EQT 272 PROBABILITY AND STATISTICSNORNADIA MOHD YAZIDINSTITUT E OF ENGINEERING MATHEMATICS (IMK)UNIVERSITI MALAYSIA PERLIS

CHAPTER 1: PROBABILITY1.1 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

1.7 Independence

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.

1.1 INTRODUCTION

1.1 INTRODUCTION

WHY DO COMPUTER ENGINEERS NEED TO STUDY PROBABILITY???????

1. Signal processing2. Computer memories3. Optical communication systems4. Wireless communication systems5. Computer network traffic

1.2. SAMPLE SPACE AND ALGEBRA OF SETS

The mathematical basis of probability is the theory of sets.

• An experiment that can result in different outcomes, even though it is repeated in the same manner every time.

Random experiments

• collection of elements or componentsSets

• The set of all possible outsomes of random experiment.Sample space, S

• a subset of the sample spaceEvents

EXAMPLE 1.1

Suppose that three items are selected at random from a manufacturing process. Each item is inspected and classified defective, D, or nondefective, N.

• Sample space: S ={DDD, DDN, DND, DNN, NDD, NDN, NND,

NNN}

EXAMPLE

1.2 SAMPLE SPACE AND ALGEBRA OF SETS 

AB

C

S

Venn diagram

Used to depicts all the possible outcomes for an experiment.

• 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

GENERAL SET THEORYUnion / “Or” Statement:

A B

• 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

GENERAL SET THEORYIntersection / “And” Statement:

A B

• 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

GENERAL SET THEORYComplement:

• When A and B have no outcomes in common, they are said to be

mutually exclusive / disjoint sets.

GENERAL SET THEORYMutually Exclusive / Disjoint:

EXAMPLE

Given the following sets;A= {2, 4, 6, 8, 10}B= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}C= {1, 3, 5,7,9, 11,….}, the set of odd

numbers

Find , and BA BA C

ANSWER

1,2,3,4,5,6,7,8,9,10A B

2,4,6,8,10A B

2,4,6,8,10,... the set of even numbers.C

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 that the event can occur ( )

total number of outcomes

( )( )

A AP A

S

n An S

1.1 INTRODUCTION TO PROBABILITY

1.4 PROPERTIES OF PROBABILITY

1) 0 ( ) 12) ( ) ( ) 13) ( ) ( ) ( )4) ( ) ( ) ( )5) ( ) 1 ( )6) (( ) ) ( )7) (( ) ) ( )8) ( ( )) ( )9) ( ) [( ) ( )]

P AP A P AP A B P A P A BP A B P B P A BP A B P A BP A B P A BP A B P A BP A A B P A BP B P A B A B

A B A BA B

S

AB

Two fair dice are thrown. Determine a) the sample space of the experimentb) 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)

EXAMPLE

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)

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)}

( ) 6 1d)   ( ) 36 6( ) 15 5 ( ) 36 12

n AP An Sn BP Bn S

Consider randomly selecting a UniMAP Master Degreeinternational 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 = 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?

 

EXAMPLE

( )P A B

SOLUTION

'

a) ( )  – ( )  = 0.5 0.4 – 0.25  0.65

b) ( ) =1 ( ) 1– 0.65 0.35

P A B P A P B P A B

P A B P A B

1.5 CONDITIONAL PROBABILITY

• 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 BP A BP B

( )(B | A)(A)

P A BPP

A study of 100 students who get A in Mathematics in SPM examination was done by UniMAP first year students. The results are given in the table :

EXAMPLE 1.4

Area/Gender

Male (C) Female (D) Total

Urban (A) 35 10 45Rural (B) 25 30 55

Total 60 40 100If 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?

SOLUTION EXAMPLE 1.4

( ) / ( )

35 100 = 0.583360 100

P A C P A C P C

Probability of male students from urban areaP A C

1.6 INDEPENDENCE

Definition : Two events in independent if and only if the probability of

event B is not influenced or changed by the occurrence of event A, or vice versa

Two events A and B are said to be independent if and only if either

Otherwise, the events are said to be dependent.

( | ) ( )or

( | ) ( )

P A B P A

P B A P B

Example :-Suppose there are two children have eye brown color. Since the eye color of a child is affected by the genetic of parents and not affect by the other child, it is reasonable to assume that event A : the first child has brown eyes and B: the second child has brown eyes, are independent.

EXAMPLE 1.5

Survey of 1000 adults, the respondents were classified according to whether they currently had a child in college and whether the loan burden for college is too high or the right amount.

Are events A and D independent?

Too High (A)

Right Amount (B)

Too Little (C)

Child in College (D)

0.35 0.08 0.01

No Child in College (E)

0.25 0.20 0.11

SOLUTION 1.

Since the two probabilities are not same, events A and D are dependent.2.

