Download B - Portal UniMAP

EQT 272
PROBABILITY
AND STATISTICS
FARAH ADIBAH ADNAN
INSTITUT E OF ENGINEERING MATHEMATICS
(IMK)
UNIVERSITI MALAYSIA PERLIS
Free Powerpoint Templates
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
Free Powerpoint Templates
1.7 Independence
Page 2
What is probability???
Is the measure of how likely something will occur.
In general:
• Probability is a measure of the likelihood
of an event A occurring in any one experiment
or trial.
• Denoted by P (A).
• Probability of all outcomes sums to 1.
• The probability value is between 0 and 1.
number of ways thattheevent A can occur ( A)
totalnumber of outcomes( S )
n( A)

Free Powerpoint Templates
n( S )
P( A) 
Page 3
Probability Line:
The mathematical basis of probability is the
theory of sets.
• Sets
A set is a collection of things/elements.
Free Powerpoint Templates
Page 6
• Sample Spaces, S
A sample space consists of all the possible
outcomes of an experiment.
• Events
An event is a collection of one or more outcomes
and a subset of the sample space.
• Experiment
An action where the result is uncertain.
Eg: Toss a die, record a person’s blood type
Free Powerpoint Templates
Page 7
• Experiment:
Tossing a die
• Sample space:
S ={1, 2, 3, 4, 5, 6}
• Events:
A: Observe an odd number
A={2,4,6}
B: Observe a number less than 4
B={1,2,3}
Venn diagram
S
A
B
Used to depicts all the possible outcomes for an
Free Powerpoint Templates
experiment.
Page 10
Union / “Or” Statement:
•
•
•
The union of events A and B - the set of all elements that
belong to A or B or both.
Meaning: joining, addition.
Denoted as A  B
Example:
Find the union of the sets A  {1,3,5}, B  {1, 2,3}
Free Powerpoint Templates
Page 11
Intersection / “And” Statement:
• The intersection of events A and B - the set of all elements that
belong to both A and B
• Meaning: overlap, things in common.
• Denoted by A  B .
Example:
Find the intersection of the sets A  {1,3,5}, B  {1, 2,3}
Free Powerpoint Templates
Page 12
Complement:
• The complement of the event A - the event that contains all of
the elements that do not belong to an event A.
• Meaning: not A.
• Denoted by A .
Example:
Let the sample space, S  {1, 2,3, 4} . Let set A  {1, 2} and set
Page 13
B  {2,3} .Find A . Free Powerpoint Templates
Mutually Exclusive / Disjoint:
• When A and B have no outcomes in common, they are said to be
mutually exclusive / disjoint sets.
Free Powerpoint Templates
Page 14
Example:
Solution:
Mutually exclusive events A and
B Powerpoint
Non-Mutually
exclusive events A and C
Free
Templates
Page 15
Exercise
• 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
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
Exercise
• 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.
Exercise
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.
1) 0  P ( A)  1
2) P ( A)  1 – P  A 
3) P ( A  B )  P  A   P  B  – P ( A  B )
4) P ( A)  P ( A)  1
S
5) P ( A  B)  P ( A)  P ( A  B )
B
6) P ( A  B )  P ( B )  P ( A  B )
A
7) P ( A  B)  1  P ( A  B )
8) P (( A  B ))  P ( A  B)
A B
9) P (( A  B ))  P ( A  B)
A  B
A  B

