07_BayesianClassifier.pptx

Dr. Debasis Samanta
Associate Professor
Department of Computer Science & Engineering
Lecture #7
Classification: Naïve Bayes’ Classifier

An interesting fact..
2
CS 61061: Data Analytics
No number before 1,000 contains the letter A.
But there are plenty of E’s, I’s, O’s, U’s, and Y’s.
@DSamanta, IIT Kharagpur

Today’s discussion…
 Introduction to Classification
 Classification Techniques
 Supervised and unsupervised classifications
 Formal statement of supervised classification technique
 Bayesian Classifier
 Principle of Bayesian classifier
 Bayes’ theorem of probability
 Naïve Bayesian Classifier
 Addressing some issues with Naïve Bayesian Classifier
 Case study
@DSamanta, IIT Kharagpur 3

A Simple Quiz: Identify the objects

Introduction to Classification
Example 7.1
 Teacher classify students as A, B, C, D and F based on their marks. The
following is one simple classification rule:
Mark ≥ 𝟗𝟎 : A
𝟗𝟎 > Mark ≥ 𝟖𝟎 : B
8𝟎 > Mark ≥ 𝟕𝟎 : C
𝟕𝟎 > Mark ≥ 𝟔𝟎 : D
6𝟎 > Mark : F
Note:
Here, we apply the above rule to a specific data
(in this case a table of marks).

Examples of Classification in Data Analytics
 Life Science: Predicting tumor cells as benign or malignant
 Security: Classifying credit card transactions as legitimate or
fraudulent
 Prediction: Weather, voting, political dynamics, etc.
 Entertainment: Categorizing news stories as finance, weather,
entertainment, sports, etc.
 Social media: Identifying the current trend and future growth

Classification : Definition
 Classification is a form of data analysis to extract models describing
important data classes.
 Essentially, it involves dividing up objects so that each is assigned to one of
a number of mutually exhaustive and exclusive categories known as
classes.
 The term “mutually exhaustive and exclusive” simply means that each object
must be assigned to precisely one class
 That is, never to more than one and never to no class at all.

Classification Techniques
 Classification consists of assigning a class label to a set of unclassified
cases.
 Supervised Classification
 The set of possible classes is known in advance.
 Unsupervised Classification
 Set of possible classes is not known. After classification we can try to assign a
name to that class.
 Unsupervised classification is called clustering.

Supervised Classification

Unsupervised Classification

Supervised Classification Technique
 Given a collection of records (training set )
 Each record contains a set of attributes, one of the attributes is the
class.
 Find a model for class attribute as a function of the values of
other attributes.
 Goal: Previously unseen records should be assigned a class as
accurately as possible.
 Satisfy the property of “mutually exclusive and exhaustive”

Illustrating Classification Tasks
Apply
Model
Induction
Deduction
Learn
Model
Model
Tid Attrib1 Attrib2 Attrib3 Class
1 Yes Large 125K No
2 No Medium 100K No
3 No Small 70K No
4 Yes Medium 120K No
5 No Large 95K Yes
6 No Medium 60K No
7 Yes Large 220K No
8 No Small 85K Yes
9 No Medium 75K No
10 No Small 90K Yes
10
Tid Attrib1 Attrib2 Attrib3 Class
11 No Small 55K ?
12 Yes Medium 80K ?
13 Yes Large 110K ?
14 No Small 95K ?
15 No Large 67K ?
10
Test Set
Learning
algorithm
Training Set

Classification Problem
 More precisely, a classification problem can be stated as below:
Given a database D = 𝑡1, 𝑡2, … . . , 𝑡𝑚 of tuples and a set of classes C =
𝑐1, 𝑐2, … . . , 𝑐𝑘 , the classification problem is to define a mapping f ∶ D → 𝐶,
Where each 𝑡𝑖 is assigned to one class.
Note that tuple 𝑡𝑖 ∈ 𝐷 is defined by a set of attributes 𝐴 = 𝐴1, 𝐴2, … . . , 𝐴𝑛 .
Definition 7.1: Classification Problem

 A number of classification techniques are known, which can be broadly
classified into the following categories:
1. Statistical-Based Methods
• Regression
• Bayesian Classifier
•
2. Distance-Based Classification
• K-Nearest Neighbours
3. Decision Tree-Based Classification
• ID3, C 4.5, CART
5. Classification using Machine Learning (SVM)
6. Classification using Neural Network (ANN)

