Bayesian Network for Farm Security System

Scenario

A farmer is concerned about protecting his orange farm from thieves during the night while he is sleeping. He has installed a sensitive alarm system that can be triggered by intruders, sheep roaming the farm, or thunder. The farmer uses a Bayesian Network to model the situation and make informed decisions about activating the alarm to ensure the safety of his oranges. The Bayesian Network includes the following random variables:

• IntruderPresence (I): Presence of intruders on the farm. States: True (intruders present), False (no intruders).
• SheepPresence (S): Presence of sheep on the farm, which could trigger the alarm. States: True (sheep present), False (no sheep).
• Thunder (T): Occurrence of thunder, which could trigger the alarm. States: True (thunder occurs), False (no thunder).
• AlarmTriggered (A): Whether the alarm is triggered. States: True (alarm triggered), False (alarm not triggered).
• FarmerWakeUp (F): Farmer's decision to wake up and defend the oranges. States: True (farmer wakes up), False (farmer remains asleep).
• OrangeSafety (O): Safety of the oranges. States: 1 (oranges secure), 0 (oranges at risk).

The farmer has collected probability data in a dataset (night_farm_dataset.csv).

Task a) Model the Bayesian Network

Construct a directed acyclic graph (DAG) for the Bayesian Network. Define the nodes (variables) and edges (dependencies) based on the scenario. Provide a brief explanation of the dependencies chosen. Draw the DAG (sketch or describe in text).

Dependency Justification

Edge	Reason
I → A	Intruders trigger the alarm
S → A	Sheep roaming can trigger the alarm
T → A	Thunder can trigger the alarm
A → F	Farmer wakes up only if alarm sounds
I → O	Intruders directly threaten orange safety
F → O	Farmer waking up protects the oranges

Result Table for Task a

Component	Description
Number of Nodes	6 (I, S, T, A, F, O)
Number of Edges	6
Graph Type	Directed Acyclic Graph (DAG)

Task b) Construct Conditional Probability Tables (CPTs)

Define the CPTs for each variable, listing states and probabilities conditioned on parent nodes. For variables with no parents, provide prior probabilities. Assume realistic probability values from the dataset if specific values are unavailable, ensuring probabilities sum to 1.

Prior Probabilities

Variable	P(True)
I	0.04999
S	0.30099
T	0.09843

P(A | I, S, T)

I	S	T	P(A=True)	Support
True	True	True	0.9065	139
True	True	False	0.9424	1,337
True	False	True	0.9190	395
True	False	False	0.9156	3,128
False	True	True	0.7983	2,786
False	True	False	0.7478	25,837
False	False	True	0.5924	6,523
False	False	False	0.0096	59,855

P(F | A)

A	P(F=True)	Support
True	0.8019	30,595
False	0.0486	69,405

P(O | I, F)

I	F	P(O=True)	Support
True	True	1.0000	3,671
True	False	0.0000	1,328
False	True	1.0000	24,237
False	False	1.0000	70,764

Result Table for Task b

CPT	Number of Entries	Source
P(I)	2	Dataset (100,000 rows)
P(S)	2	Dataset (100,000 rows)
P(T)	2	Dataset (100,000 rows)
P(A|I,S,T)	8	Dataset (100,000 rows)
P(F|A)	2	Dataset (100,000 rows)
P(O|I,F)	4	Dataset (100,000 rows)

Task c) Calculate Marginal Probabilities

Using the Bayesian Network and CPTs, calculate the marginal probabilities for each variable (I, S, T, A, F, O) with no prior evidence. Show calculations, including chain rule or marginalization.

P(A=True) Calculation

$$ P(A=True) = \sum_{i,s,t} P(I=i) \cdot P(S=s) \cdot P(T=t) \cdot P(A=True \mid I=i, S=s, T=t) $$

Term	I	S	T	Product
1	T	T	T	0.04999 × 0.30099 × 0.09843 × 0.9065 = 0.001343
2	T	T	F	0.04999 × 0.30099 × 0.90157 × 0.9424 = 0.012784
3	T	F	T	0.04999 × 0.69901 × 0.09843 × 0.9190 = 0.003161
4	T	F	F	0.04999 × 0.69901 × 0.90157 × 0.9156 = 0.028845
5	F	T	T	0.95001 × 0.30099 × 0.09843 × 0.7983 = 0.022468
6	F	T	F	0.95001 × 0.30099 × 0.90157 × 0.7478 = 0.192792
7	F	F	T	0.95001 × 0.69901 × 0.09843 × 0.5924 = 0.038719
8	F	F	F	0.95001 × 0.69901 × 0.90157 × 0.0096 = 0.005721
Total				0.3058