10) P ( A  ( A  B ))  P ( A  B )
11) If A and B are mutually exclusive events,
then P ( A  B )  0
12) If A1 and A2 are the subset of S where A1  A2 ,
then P  A1   P  A2  Free Powerpoint Templates
Page 20
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)
Free Powerpoint Templates
Page 21
Solution
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)
Free Powerpoint Templates
Page 22
Solution
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
Free Powerpoint Templates
Page 23
Consider randomly selecting a UniMAP Master Degree
international student. 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?
Free Powerpoint Templates
Page 24
Solution
a) P( A  B)  P  A   P  B  – P( A  B )
= 0.5  0.4 – 0.25  0.65
b) P( A  B) =1  P( A  B)
'
 1 – 0.65  0.35
Free Powerpoint Templates
Page 25
1.4.1 Tree diagrams
• Tree diagrams help us to u/stand
probability concepts by presenting them
visually.
• In a tree diagram,each outcome is
represented by a branch of the tree.
• A tree diagram helps to find simple
events.
Free Powerpoint Templates
Page 26
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
Free Powerpoint Templates
Page 27
First ball
Second ball
R1
Y1
R2
Y1
R1
R2
RESULTS
Y1R1
Y1R2
R1Y1
R1R2
Y1
R2Y1
R1
R2R1
R2
Free Powerpoint Templates
Page 28
Exercise
A couple has three children. Draw the three
diagram for the possible gender of the three
children. Find the probability of getting:
a) Three boys
b) One girl and 2 boys
Free Powerpoint Templates
Page 29
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
Free Powerpoint Templates
Page 30
Multiplication
Principle
Counting
rules
Permutations
Combinations
Free Powerpoint Templates
Page 31
• Is used to find the total number of
outcomes.
• If an experiment consists of 3 steps, and if
the first step can result in m outcomes,
then the second step in n outcomes, and
the third step in k outcomes, the formula
is:
Total Outcomes = m  n  k
Example:
• Suppose we toss a coin 3 times. Determine
the total outcomes for tosses of a coin.
• This experiment has 3 steps: the first toss,
the second toss, and the third toss. Each
step has 2 outcomes, head and tail. Thus,
total outcomes = 2 x 2 x 2 = 8.
The 8 outcomes for this experiment are
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT
• All possible arrangements of a collection
of things, where the order is important.
• There are basically two types of
permutation:
a) Repetition is Allowed: such as a lock.
It could be "333".
b) No Repetition: for example the first three
people in a running race. You can't be
both first and second.
a) 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
Page 35
• 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
Free Powerpoint Templates
Page 36
b) 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.
Free Powerpoint Templates
Page 37
• 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
Page 38
• The number of ways to arrange
an entire set of n distinct items is
n
Pn  n!
Free Powerpoint Templates
Page 39
• 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 )!
Free Powerpoint Templates
Page 40
• "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"
Free Powerpoint Templates
Page 41
• 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
Solution
A
B
A
C
A
B
C
B
A
AC
C
A
BC
C
A
C
B
AB
BA
CA
CB
n
3
n!
Pr 
( n  r )!
3!
P2 
( 3 2 )!
6
There are 6 ways to select and
arrange the books
in order.
Free Powerpoint
Templates
Page 44
Exercise
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?
Free Powerpoint Templates
Page 45
• A collection of things, in which the
order does not matter.
Example:
You are making a sandwich. How
many different combinations of 2
ingredients can you make with
cheese, mayo and ham?
Answer:
{cheese, mayo}, {cheese, ham} or
{mayo, ham} Free Powerpoint Templates
Page 46
• Formula:
n!
nC   n  
r r 
  r ! n  r  !
• It is often called "n choose r"
Free Powerpoint Templates
Page 47
• 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
Free Powerpoint Templates
Page 48
Solution
A
B
AB
A
C
AC
A
B
C
BC
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
important
Exercise
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.
Free Powerpoint Templates
Page 51
• 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
Page 52
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?
Free Powerpoint Templates
Page 53
Solution
P( A | C )
P( A  C )

P(C )
Probability of male
students from urban
area
35 / 90

60 / 90
 0.5833
Free Powerpoint Templates
Page 54
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
Free Powerpoint Templates
(B)
Red
Page 55
• 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  )
Free Powerpoint Templates
Page 56
Solution
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
Free Powerpoint Templates
Page 57
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
Free Powerpoint Templates
Page 58
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
Free Powerpoint Templates
Page 59
-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
Free Powerpoint Templates
j
Page 60
• 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
.
Free Powerpoint Templates
Page 61
• 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.
Free Powerpoint Templates
Page 62
• 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)
Free Powerpoint Templates
Page 63
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
Free Powerpoint Templates
Page 64
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?
Free Powerpoint Templates
Page 65
Solution
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
Free Powerpoint Templates
Page 66
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
Page 67
• 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.
Free Powerpoint Templates
Page 68
• Two events, A and B, are
independent if the fact that
A occurs does not affect the
probability of B occurring.
Free Powerpoint Templates
Page 69
• 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.
Free Powerpoint Templates
Page 70
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 
Free Powerpoint Templates
Page 71
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
Free Powerpoint Templates
Page 72