Bayesian Classifier

Bayesian Classifier
 Principle
 If it walks like a duck, quacks like a duck, then it is probably a duck

 A statistical classifier
 Performs probabilistic prediction, i.e., predicts class membership probabilities
 Foundation
 Based on Bayes’ Theorem.
 Assumptions
1. The classes are mutually exclusive and exhaustive.
2. The attributes are independent given the class.
 Called “Naïve” classifier because of these assumptions
 Empirically proven to be useful.
 Scales very well.
Bayesian Classifier

Example: Bayesian Classification
 Example 7.2: Air Traffic Data
 Let us consider a set
observation recorded in a
database
 Regarding the arrival of airplanes
in the routes from any airport to
New Delhi under certain
conditions.

Air-Traffic Data
Days Season Fog Rain Class
Weekday Spring None None On Time
Weekday Winter None Slight On Time
Weekday Winter None None On Time
Holiday Winter High Slight Late
Saturday Summer Normal None On Time
Weekday Autumn Normal None Very Late
Holiday Summer High Slight On Time
Sunday Summer Normal None On Time
Weekday Winter High Heavy Very Late
Weekday Summer None Slight On Time
Cond. to next slide…

Air-Traffic Data
Days Season Fog Rain Class
Saturday Spring High Heavy Cancelled
Weekday Summer High Slight On Time
Weekday Winter Normal None Late
Weekday Summer High None On Time
Weekday Winter Normal Heavy Very Late
Saturday Autumn High Slight On Time
Weekday Autumn None Heavy On Time
Holiday Spring Normal Slight On Time
Weekday Spring Normal None On Time
Weekday Spring Normal Heavy On Time
Cond. from previous slide…

Air-Traffic Data
 In this database, there are four attributes
A = [ Day, Season, Fog, Rain]
with 20 tuples.
 The categories of classes are:
C= [On Time, Late, Very Late, Cancelled]
 Given this is the knowledge of data and classes, we are to find most likely
classification for any other unseen instance, for example:
 Classification technique eventually to map this tuple into an accurate class.
Week Day Winter High None ???

Bayesian Classifier
 In many applications, the relationship between the attributes set and the
class variable is non-deterministic.
 In other words, a test cannot be classified to a class label with certainty.
 In such a situation, the classification can be achieved probabilistically.
 The Bayesian classifier is an approach for modelling probabilistic
relationships between the attribute set and the class variable.
 More precisely, Bayesian classifier use Bayes’ Theorem of Probability for
classification.
 Before going to discuss the Bayesian classifier, we should have a quick look
at the Theory of Probability and then Bayes’ Theorem.

Bayes’ Theorem of Probability

Simple Probability
If there are n elementary events associated with a random experiment and m of n
of them are favorable to an event A, then the probability of happening or
occurrence of A is
𝑃 𝐴 =
𝑚
𝑛
Definition 7.2: Simple Probability

Simple Probability
 Suppose, A and B are any two events and P(A), P(B) denote the
probabilities that the events A and B will occur, respectively.
 Mutually Exclusive Events:
 Two events are mutually exclusive, if the occurrence of one precludes the
occurrence of the other.
Example: Tossing a coin (two events)
Tossing a ludo cube (Six events)
Can you give an example, so that two events are not mutually exclusive?
Hint: Tossing two identical coins, Weather (sunny, foggy, warm)

Simple Probability
 Independent events: Two events are independent if occurrences of one
does not alter the occurrence of other.
Example: Tossing both coin and ludo cube together.
(How many events are here?)
Can you give an example, where an event is dependent on one or more
other events(s)?
Hint: Receiving a message (A) through a communication channel (B)
over a computer (C), rain and dating.