P(F=True) Calculation

$$ P(F=\text{True}) = P(F=\text{True} \mid A=\text{True}) \times P(A=\text{True}) + P(F=\text{True} \mid A=\text{False}) \times P(A=\text{False}) $$

$$ = 0.8019 \times 0.3058 + 0.0486 \times (1 - 0.3058) = 0.245237 + 0.033756 = 0.2790 $$

P(O=True) Calculation

$$ P(O=\text{True}) = \sum_{i,f} P(O=\text{True} \mid I=i, F=f) \cdot P(I=i) \cdot P(F=f) $$

$$ = (1.0 \times 0.04999 \times 0.2790) + (0.0 \times 0.04999 \times 0.7210) + (1.0 \times 0.95001 \times 0.2790) + (1.0 \times 0.95001 \times 0.7210) $$

$$ = 0.013947 + 0 + 0.265046 + 0.684964 = 0.9640 $$

Result Table for Task c

Variable	Marginal Probability
P(I=True)	0.0500
P(S=True)	0.3010
P(T=True)	0.0984
P(A=True)	0.3058
P(F=True)	0.2790
P(O=True)	0.9640

Task d) Perform Probabilistic Inference

Answer the following:

• Compute P(A = True | S = True), the probability that the alarm is triggered due to sheep presence. Show calculations, accounting for other variables.
• Compute P(F = True | I = True), the probability that the farmer wakes up given an intruder. Show calculations, including summations or Bayes' theorem.

Query 1: P(A=True | S=True)

$$ P(A=\text{True} \mid S=\text{True}) = \frac{P(A=\text{True}, S=\text{True})}{P(S=\text{True})} $$

$$ P(A=\text{True}, S=\text{True}) = \sum_{i,t} P(I=i) \times P(T=t) \times P(S=\text{True} \mid I=i, T=t) \times P(A=\text{True} \mid I=i, S=\text{True}, T=t) $$

I	T	P(S=True|I, T)	Contribution
True	True	0.2603	0.04999 × 0.09843 × 0.2603 × 0.9065 = 0.001161
True	False	0.2994	0.04999 × 0.90157 × 0.2994 × 0.9424 = 0.012718
False	True	0.2993	0.95001 × 0.09843 × 0.2993 × 0.7983 = 0.022340
False	False	0.3015	0.95001 × 0.90157 × 0.3015 × 0.7478 = 0.193125

$$ P(A=\text{True}, S=\text{True}) = 0.001161 + 0.012718 + 0.022340 + 0.193125 = 0.229345 $$

$$ P(S=\text{True}) = 0.300990 $$

$$ P(A=\text{True} \mid S=\text{True}) = \frac{0.229345}{0.300990} = 0.7620 $$

Query 2: P(F=True | I=True)

$$ P(F=\text{True} \mid I=\text{True}) = \frac{P(F=\text{True}, I=\text{True})}{P(I=\text{True})} $$

$$ P(F=\text{True}, I=\text{True}) = \sum_{s,t,a} P(I=\text{True}) \cdot P(S=s) \cdot P(T=t) \cdot P(A=a \mid I=\text{True}, S=s, T=t) \cdot P(F=\text{True} \mid A=a) $$

After summing over all 8 combinations:

$$ P(F=\text{True}, I=\text{True}) = 0.037180 $$

$$ P(I=\text{True}) = 0.049990 $$

$$ P(F=\text{True} \mid I=\text{True}) = \frac{0.037180}{0.049990} = 0.7437 $$

Result Table for Task d

Query	Result	Interpretation
P(A=True | S=True)	0.7620	Given sheep presence, alarm triggers with 76.20% probability
P(F=True | I=True)	0.7437	Given intruder presence, farmer wakes up with 74.37% probability

Learning Objectives

• Understand Bayesian Network structure and probabilistic reasoning.
• Model dependencies and perform inference for marginal and conditional probabilities.
• Interpret results in a real-world context.

Final Results Summary Table

Task	Key Result
Task a	DAG with 6 nodes, 6 edges
Task b	6 CPTs extracted from 100,000 rows
Task c	Marginals: P(A)=0.3058, P(F)=0.2790, P(O)=0.9640
Task d	P(A|S)=0.7620, P(F|I)=0.7437

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This code was developed as part of a Machine Learning course assignment.