be independent
if and only if either
P ( A | B )  P ( A)
or
P ( B | A)  P ( B )
Otherwise, the events are said to be dependent.
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• Two events, A and B, are
independent if the fact that
A occurs does not affect the
probability of B occurring.
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• Some other examples of independent
events are:
Landing on heads after tossing a coin
AND rolling a 5 on a single 6-sided die.
Choosing a marble from a jar AND
landing on heads after tossing a coin.
Choosing a 3 from a deck of cards,
replacing it, AND then choosing an ace
as the second card.
Rolling a 4 on a single 6-sided die,
AND then rolling a 1 on a second roll of
the die.
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Multiplicative Rule of Probability:
The probability that both two events A and B, occur is
P( A  B)  P  A  P  B | A 
 P  B P  A | B
If A and B are independent,
P( A  B)  P  A  P  B 
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3
1
Suppose that P( A)  and P( B)  . Are events A and B independent or
5
3
mutually exclusive if ,
1
a) P( A  B) 
5
14
b) P( A  B) 
15
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