Joint Probability
If P(A) and P(B) are the probability of two events, then
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵
If A and B are mutually exclusive, then 𝑃 𝐴 ∩ 𝐵 = 0
If A and B are independent events, then 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 . 𝑃(𝐵)
Thus, for mutually exclusive events
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵
Definition 7.3: Joint Probability

Conditional Probability
29
If events are dependent, then their probability is expressed by conditional
probability. The probability that A occurs given that B is denoted by 𝑃(𝐴|𝐵).
Suppose, A and B are two events associated with a random experiment. The
probability of A under the condition that B has already occurred and 𝑃(𝐵) ≠ 0 is
given by
𝑃 𝐴 𝐵 =
Number of events in 𝐵 which are favourable to 𝐴
Number of events in 𝐵
=
Number of events favourable to 𝐴 ∩ 𝐵
Number of events favourable to 𝐵
=
𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵)
Definition 7.2: Conditional Probability

30
𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 . 𝑃 𝐵 𝐴 , 𝑖𝑓 𝑃 𝐴 ≠ 0
or 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐵 . 𝑃 𝐴 𝐵 , 𝑖𝑓 𝑃(𝐵) ≠ 0
For three events A, B and C
𝑃 𝐴 ∩ 𝐵 ∩ 𝐶 = 𝑃 𝐴 . 𝑃 𝐵 . 𝑃 𝐶 𝐴 ∩ 𝐵
For n events A1, A2, …, An and if all events are mutually independent to each other
𝑃 𝐴1 ∩ 𝐴2 ∩ … … … … ∩ 𝐴𝑛 = 𝑃 𝐴1 . 𝑃 𝐴2 … … … … 𝑃 𝐴𝑛
Note:
𝑃 𝐴 𝐵 = 0 if events are mutually exclusive
𝑃 𝐴 𝐵 = 𝑃 𝐴 if A and B are independent
𝑃 𝐴 𝐵 ⋅ 𝑃 𝐵 = 𝑃 𝐵 𝐴 ⋅ 𝑃(𝐴) otherwise,
P A ∩ B = P(B ∩ A)
Corollary 7.1: Conditional Probability

 Generalization of Conditional Probability:
P A B =
P(A ∩ B)
P(B)
=
P(B ∩ A)
P(B)
=
P(B|A)∙P(A)
P(B)
∵ P A ∩ B = P(B|A) ∙P(A) = P(A|B) ∙P(B)
By the law of total probability : P(B) = P B ∩ A ∪ B ∩ A , where A denotes
the compliment of event A. Thus,
P A B =
P(B|A) ∙ P(A)
P B ∩ A ∪ B ∩ A
=
P B A ∙P(A)
P B A ∙P A +P(B│A)∙P(A)

In general,
P A D =
P(A) ∙ P D A
P A ∙ P D A + P B ∙ P D B + P(C) ∙ P(D│C)

Total Probability
CS 40003: Data Analytics 33
Let 𝐸1, 𝐸2, … … 𝐸𝑛 be n mutually exclusive and exhaustive events associated with a
random experiment. If A is any event which occurs with 𝐸1 𝑜𝑟 𝐸2 𝑜𝑟 … … 𝐸𝑛 , then
𝑃 𝐴 = 𝑃 𝐸1 . 𝑃 𝐴 𝐸1 + 𝑃 𝐸2 . 𝑃 𝐴 𝐸2 + ⋯ … … … . +𝑃 𝐸𝑛 . 𝑃(𝐴|𝐸𝑛)
Definition 7.3: Total Probability

Example 7.3
A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls.
One bag is selected at random. From the selected bag, one ball is drawn. What is the
probability that the ball drawn is red?
This problem can be answered using the concept of Total Probability
𝐸1 =Selecting bag I
𝐸2 =Selecting bag II
A = Drawing the red ball
Thus, 𝑃 𝐴 = 𝑃 𝐸1 . 𝑃 𝐴|𝐸1 + 𝑃 𝐸2 . 𝑃(𝐴|𝐸2)
where, 𝑃 𝐴|𝐸1 = Probability of drawing red ball when first bag has been chosen
and 𝑃 𝐴|𝐸2 = Probability of drawing red ball when second bag has been chosen
Total Probability: An Example

Example 7.4:
A bag (Bag I) contains 4 red and 3 black balls. A second bag (Bag II) contains 2 red and 4
black balls. You have chosen one ball at random. It is found as red ball. What is the
probability that the ball is chosen from Bag I?
Here,
𝐸1 = Selecting bag I
𝐸2 = Selecting bag II
A = Drawing the red ball
We are to determine P(𝐸1|A). Such a problem can be solved using Bayes' theorem of
probability.
Reverse Probability

