Intelligent Systems II · TEE · papers

Q1(a) Joint probability table [5]

Final examination · 24 Nov 2025

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

Question. A supermarket recorded the joint distribution of Product type (Y: premium = 1, regular = 0), Payment mode (X: credit = 1, debit = 0) and Gender (Z: female = 1, male = 0). (a) If a customer bought a premium item, find the probability that (1) they are female, (2) they paid by credit card. (b) Find (1) P(premium) and (2) P(credit card). (c) Given a male customer paid by debit card, find the probability he bought a regular item. [5 marks]

Answer

Given

The printed eight entries are reproduced below. Their total is 1.002, not 1. Therefore the table is inconsistent as a probability distribution.

Credit, femaleCredit, maleDebit, femaleDebit, male
Premium (Y=1)0.080.100.0720.05
Regular (Y=0)0.100.150.250.20

To find

The five requested probabilities, stating how the inconsistent data are interpreted.

Formula and why it applies

For this worked solution, treat all printed entries as nonnegative weights and normalise by their total 1.002. This preserves the printed ratios. It is a declared interpretation, not a claim about the examiner’s intended correction.

P(E)=\frac{\text{sum of weights for }E}{\text{sum of all weights}}LaTeX: P(E)=\frac{\text{sum of weights for }E}{\text{sum of all weights}}
P(A\mid E)=\frac{\text{sum of weights for }A\cap E}{\text{sum of weights for }E}LaTeX: P(A\mid E)=\frac{\text{sum of weights for }A\cap E}{\text{sum of weights for }E}

Solution

Step 1. Validate the full table
0.08+0.10+0.072+0.05+0.10+0.15+0.25+0.20=1.002LaTeX: 0.08+0.10+0.072+0.05+0.10+0.15+0.25+0.20=1.002

Use total weight 1.002 consistently for unconditional probabilities.

Step 2. Premium row total
w(premium)=0.08+0.10+0.072+0.05=0.302LaTeX: w(premium)=0.08+0.10+0.072+0.05=0.302
Step 3. Female given premium
P(female\mid premium)=\frac{0.08+0.072}{0.302}=\frac{0.152}{0.302}\approx0.503311LaTeX: P(female\mid premium)=\frac{0.08+0.072}{0.302}=\frac{0.152}{0.302}\approx0.503311
Step 4. Credit given premium
P(credit\mid premium)=\frac{0.08+0.10}{0.302}=\frac{0.18}{0.302}\approx0.596026LaTeX: P(credit\mid premium)=\frac{0.08+0.10}{0.302}=\frac{0.18}{0.302}\approx0.596026
Step 5. Normalised premium marginal
P(premium)=\frac{0.302}{1.002}\approx0.301397LaTeX: P(premium)=\frac{0.302}{1.002}\approx0.301397
Step 6. Normalised credit marginal
P(credit)=\frac{0.08+0.10+0.10+0.15}{1.002}=\frac{0.43}{1.002}\approx0.429142LaTeX: P(credit)=\frac{0.08+0.10+0.10+0.15}{1.002}=\frac{0.43}{1.002}\approx0.429142
Step 7. Regular given male and debit
P(regular\mid male,debit)=\frac{0.20}{0.05+0.20}=0.8LaTeX: P(regular\mid male,debit)=\frac{0.20}{0.05+0.20}=0.8
Final answer. Under common normalisation: (a) 0.503311 and 0.596026; (b) 0.301397 and 0.429142; (c) 0.8.

Alternative correction: if 0.072 is corrected to 0.070, the total becomes 1 and the corresponding answers are (a) 0.50 and 0.60; (b) 0.30 and 0.43; (c) 0.8. This agrees with the interpretation used in the supplied key. State the correction; do not call raw totals 0.302 and 0.43 valid marginals of the original inconsistent table.