Since , events A and D are dependent.

0.35

0.60

0.44

0.60(0.44) 0.264

P A D

P A

P D

P A P D

0.35 0.800.44

P A DP A D

P D

P A D P A

MULTIPLICATIVE RULE OF PROBABILITY:

The probability that both two events and , occur is  ( ) |                  

  |   

If and are independent,   ( )

A BP A B P A P B A

P B P A B

A BP A B P A P B

Mutually Exclusive VS Independent

Mutually Exclusive Independent

Definitiom Two events cant occur together

Occurrence 1 event does not effect the occurrence of another event

Multiplication RuleAdditional Rule

Extra

0P A B P A B P A P B

P A B P A P B P A B P A P B P A B

P A B P A

P B A P B

EXAMPLE 1.6

3 1Suppose that ( ) and ( ) . Are events and independent or 5 3

mutually exclusive if ,

1a) ( )514b) ( )15

P A P B A B

P A B

P A B

SOLUTION

3 1Suppose that ( ) and ( ) . Are events and independent or 5 3

mutually exclusive if ,

1a) ( ) - Independent (Not mutually exclusive)5

3 1 1( )5 3 5

14b) ( ) - Mutually Exclusive15

P A P B A B

P A B

P A B P A P B

P A B

P

3 1 14( )5 3 15

A B P A P B

1.6 BAYES’ THEOREM

1 2 If   , ,..., is a partition of a sample space, then the of events  conditional on an event can be obtained

from the probabilities and  | using the formula,

|

n

i

i i

i

A A AA B

P A P B A

P A

posteriorprobabilities

1

| |

|

i i i i in

j jj

P A B P A P B A P A P B AB

P B P B P A P B A

-Used to revise previously calculated probabilities based on new information.

-Extension of conditional probability

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.

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.

Probability that the conversation was held with a woman, given the fact that the person had long hair

)|( longwomenP

)()|()()|()()|(

manPmanlongPwomenPwomenlongPwomenPwomenlongP

5.0)(,5.0)( manPwomenP

15.0)|(,75.0)|(

manlongPwomenlongP

)()|()()|()()|(

manPmanlongPwomenPwomenlongPwomenPwomenlongP

)5.0(15.0)5.0(75.0)5.0(75.0

83.0

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?

EXAMPLE

)detailed|successful(P

)failure()failure|detailed()success()success|detailed()success()success|detailed(

PPPPPP

6.0)(,4.0)( failurePsuccessP

2.0)failure|detail(,6.0)success|detail(

PP

)6.0(2.0)4.0(6.0)4.0(6.0

6667.0

SOLUTION

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"?

Tree diagrams help us to understand 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.

1.4 TREE DIAGRAMS AND COUNTING TECHNIQUE

1.4.1 Tree diagrams

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.

EXAMPLE

Y1 R1

R2

First ball

Second ball

RESULTS

Y1

R1

R2

R1

R2

Y1

R2

Y1

R1

Y1R1

Y1R2

R1Y1

R1R2

R2Y1

R2R1

We can use counting techniques or counting rules to

1.4 TREE DIAGRAMS AND COUNTING TECHNIQUE

1.4.2 Counting technique

# find the number of ways to accomplish the experiment

# find the number of simple events.

# find the number of outcomes

Counting rules

Multiplication Principle

Permutations

Combinations

MULTIPLICATION Assume an operation can be described as a sequence of k steps, the number of ways completing step 1 is n1

the number of ways completing step 2 is n2 the number of ways completing step 3 is n3

Example

The design for a website is to consist of 4 colours, 3 fonts, and 3 position for an image. In order to identify the possible design, multiplication rule can be applied.

Possible designs 4 3 336

1 2 3 kTotal number of ways completing operation =n n n ... n

A permutaion of elements is an ordered sequence of the elements.

All possible arrangements of a collection of things, where the order is important.

There are basically two types of permutation:1) Repetition is Allowed2) No Repetition

Permutations

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: n × n × ... (r times) = nr

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

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.

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.

The number of ways to arrange an entire set of n distinct items is

Permutations

!nPnn

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.

Permutations

)!(!rn

nPrn

"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"

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.

EXAMPLE

A B C

A B

A C

A CB

AB

AC

A CBC

AB

AC

BC

BA

CA

CB

SOLUTION

)!(!rn

nPrn

6

)!23(!3

23

P

There are 6 ways to select and arrange the books in order.

Combinations• 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}

Combinations• Formula:

• It is often called "n choose r"

!

! !n nnCr r r n r

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.

EXAMPLE

A B C

A B

A C

A CB

AB

AC

BC

SOLUTION

3

)!23(!2!3

23

C

There are 3 ways to select and arrange the books when the order is not important

!

! !n nnCr r r n r