Using a decision tree (ID3), the new tuple (Homeowner = Yes, Status = Employed, Income = Average) is classified as Defaulted = Yes.
The data has 10 records: Defaulted = Yes for records 5, 8, 10 (3) and No for the other 7.
Step 1: Entropy of the whole set
\[E(S) = -0.3\log_2 0.3 - 0.7\log_2 0.7 = 0.881\]
Step 2: Information gain of each attribute
- Homeowner: Yes = 3 records (all No), entropy 0; No = 7 records (3 Yes, 4 No), entropy 0.985.
\(E = 0.7\times0.985 = 0.690\); Gain \(= 0.881 - 0.690 = 0.191\)
- Status: Employed = 4 (2 Yes, 2 No), entropy 1; Business = 4 (all No), 0; Unemployed = 2 (1 Yes, 1 No), 1.
\(E = 0.4(1) + 0.4(0) + 0.2(1) = 0.6\); Gain \(= 0.281\)
- Income: High = 3 (all No), 0; Low = 3 (all No), 0; Average = 4 (3 Yes, 1 No), entropy 0.811.
\(E = 0.4\times0.811 = 0.325\); Gain \(= 0.557\)
Income has the highest gain, so it is the root.
Step 3: Split the Income = Average branch (records 2, 5, 8, 10)
- All four have Homeowner = No, so Homeowner gives no gain.
- Status separates them perfectly: Business (record 2) → No; Unemployed (record 5) → Yes; Employed (records 8, 10) → Yes.
Final tree
Income?
|-- High -> No
|-- Low -> No
|-- Average -> Status?
|-- Business -> No
|-- Employed -> Yes
|-- Unemployed -> Yes
Classify the new tuple: Income = Average → Status = Employed → Defaulted = Yes.
Note: if Naive Bayes (without Laplace smoothing) is used instead, the answer becomes No, because no defaulter in the data is a homeowner, so \(P(\text{Homeowner = Yes}\mid\text{Yes}) = 0\). This tuple is an unseen combination, so different classifiers can disagree; state which method you use.
