HMM POS Tagging MCQs (Advanced) | Viterbi, Baum-Welch & NLP Concepts

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20. If transitions are uniform/random, HMM POS tagger becomes:

A. Fully deterministic
B. Distribution-free classifier
C. Equivalent to unigram emission selector
D. Memory-based classifier

Correct Answer: C

With uniform transitions, tagging depends only on P(word|tag), i.e., emission probabilities.

With uniform transitions, an HMM POS tagger reduces to a unigram model that tags each word independently using emission probabilities only.

Step-by-step Explanation

An HMM POS tagger assigns part-of-speech tags using two probabilities:

1. Transition probability - P(t_i | t_i−1)
   → How likely a tag follows the previous tag


2. Emission probability - P(w_i | t_i)
   → How likely a word is generated by a tag

During decoding using the Viterbi algorithm, the model maximizes:

P(t_i | t_i−1) × P(w_i | t_i)

What does uniform / random transitions mean?

Uniform transitions imply:

P(t_i | t_i−1) = constant for all tag pairs

• Transition probabilities do not prefer any particular tag sequence
• They contribute the same value for every possible path

Therefore, transition probabilities no longer influence the tagging decision.

What remains?

Only the emission probabilities matter:

arg max_ti P(w_i | t_i)

This is exactly what a unigram POS tagger does:

• Assigns each word the tag with the highest emission probability
• Ignores contextual information entirely

21. Unknown word tagging accuracy is highest when model learns:

A. Prefix/suffix distribution per POS
B. Word frequency only
C. Stopword probability
D. Character bigrams only

Correct Answer: A

This question is about how POS taggers handle unknown (out-of-vocabulary) words—words that were not seen during training.

In Part-of-Speech (POS) tagging, unknown words present a fundamental challenge—they don't appear in the training corpus, so the model cannot rely on learned word-to-tag associations. The solution lies in morphological features, particularly prefix and suffix distributions linked to grammatical categories. Morphological cues like -ly, -ness, -tion strongly correlate with POS tags.

Prefix/suffix distribution per POS

Why? Many parts of speech follow strong morphological patterns:

• -tion, -ness → Noun
• -ly → Adverb
• -ing, -ed → Verb
• un-, re-, pre- → Verbs / Adjectives

By learning which prefixes and suffixes are likely for each POS, the model can:

• Infer the POS of new (unknown) words it has never seen

This is the most effective and widely used approach in POS tagging models such as HMMs, CRFs, and neural taggers.

Therefore, unknown word tagging accuracy is highest when the model learns prefix/suffix distributions per POS.

22. Training HMM with labeled POS corpus is:

A. Unsupervised
B. Supervised
C. Reinforcement-based
D. Zero-shot

Correct Answer: B

Both words and tags are known, so probabilities are estimated directly.

23. If only words are available (no tags), HMM must be trained using:

A. Viterbi only
B. Forward-Backward (Baum-Welch)
C. MEMM
D. CRF

Correct Answer: B

Baum–Welch uses EM to estimate hidden states from unlabeled data.

This question tests your understanding of the three fundamental HMM problems and when to apply each algorithm.

This question asks about the Learning problem of three HMM problems. As per the Learning problem, we are given only the observation sequence without tags, and we need to find the model parameters with the help of Forward-Backward (Baum-Welch) algorithm. This is unsupervised learning.

24. Error propagation in Viterbi decoding is more likely when:

A. Transition matrix is dense
B. Emission probability is sharply peaked for rare word
C. All tags have equal probability
D. Frequent tags dominate data

Correct Answer: B

A wrong but strong emission can lock Viterbi into an incorrect path.

More information:

The question is about Viterbi decoding. Viterbi decoding is used in POS tagging and other sequence labeling tasks. In these tasks, each tag depends on the previous tag.

What is error propagation?

In Viterbi decoding: The algorithm selects the best path step by step. If a wrong tag is chosen early, that wrong choice affects the next tags. As a result, more errors occur later. This spreading of mistakes is called error propagation.

Why does error propagation happen in Viterbi algorithm?

Viterbi uses dynamic programming. It keeps only one best path, not many alternatives. It depends on: Transition probabilities (from one tag to the next) and Emission probabilities (tag to word). If the model is too confident about a wrong tag, the error continues through the sentence.

Why option B is correct?

