Intelligent Systems II · TEE · mocks

Q3(a) Bayesian network: fever given no flu [10]

Re-examination · 31 Jan 2026

100-mark timed practice using the actual Re-exam 2026 paper. Q1 compulsory; any four of Q2–Q7. Worked solutions are below the questions.

Question. Weather → Flu, Weather → Cough, Flu and Cough → Fever. P(Weather) = 0.75; P(Flu | W) = 0.2, P(Flu | ¬W) = 0.5; P(Cough | W) = 0.70, P(Cough | ¬W) = 0.25; P(Fever | Flu, Cough) = 0.60, P(Fever | Flu, ¬Cough) = 0.02, P(Fever | ¬Flu, Cough) = 0.60, P(Fever | ¬Flu, ¬Cough) = 0.01. Calculate the probability of a person having fever given he doesn't have flu. [10 marks]

Answer

The following table enumerates the values of Weather and Cough that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.

Given

W→F, W→C, (F,C)→V. The complete probabilities are in the question. Evidence F=false.

To find

P(V=true | F=false), summing hidden W and C.

Formula and why it applies

Use the network joint and normalise by the no-Flu evidence.

P(V\mid\neg F)=\frac{\sum_{w,c}P(w)P(\neg F\mid w)P(c\mid w)P(V\mid\neg F,c)}{\sum_wP(w)P(\neg F\mid w)}LaTeX: P(V\mid\neg F)=\frac{\sum_{w,c}P(w)P(\neg F\mid w)P(c\mid w)P(V\mid\neg F,c)}{\sum_wP(w)P(\neg F\mid w)}

Solution

Swipe across the diagram to read its labels ↔
Re-exam 2026 Q3a: weather–flu–cough–fever network
Diagram: Re-exam 2026 Q3a: weather–flu–cough–fever network. Labels: Re-exam 2026 Q3a: weather–flu–cough–fever network | Weather | P(W) = 0.75 | Flu | P(F|W) = 0.2, P(F|¬W) = 0.5 | Cough | P(C|W) = 0.7, P(C|¬W) = 0.25 | Fever | P(V|F,C): .60 · .02 · .60 · .01 | (TT · TF · FT · FF)
Step 1. Chain rule
P(W,F,C,V)=P(W)\,P(F\mid W)\,P(C\mid W)\,P(V\mid F,C)LaTeX: P(W,F,C,V)=P(W)\,P(F\mid W)\,P(C\mid W)\,P(V\mid F,C)
Step 2. Sum over W and C with F = false, V = true
WCP(W)P(¬F|W)P(C|W)P(V|¬F,C)Product
TT0.750.80.700.600.252
TF0.750.80.300.010.0018
FT0.250.50.250.600.01875
FF0.250.50.750.010.0009375
P(V, ¬F)0.2734875
Step 3. Evidence
P(\lnot F)=0.75(0.8)+0.25(0.5)=0.6+0.125=0.725LaTeX: P(\lnot F)=0.75(0.8)+0.25(0.5)=0.6+0.125=0.725
Step 4. Divide
P(V\mid\lnot F)=\frac{0.2734875}{0.725}=0.3772LaTeX: P(V\mid\lnot F)=\frac{0.2734875}{0.725}=0.3772
Final answer. P(fever | no flu) ≈ 0.377 (fever without flu comes mostly from cough).
Key correction. The key uses P(Flu) = 0.05 and P(Fever | ¬Flu) = 0.2, which are not in this network, and answers 0.2. Marking scheme (still applies): write the equation with hidden variables (2), take true/false cases (3), substitute (2), answer (3).