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ARTIBA ARTIFICIAL INTELLIGENCE ENGINEER CERTIFICATION EXAM PREP 2026 FULL EXAMINATION

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ARTIBA ARTIFICIAL INTELLIGENCE ENGINEER CERTIFICATION EXAM PREP 2026 FULL EXAMINATION

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ARTIBA ARTIFICIAL INTELLIGENCE ENGINEER CERTIFICAT
Course
ARTIBA ARTIFICIAL INTELLIGENCE ENGINEER CERTIFICAT

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ARTIBA ARTIFICIAL INTELLIGENCE ENGINEER
CERTIFICATION EXAM PREP 2026 FULL
EXAMINATION

◉ Least Constraining Value. Answer: Given a variable, choose the
value that constrains the other unassigned variables the least.


◉ Forward Checking. Answer: When a variable is assigned a value,
establish arc consistency - discarding states which don't meet the
constraints.


◉ Min-Conflicts Heuristic. Answer: Choose the value that violates
the smallest number of constraints.


◉ Probabilistic Inference. Answer: Computing the posterior
probability of a given proposition given observed evidence.


◉ Coarticulation. Answer: The articulation of two or more speech
sounds together, so that one influences the other.


◉ Exhaustive. Answer: The sample set contains all possible
outcomes.

,◉ Mutually Exclusive. Answer: Only one outcome can exist at any
given time.


◉ Proposition. Answer: A statement to be evaluated as true or false.


◉ Unconditional Probability / Prior Probability / Prior. Answer:
Probability value in the absence of any other information.


◉ Conditional Probability / Posterior Probability / Posterior.
Answer: Probability value in the presence of additional information
(evidence).


◉ Conditional Probability: Formula. Answer:


◉ Domain. Answer: Set of possible values the variable can take on.


◉ Probability Distribution: One Variable Example. Answer:
P(Weather) = ⟨0.6, 0.1, 0.29, 0.01⟩


◉ Probability Distribution: Multiple Variables Example. Answer:
P(Weather, Cavity): a 4×2 table of probabilities - joint probability
distribution of Weather and Cavity

, ◉ Joint Probability Distribution. Answer: The probability
distribution determining the probabilities of outcomes involving two
or more random variables.


◉ Full Joint Probability Distribution. Answer: The joint distribution
for all of the random variables.


◉ Marginalisation. Answer: Extracting the distribution over a single
variable or a subset of variables.


◉ Expected Value: Formula. Answer:


◉ Chain Rule Formula (n=3). Answer: P(a, b, c) = P(a | b, c) P(b | c)
P(c)


◉ Bayes' Rule. Answer:


◉ P(a, b) Equivalent. Answer: P(a, b) = P(a | b) P(b) = P(b | a) P(a) =
P(b, a)


◉ P(Y | X, e) Equation. Answer: P(X | Y, e) P(Y | e)
-------------------
P(X | e)

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