Bayes’ Theorem
Let 𝐸1, 𝐸2, … … 𝐸𝑛 be n mutually exclusive and exhaustive events associated with a
random experiment. If A is any event which occurs with 𝐸1 𝑜𝑟 𝐸2 𝑜𝑟 … … 𝐸𝑛 , then
𝑃(𝐸𝑖 𝐴 =
𝑃 𝐸𝑖 . 𝑃(𝐴|𝐸𝑖)
𝑖=1
𝑛
𝑃 𝐸𝑖 . 𝑃(𝐴|𝐸𝑖)
Theorem 7.4: Bayes’Theorem

Prior and Posterior Probabilities
 P(A) and P(B) are called prior probabilities
 P(A|B), P(B|A) are called posterior probabilities
Example 8.6: Prior versus Posterior Probabilities
 This table shows that the event Y has two outcomes
namely A and B, which is dependent on another event X
with various outcomes like 𝑥1, 𝑥2 and 𝑥3.
 Case1: Suppose, we don’t have any information of the
event A. Then, from the given sample space, we can
calculate P(Y = A) =
5
10
= 0.5
•
 Case2: Now, suppose, we want to calculate P(X =
𝑥2|Y =A) =
2
5
= 0.4 .
The later is the conditional or posterior probability, where
as the former is the prior probability.
X Y
𝑥1 A
𝑥2 A
𝑥3 B
𝑥3 A
𝑥2 B
𝑥1 A
𝑥1 B
𝑥3 B
𝑥2 B
𝑥2 A

Naïve Bayesian Classifier
 Suppose, Y is a class variable and X = 𝑋1, 𝑋2, … . . , 𝑋𝑛 is a set of attributes,
with instance of Y.
 The classification problem, then can be expressed as the class-conditional
probability
𝑃 𝑌 = 𝑦𝑖| 𝑋1 = 𝑥1 AND 𝑋2 = 𝑥2 AND … . . 𝑋𝑛 = 𝑥𝑛
INPUT (X) CLASS(Y)
… … …
… … … …
𝑥 1 , 𝑥 2 , … , 𝑥 𝑛 𝑦 𝑖
… … … …

 Naïve Bayesian classifier calculate this posterior probability using Bayes’
theorem, which is as follows.
 From Bayes’ theorem on conditional probability, we have
𝑃 𝑌 𝑋 =
𝑃(𝑋|𝑌)∙𝑃(𝑌)
𝑃(𝑋)
=
𝑃(𝑋|𝑌) ∙ 𝑃(𝑌)
𝑃 𝑋 𝑌 = 𝑦1 ∙ 𝑃 𝑌 = 𝑦1 + ⋯ + 𝑃 𝑋 𝑌 = 𝑦𝑘 ∙ 𝑃 𝑌 = 𝑦𝑘
where,
𝑃 𝑋 = 𝑖=1
𝑘
𝑃(𝑋|𝑌 = 𝑦𝑖) ∙ 𝑃(Y = 𝑦𝑖)
Note:
 𝑃 𝑋 is called the evidence (also the total probability) and it is a constant.
 The probability P(Y|X) (also called class conditional probability) is therefore
proportional to P(X|Y)∙ 𝑃(𝑌).
 Thus, P(Y|X) can be taken as a measure of Y given that X.
P(Y|X) ≈ 𝑃 𝑋 𝑌 ∙ 𝑃(𝑌)

 Suppose, for a given instance of X (say x = (𝑋1 = 𝑥1) and ….. (𝑋𝑛= 𝑥𝑛)).
 There are any two class conditional probabilities namely P(Y= 𝑦𝑖|X=x)
and P(Y= 𝑦𝑗 | X=x).
 If P(Y= 𝑦𝑖 | X=x) > P(Y= 𝑦𝑗 | X=x), then we say that 𝑦𝑖 is more stronger
than 𝑦𝑗 for the instance X = x.
 The strongest 𝑦𝑖 is the classification for the instance X = x.

