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Illustrate any one classification technique for the above data set. Show how can we classify a new tuple, with (Homeowner=Yes, Status=Employed, Income=Average)
Id House Owner Status Income Defaulted
1 Yes Employed High No
2 No Business Average No
3 No Employed Low No
4 Yes Business High No
5 No Unemployed Average Yes
6 No Business Low No
7 Yes Employed High No
8 No Unemployed Average Yes
9 No Business Low No
10 No Employed Average Yes

Illustrate any one classification technique for the above data set. Show how can we classify a new tuple, with (Homeowner=Yes, Status=Employed, Income=Average)

1 Answer
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Naive Bayes Classifier Method

Here, we want to classify a Homeowner = Yes, Status = Employed, Income = Average and predict the class label for this sample.

  • The above sample dataset does not contain the class label for this sample.
  • Hence, by using Naïve Bayes Classifier we can find out the class table value for the given sample.
  • To do this we need to calculate the below probabilities:

First, calculates the Probabilities for $P(x | Yes)$

$$P(Yes) = \frac{3}{10} $$

$$ P(Homeowner = Yes│Yes) = \frac{0}{3} = 0 $$

$$ P(Status = Employed│Yes) = \frac{1}{3} $$

$$ P(Income = Average│Yes) = \frac{3}{3} = 1 $$


Second, calculates the Probabilities for $P(x | No)$

$$P(No) = \frac{7}{10}$$

$$P(Homeowner = Yes│No) = \frac{3}{7}$$

$$P(Status = Employed│No) = \frac{3}{7}$$

$$P(Income = Average │No) = \frac{1}{7}$$


According to Naïve Bayes Classifier

$$P(C│X)=P(x1│C) × P(x2│C) ×…………P(xn│C) × P(C)$$


Therefore, Probability for Class label value Yes we have

$$Yes = P(X│Yes) × P(Yes)$$

$$Yes = P(Homeowner = Yes│Yes) × P(Status = Employed│Yes) × P(Income = Average│Yes) × P(Yes)$$

$$= 0 ×\frac{1}{3} × 1 × \frac{3}{10} = 0$$

If one of the conditional probabilities is zero, then the entire expression becomes zero. This can be seen in the above scenario.


Therefore, Probability for Class label No we have

$$ No =P(X│No) × P(No)$$

$$No = P(Homeowner = Yes│No) × P(Status = Employed│No) × P(Income = Average │No) × P(No)$$

$$= \frac{3}{7} × \frac{3}{7} × \frac{1}{7} × \frac{7}{10} = 0.018$$


As

$$P(X│No).P(No) \gt P(X│Yes).P(Yes)$$

$$because,\ 0.018 \gt 0$$

Therefore, this sample (Homeowner = Yes, Status = Employed, Income = Average) gets classified under Class label ’No’.

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