Intelligent Systems II · TEE · papers

Q3(a) Bayesian-network conditional cough probability [10]

Final / re-examination · 28 Apr 2025

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

Question. Q3(a) [10 marks · 2025-04-28]

Original graph: Smoking→Lung Cancer→Cough; Pneumonia→Cough and Pneumonia→Fever.

Smoking S	P(S)
T	0.3
F	0.7

Smoking S	P(C=T\|S)	P(C=F\|S)
T	0.1	0.9
F	0.01	0.99

Pneumonia P	P(P)
T	0.3
F	0.7

C	P	P(Co=T\|C, P)	P(Co=F\|C, P)
T	T	0.95	0.05
T	F	0.8	0.2
F	T	0.7	0.3
F	F	0.1	0.9

P	P(Fever=T\|P)	P(Fever=F\|P)
T	0.9	0.1
F	0.01	0.99

Given the above tables for the Bayesian network, calculate Probability of a person having cough given he doesn’t have lung cancer but Smokes.

Answer

Given

Evidence: C=F (no lung cancer), S=T (smokes). Query: Co=T. Pneumonia is not observed; sum over its two states.

To find

P(Co=T | C=F, S=T).

Formula and why

P(Co\mid\neg C,S)=\sum_{p\in\{T,F\}}P(Co\mid\neg C,S,P=p)P(P=p\mid\neg C,S)

The network says cough depends on its parents C and P. Once they are known, smoking provides no additional direct information about cough. The pneumonia root is independent of S and C in this network when no descendant evidence such as cough/fever is conditioned on.

Step 1. Write the network factorization

P(S,C,P,Co,F)=P(S)P(C\mid S)P(P)P(Co\mid C,P)P(F\mid P)

This factorization expresses the model, not a general medical truth. Fever is unobserved and its two conditional probabilities sum to 1.

Step 2. Select the correct cough rows

Hidden pneumonia state	Cough probability with C=F	Pneumonia weight
P=T	P(Co=T\|C=F, P=T)=0.70	0.30
P=F	P(Co=T\|C=F, P=F)=0.10	0.70

Do not use 0.95 or 0.80: those rows assume lung cancer is present, which contradicts the given evidence.

Step 3. Multiply each conditional probability by its weight

0.70\times0.30=0.21,\qquad0.10\times0.70=0.07

Step 4. Add the mutually exclusive pneumonia cases

P(Co=T\mid C=F,S=T)=0.21+0.07=\boxed{0.28}

Final answer:28%. We do not multiply the final answer byP(smoking)=0.3: smoking is already a condition, not an additional joint event.

Step 5. Verify using joint-over-evidence

P(\neg C,S)=0.9\times0.3=0.27

P(Co,\neg C,S)=0.27\times0.28=0.0756

P(Co\mid\neg C,S)=0.0756/0.27=0.28