 Example: With reference to the Air Traffic Dataset mentioned earlier, let
us tabulate all the posterior and prior probabilities as shown below.
Class
Attribute On Time Late Very Late Cancelled
Day
Weekday 9/14 = 0.64 ½ = 0.5 3/3 = 1 0/1 = 0
Saturday 2/14 = 0.14 ½ = 0.5 0/3 = 0 1/1 = 1
Sunday 1/14 = 0.07 0/2 = 0 0/3 = 0 0/1 = 0
Holiday 2/14 = 0.14 0/2 = 0 0/3 = 0 0/1 = 0
Season
Spring 4/14 = 0.29 0/2 = 0 0/3 = 0 0/1 = 0
Summer 6/14 = 0.43 0/2 = 0 0/3 = 0 0/1 = 0
Autumn 2/14 = 0.14 0/2 = 0 1/3= 0.33 0/1 = 0
Winter 2/14 = 0.14 2/2 = 1 2/3 = 0.67 0/1 = 0

Class
Attribute On Time Late Very Late Cancelled
Fog
None 5/14 = 0.36 0/2 = 0 0/3 = 0 0/1 = 0
High 4/14 = 0.29 1/2 = 0.5 1/3 = 0.33 1/1 = 1
Normal 5/14 = 0.36 1/2 = 0.5 2/3 = 0.67 0/1 = 0
Rain
None 5/14 = 0.36 1/2 = 0.5 1/3 = 0.33 0/1 = 0
Slight 8/14 = 0.57 0/2 = 0 0/3 = 0 0/1 = 0
Heavy 1/14 = 0.07 1/2 = 0.5 2/3 = 0.67 1/1 = 1
Prior Probability 14/20 = 0.70 2/20 = 0.10 3/20 = 0.15 1/20 = 0.05

Instance:
Case1: Class = On Time : 0.70 × 0.64 × 0.14 × 0.29 × 0.07 = 0.0013
Case2: Class = Late : 0.10 × 0.50 × 1.0 × 0.50 × 0.50 = 0.0125
Case3: Class = Very Late : 0.15 × 1.0 × 0.67 × 0.33 × 0.67 = 0.0222
Case4: Class = Cancelled : 0.05 × 0.0 × 0.0 × 1.0 × 1.0 = 0.0000
Case3 is the strongest; Hence correct classification is Very Late
Week Day Winter High Heavy ???

Note: 𝒑𝒊 ≠ 𝟏, because they are not probabilities rather proportion values (to
posterior probabilities)
Input: Given a set of k mutually exclusive and exhaustive classes C =
𝑐1, 𝑐2, … . . , 𝑐𝑘 , which have prior probabilities P(C1), P(C2),….. P(Ck).
There are n-attribute set A = 𝐴1, 𝐴2, … . . , 𝐴𝑛 , which for a given instance have
values 𝐴1 = 𝑎1, 𝐴2 = 𝑎2,….., 𝐴𝑛 = 𝑎𝑛
Step: For each 𝑐𝑖 ∈ C, calculate the class condition probabilities, i = 1,2,…..,k
𝑝𝑖 = 𝑃 𝐶𝑖 × 𝑗=1
𝑛
𝑃(𝐴𝑗 = 𝑎𝑗|𝐶𝑖)
𝑝𝑥 = max 𝑝1, 𝑝2, … . . , 𝑝𝑘
Output: 𝐶𝑥 is the classification
Algorithm: Naïve Bayesian Classification

Pros and Cons
 The Naïve Bayes’ approach is a very popular one, which often works well.
 However, it has a number of potential problems
 It relies on all attributes being categorical.
 If the data is less, then it estimates poorly.

Approaches to overcome the limitations in Naïve Bayesian
Classification
 Estimating the posterior probabilities for continuous attributes
 In real life situation, all attributes are not necessarily be categorical, In
fact, there is a mix of both categorical and continuous attributes.
 There are two approaches to deal with continuous attributes in Bayesian
classifier.
Approach 1:
We can discretize each continuous attributes and then replace the
continuous values with its corresponding discrete intervals.

 Estimating the posterior probabilities for continuous attributes
Approach 2:
We can assume a certain form of probability distribution for the continuous
variable and estimate the parameters of the distribution using the training
data. A Gaussian distribution is usually chosen to represent the posterior
probabilities for continuous attributes. A general form of Gaussian
distribution will look like
P x: μ, σ2 =
1
2πσ
e− x−μ 2
2σ2 𝑑𝑥
where, μ and σ2
denote mean and variance, respectively.

For each class Ci, the posterior probabilities for attribute Aj (it is the numeric
attribute) can be calculated following Gaussian normal distribution as
follows.
P Aj = x|Ci =
1
2πσij
e−
(x − μ𝑖𝑗)2
2σij
2
𝑑𝑥
Here, the parameter μij can be calculated based on the sample mean of
attribute values of Aj for the training records that belong to the class Ci.
Similarly, σij
2
can be estimated from the calculation of variance of such
training records.

