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

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q3a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q3a/

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

**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.



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



#### Solution

Swipe across the diagram to read its labels ↔

![Re-exam 2026 Q3a: weather–flu–cough–fever network](https://deep-learning-study-room.vercel.app/is2/fig/bn_flu.1d732f0fb6.svg)

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



$$
P(W,F,C,V)=P(W)\,P(F\mid W)\,P(C\mid W)\,P(V\mid F,C)
$$



##### Step 2. Sum over W and C with F = false, V = true

| W | C | P(W) | P(¬F|W) | P(C|W) | P(V|¬F,C) | Product |
| --- | --- | --- | --- | --- | --- | --- |
| T | T | 0.75 | 0.8 | 0.70 | 0.60 | 0.252 |
| T | F | 0.75 | 0.8 | 0.30 | 0.01 | 0.0018 |
| F | T | 0.25 | 0.5 | 0.25 | 0.60 | 0.01875 |
| F | F | 0.25 | 0.5 | 0.75 | 0.01 | 0.0009375 |
| **P(V, ¬F)** | | | | | | **0.2734875** |

##### Step 3. Evidence



$$
P(\lnot F)=0.75(0.8)+0.25(0.5)=0.6+0.125=0.725
$$



##### Step 4. Divide



$$
P(V\mid\lnot F)=\frac{0.2734875}{0.725}=0.3772
$$



**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).