Intelligent Systems II · TEE · papers

Q4(b) Naive Bayes: job offer [10]

Re-examination · 31 Jan 2026

Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged)

Question. Predict whether a candidate accepts a job offer from Experience (Fresher/Mid/Senior), Expected Salary (Low/Medium/High) and Company Reputation (Good/Average). Data: (Fresher, Low, Good, Yes), (Mid, High, Good, Yes), (Senior, High, Average, No), (Fresher, Medium, Average, Yes), (Senior, Low, Good, No), (Mid, Medium, Good, Yes), (Fresher, High, Average, No), (Senior, Medium, Average, No). New candidate: Mid, Low, Good. [10 marks]

Answer

Given

The full training data are in Step 1. Query: Experience=Mid, Salary=Low, Reputation=Good. The main calculation uses unsmoothed counts.

To find

Both class scores and the predicted class.

Formula and why it applies

score(C)=P(C)\prod_jP(a_j\mid C)

C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.

Solution

Step 1. Write the training rows as a table

Row	Experience	Salary	Reputation	Class
1	Fresher	Low	Good	Yes
2	Mid	High	Good	Yes
3	Senior	High	Average	No
4	Fresher	Medium	Average	Yes
5	Senior	Low	Good	No
6	Mid	Medium	Good	Yes
7	Fresher	High	Average	No
8	Senior	Medium	Average	No

Step 2. Separate the two class groups

Class	Rows in this group	Class count	Prior
Yes	1, 2, 4, 6	4	4/8 = 0.5
No	3, 5, 7, 8	4	4/8 = 0.5

Step 3. For class Yes, count Experience=Mid

Look only at class Yes rows 1, 2, 4, 6. The matching rows are 2, 6.

P(\text{Experience}=\text{Mid}\mid C=\text{Yes})=\frac{2}{4}=0.5

Step 4. For class Yes, count Salary=Low

Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1.

P(\text{Salary}=\text{Low}\mid C=\text{Yes})=\frac{1}{4}=0.25

Step 5. For class Yes, count Reputation=Good

Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1, 2, 6.

P(\text{Reputation}=\text{Good}\mid C=\text{Yes})=\frac{3}{4}=0.75

Step 6. Multiply the Yes score one factor at a time

P(Yes)=4/8=0.5

0.5\times0.5=0.25

0.25\times0.25=0.0625

0.0625\times0.75=0.046875

Step 7. For class No, count Experience=Mid

Look only at class No rows 3, 5, 7, 8. No row in this class has that feature value.

P(\text{Experience}=\text{Mid}\mid C=\text{No})=\frac{0}{4}=0

Step 8. For class No, count Salary=Low

Look only at class No rows 3, 5, 7, 8. The matching rows are 5.

P(\text{Salary}=\text{Low}\mid C=\text{No})=\frac{1}{4}=0.25

Step 9. For class No, count Reputation=Good

Look only at class No rows 3, 5, 7, 8. The matching rows are 5.

P(\text{Reputation}=\text{Good}\mid C=\text{No})=\frac{1}{4}=0.25

Step 10. Multiply the No score one factor at a time

P(No)=4/8=0.5

0.5\times0=0

0\times0.25=0

0\times0.25=0

Step 11. Compare the scores

score(Yes)=0.046875\qquad score(No)=0

The larger score belongs to Yes. Normalise only if a posterior probability is requested: divide each score by their total 0.046875.

Final answer. Predict Yes. Yes score=0.046875; No score=0.

Official key note. The key gives No = 0.0068 and P(Yes) = 0.873. That No value comes from Laplace smoothing (1/7 × 2/7 × 2/6 × 0.5), but the key did not smooth the Yes side. If you smooth, smooth both: Yes = 0.5 × 3/7 × 2/7 × 4/6 = 0.0408, No = 0.0068, P(Yes) = 0.857. The prediction is Yes either way.

Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.