 The M-estimation is to deal with the potential problem of Naïve
Bayesian Classifier when training data size is too poor.
 If the posterior probability for one of the attribute is zero, then the
overall class-conditional probability for the class vanishes.
 In other words, if training data do not cover many of the attribute values,
then we may not be able to classify some of the test records.
 This problem can be addressed by using the M-estimate approach.

M-estimate Approach
 M-estimate for a posterior probability calculation is stated as follows:
P Aj=x|Ci =
𝑛𝑐𝑖
+𝑚𝑝
𝑛+𝑚
where, n = total number of instances from class C𝑖
𝑛𝑐𝑖
= number of training examples from class C𝑖 that take the
value Aj=x
m = it is a parameter known as the equivalent sample size, and
p = is a user specified parameter.
Note:
If n = 0, that is, if there is no training data available, then 𝑃 Aj=x|C𝑖 = p,
Thus, it gives a non=zero value, in case of absence of a sample value.

Case Study - 1
Example 7.5
age income student
credit_rating
buys_compu
<=30 high no fair no
<=30 high no excellent no
31…40 high no fair yes
>40 medium no fair yes
>40 low yes fair yes
>40 low yes excellent no
31…40 low yes excellent yes
<=30 medium no fair no
<=30 low yes fair yes
>40 medium yes fair yes
<=30 medium yes excellent yes
31…40 medium no excellent yes
31…40 high yes fair yes
>40 medium no excellent no
Class:
C1:buys_computer = ‘yes’
C2:buys_computer = ‘no’
Data instance
X = (age <=30,
Income = medium,
Student = yes
Credit_rating = fair)

Solution to Case Study -1
 P(Ci): P(buys_computer = “yes”) = 9/14 = 0.643
P(buys_computer = “no”) = 5/14= 0.357
 Compute P(X|Ci) for each class
P(age = “<=30” | buys_computer = “yes”) = 2/9 = 0.222
P(age = “<= 30” | buys_computer = “no”) = 3/5 = 0.6
P(income = “medium” | buys_computer = “yes”) = 4/9 = 0.444
P(income = “medium” | buys_computer = “no”) = 2/5 = 0.4
P(student = “yes” | buys_computer = “yes) = 6/9 = 0.667
P(student = “yes” | buys_computer = “no”) = 1/5 = 0.2
P(credit_rating = “fair” | buys_computer = “yes”) = 6/9 = 0.667
P(credit_rating = “fair” | buys_computer = “no”) = 2/5 = 0.4
 X = (age <= 30 , income = medium, student = yes, credit_rating = fair)
P(X|Ci) : P(X|buys_computer = “yes”) = 0.222 × 0.444 × 0.667 × 0.667 = 0.044
P(X|buys_computer = “no”) = 0.6 × 0.4 × 0.2 × 0.4 = 0.019
P(X|Ci)*P(Ci) : P(X|buys_computer = “yes”) * P(buys_computer = “yes”) = 0.028
P(X|buys_computer = “no”) * P(buys_computer = “no”) = 0.007
Therefore, X belongs to class (“buys_computer = yes”)

Case Study - 2
Example 7.6
WEATHER TEMPERATURE HUMIDITY WINDY PLAY GOLF
1 Rainy Hot High False No
2 Rainy Hot High True No
3 Overcast Hot High False Yes
4 Sunny Mild High False Yes
5 Sunny Cool Normal False Yes
6 Sunny Cool Normal True No
7 Overcast Cool Normal True Yes
8 Rainy Mild High False No
9 Rainy Cool Normal False Yes
10 Sunny Mild Normal False Yes
11 Rainy Mild Normal True Yes
12 Overcast Mild High True Yes
13 Overcast Hot Normal False Yes
14 Sunny Mild High True No

Case Study - 2
WEATHER TEMPERATURE HUMIDITY WINDY
Rainy Hot High False

Reference
 The detail material related to this lecture can be found in
Data Mining: Concepts and Techniques, (3rd Edn.), Jiawei Han, Micheline Kamber,
Morgan Kaufmann, 2015.
Introduction to Data Mining, Pang-Ning Tan, Michael Steinbach, and Vipin Kumar,
Addison-Wesley, 2014

Any question?
56
@DSamanta, IIT Kharagpur CS 61061: Data Analytics

07_BayesianClassifier.pptx

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