# Q2(a) Bayesian network joint probabilities [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-final/q2a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-final--q2a/

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

**Question. Q2(a)** [10 marks · 2024-04-17]

Given the task of modelling a student’s marks for an exam, use the Bayesian network and conditional probability tables below. Exam E: 0 easy, 1 difficult. IQ I: 0 low, 1 high. Aptitude S: 0 low, 1 high. (a) Calculate the probability of a difficult exam, low IQ, low aptitude, passing and securing admission. (b) Calculate the probability of high IQ and aptitude, an easy exam, failing and no admission.

Swipe across the diagram to read its labels ↔

![Redrawn transition diagram. Each arrow goes from the current state to the next state; labels are probabilities.](https://deep-learning-study-room.vercel.app/is2/fig/legacy24-exam-bn.9070a35002.svg)

Diagram: Redrawn transition diagram. Each arrow goes from the current state to the next state; labels are probabilities.. Labels: legacy24-exam-bn | Exam E | IQ I | Marks M | Aptitude S | Admit A

Redrawn transition diagram. Each arrow goes from the current state to the next state; labels are probabilities.

| Root | P(0) | P(1) |
| --- | --- | --- |
| E | 0.7 | 0.3 |
| I | 0.8 | 0.2 |

| I | E | P(M=0) | P(M=1) |
| --- | --- | --- | --- |
| 0 | 0 | 0.6 | 0.4 |
| 0 | 1 | 0.9 | 0.1 |
| 1 | 0 | 0.5 | 0.5 |
| 1 | 1 | 0.8 | 0.2 |

| I | P(S=0) | P(S=1) |
| --- | --- | --- |
| 0 | 0.75 | 0.25 |
| 1 | 0.4 | 0.6 |

| M | P(A=0) | P(A=1) |
| --- | --- | --- |
| 0 | 0.6 | 0.4 |
| 1 | 0.9 | 0.1 |

### Answer

#### Given

All five variables and their CPTs are reproduced above.

#### To find

The two complete joint-event probabilities.

#### Formula and why it applies

For a Bayesian network, multiply each node’s probability conditional on its parents. Roots have no parents. Do not multiply an extra marginal for a node whose conditional factor is already present.



$$
P(E,I,M,S,A)=P(E)P(I)P(M\mid E,I)P(S\mid I)P(A\mid M)
$$



#### Solution

**Source / assumption note.** The source defines codes for E, I and S but does not explicitly define pass/admission coding. The calculations below state the conventional assumption M=1 means pass and A=1 means admitted; 0 means fail/not admitted. If the examiner defines reversed coding, select the corresponding CPT columns instead.

##### Step 1. Write event (a) in symbols



$$
(E,I,M,S,A)=(1,0,1,0,1)
$$



##### Step 2. Select one CPT entry for each node

| Factor | Why this entry | Value |
| --- | --- | --- |
| P(E=1) | Difficult | 0.3 |
| P(I=0) | Low IQ | 0.8 |
| P(M=1|E=1, I=0) | Pass, difficult, low IQ | 0.1 |
| P(S=0|I=0) | Low aptitude given low IQ | 0.75 |
| P(A=1|M=1) | Admission given pass | 0.1 |

##### Step 3. Multiply without dropping root probabilities



$$
0.3\times0.8=0.24\quad;\quad0.24\times0.1=0.024
$$





$$
0.024\times0.75=0.018\quad;\quad0.018\times0.1=0.0018
$$



##### Step 4. Write and solve event (b)



$$
(E,I,M,S,A)=(0,1,0,1,0)
$$





$$
0.7\times0.2\times0.5\times0.6\times0.6=0.0252
$$



Intermediate products: 0.7×0.2=0.14; ×0.5=0.07; ×0.6=0.042; ×0.6=0.0252. These are joint probabilities of the whole described event. If E/I/S were explicitly conditioned on, their factors would cancel; the conditional answers would be 0.1×0.1=0.01 and 0.5×0.6=0.30.

**Final answer.** Under the declared coding and joint-event reading: (a) 0.0018, (b) 0.0252.