# Q1(a) Disease, fever and test joint table [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q1a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q1a/

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

**Question. For a disease Y with fever X as a symptom, a test Z is conducted. Joint probabilities: Disease (Y=1): fever & test+ 0.576, fever & test− 0.144, no fever & test+ 0.072, no fever & test− 0.016; No disease (Y=0): 0.008, 0.064, 0.012, 0.108. (1) If fever occurs and the test is positive, what is the probability of disease? (2) What is the probability that a random person tests positive? (3) Given no fever, what is the probability of still having the disease?** [5 marks]

### Answer

#### Given

The complete joint table or cell values printed in this question. The rows and columns identify the event combinations.

#### To find

All requested marginal and conditional probabilities.

#### Formula and why it applies

Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum.



$$
P(A\mid E)=\frac{P(A\cap E)}{P(E)},\quad P(E)>0
$$



#### Solution

##### Step 1. (1)



$$
P(Y{=}1\mid X{=}1,Z{=}1)=\frac{0.576}{0.576+0.008}=\frac{0.576}{0.584}=0.986
$$



##### Step 2. (2) Add every test-positive cell



$$
P(Z{=}1)=0.576+0.072+0.008+0.012=0.668
$$



##### Step 3. (3)



$$
P(X{=}0)=0.072+0.016+0.012+0.108=0.208,\qquad P(Y{=}1\mid X{=}0)=\frac{0.072+0.016}{0.208}=\frac{0.088}{0.208}=0.423
$$



**Final answer.** (1) 0.986; (2) 0.668; (3) 0.423.

**Key correction.** The key writes 0.088/0.076 = 1.158 for part (3). A probability cannot exceed 1; the denominator must be P(no fever) = 0.208.