3. Naïve Bayes ModelsDescribe a classification task from your experience and represent it as a Naïve Bayes model. Makesure to explicitly specify both the labels and the features. You do not need to specify the values of theparameters (those would come from data!). To what extent are the independence assumptions madeby a Naïve Bayes model reasonable in your task?

Consider a dataset as follows: Outlook Temperature Humidity Windy Play Golf Rainy Hot High False No Rainy Hot High True No Overcast Hot High False Yes Sunny Mild High False Yes Sunny Cool Normal False Yes Sunny Cool Normal True No Overcast Cool Normal True Yes Rainy Mild High False No Rainy Cool Normal False Yes Sunny Mild Normal False Yes Rainy Mild Normal True Yes Overcast Mild High True Yes Overcast Hot Normal False Yes Sunny Mild High True No The above dataset contains 14 rows and 5 columns. The columns Outlook, Temperature, Humidity, and Windy are independent features while Play Golf is the dependent feature or target variable. The target variable "Play Golf" contains two different labels: Yes and No. So, binary classification is applicable here.Naive Bayes Method Assumptions: Every feature makes an independent offering to the output result. This means the value of the temperature being "Hot" does not depend on the humidity being "Normal" or the outlook being "Rainy" or the presence or absence of winds. Every feature variable has equal importance which means only Temperatur and Humidity cannot predict the outcome correctly. The Bayes theorem is mathematically represented as follows: P(A|B)=P(B|A)*P(A)P(B) where P(A) is the prior probability of A, P(A|B) is a posteriori probability of B. Applying the Bayes theorem to the above dataset: P(Y|X)=P(X|Y)*P(X)P(Y) Consider Y as a class variable and X as a dependent vector of size n such that X=(x1, x2, x3,......,xn). If two events A and B are independent, then their probability is written as follows: P(A,B)=P(A)*P(B) Since every feature in the Naive Bayes method is independent, therefore it can be written as follows: P(y|x1, x2, x3,........,xn)=P(x1|Y)P(x2|Y).....P(xn|Y)P(Y)P(x1)P(x2).........P(xn).Probability calculation for every column: Outlook Yes No P(Yes) P(No) Sunny 2 3 2/9 3/5 Overcast 4 0 4/9 0/5 Rainy 3 2 3/9 2/5 Total 9 5 1 1 Temperature Yes No P(Yes) P(No) Hot 2 2 2/9 2/5 Mid 4 2 4/9 2/5 Cool 3 1 3/9 1/5 Total 9 5 1 1 Humidity Yes No P(Yes) P(No) High 3 4 3/9 4/5 Normal 6 1 6/9 1/5 Total 9 5 1 1 Wind Yes No P(Yes) P(No) False 6 2 6/9 2/5 True 3 3 3/9 3/5 Total 9 5 1 1 Play Golf P(Yes)/P(No) Yes 9 9/14 No 5 5/14 Total 14 1Consider a tuple today with a new set of features: today=(Sunny, Hot, Normal, False) where Outlook=Sunny, Temperature=Hot, Humidity=Normal and Wind=False Predict whether today is suitable to Play Golf using the Naive Bayes method: The probability of playing golf today is as follows: P(Yes|today)=P(SunnyOutlook|Yes)P(HotTemperature|Yes))P(NormalHumidity|Yes)P(FalseWind|Yes)P(Yes)P(today) The probability of not playing golf today is as follows: P(No|today)=P(SunnyOutlook|No)P(HotTemperature|No))P(NormalHumidity|No)P(FalseWind|No)P(No)P(today) As P(today) is present in the denominators of both the probabilities, P(today) is ignored and finds proportional probabilities as follows: P(Yes|today)∝29*29*69*69*914≈0.0141 P(No|today)∝35*25*15*25*514≈0.0068 P(Yes|today)+P(No|today)=1 Therefore, after normalization: P(Yes|today)=0.01410.0141+0.0068=0.67 P(No|today)=0.00680.0141+0.0068=0.33 As P(Yes|today)>P(No|today), the prediction value for the target column is "Yes". The above method is applicable to discrete data only. For continuous data, assumptions have to be made regarding the distribution of values of each feature.