Intelligent Systems II · TEE · mocks

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

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. 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.