For a rare or unknown word, the model assigns: Very high probability to one tag and Very low probability to others. If that high-probability tag is wrong, Viterbi commits strongly to it and the alternative paths are discarded. As a result, future tags are forced to follow this wrong tag via transitions

25. Modern POS taggers outperform HMM mainly because:

A. HMM is non-probabilistic
B. Neural models capture long-context + subword info
C. HMM has no decoding algorithm
D. HMM only works on small corpora

Correct Answer: B

Neural models capture global dependencies beyond Markov assumptions.

26. High P(NN | DT) indicates:

A. Noun unlikely after determiner
B. Noun likely after determiner
C. Determiner depends on word length
D. Transition invalid

Correct Answer: B

Determiner → noun is a common English syntactic pattern.

P(NN | DT) means, "The probability that the next tag is a Noun (NN), given that the current tag is a Determiner (DT)".

High probability indicates that Determiners are usually followed by Nouns in language data. Example: "the book", "a pen", "this idea", etc.

27. Sentence probability in an HMM is:

A. Sum of emission probabilities
B. Sum of transitions
C. Sum of products of transition and emission probabilities across all possible state sequences
D. Ratio of emissions

Correct Answer: C

HMM probability is the product over all transition and emission terms.

Sentence probability in HMM = the sum of probabilities of all possible hidden state sequences that could generate the observed sentence, where each path's probability is the product of its transition and emission probabilities.

This probabilistic framework is what makes HMMs powerful for sequence modeling in NLP and speech processing—it elegantly handles the uncertainty of hidden states while maintaining computational efficiency through dynamic programming.

28. Viterbi backpointer table stores:

A. Loss function values
B. Most likely previous tag
C. Word embeddings
D. Vocabulary index

Correct Answer: B

Backpointers reconstruct the optimal tag sequence.

The Viterbi backpointer table is a table that stores where each best probability came from during Viterbi decoding. In simple words, it remembers which previous state (tag) gave the best path to the current state.

Why do we need a backpointer table?

Viterbi decoding has two main steps: (1) Forward pass (Compute the best probability of reaching each state at each time step, Store these probabilities in the Viterbi table.), (2) Backtracking (Recover the best sequence of tags, For this, we must know which state we came from). The backpointer table makes backtracking possible.

Simple Viterbi Example

Sentence: "Time flies"

Possible Tags:

• NN (Noun)
• VB (Verb)

Viterbi Probability Table

Time	NN	VB
t = 1	0.3	0.7
t = 2	0.6	0.4

Backpointer Table

Time	NN	VB
t = 2	VB	NN

Interpretation

• Best path to NN at t = 2 came from VB at t = 1.
• Best path to VB at t = 2 came from NN at t = 1.

29. Gaussian emission HMMs are preferred in speech tagging because:

A. Text is discrete
B. Speech features are continuous
C. POS-tags depend on semantics
D. They remove smoothing

Correct Answer: B

Acoustic signals are continuous-valued, well-modeled by Gaussians.

Fundamental Distinction: Discrete vs. Continuous Observations

The choice between discrete and Gaussian (continuous) HMM emission distributions depends entirely on the nature of the observations being modeled.

Discrete HMMs represent observations as discrete symbols from a finite alphabet—such as words in part-of-speech tagging or written text. When observations are discrete, emission probabilities are modeled as categorical distributions over symbol categories.

Continuous (Gaussian) HMMs represent observations as continuous-valued feature vectors. When observations are real-valued, discrete emission probabilities are not applicable; instead, the probability density is modeled using continuous distributions such as Gaussians or Gaussian Mixture Models (GMMs).

Why Gaussian Emission HMMs Fit Speech Data

In speech tagging or speech recognition, the observed data are acoustic features, not words.

Hidden Markov Models require an emission model to represent:

P(observation | state)

• In text POS tagging → observations are discrete words
• In speech tagging → observations are continuous feature vectors

Therefore, Gaussian (or Gaussian Mixture) distributions are ideal for modeling continuous acoustic data.

Gaussian-emission HMMs model:

P(x_t | state_t)

where x_t is a continuous acoustic feature vector.

30. HMM POS tagging underperforms neural models mainly because it:

A. Requires GPUs
B. Models short context only
C. Cannot generate emissions
D. Lacks training algorithms

Correct Answer: B

HMMs rely on local Markov assumptions, unlike deep contextual models.

HMMs underperform because they rely on short-context Markov assumptions, while neural models capture long-range and global linguistic information.

HMM-based POS taggers rely on the Markov assumption, typically using bigram or trigram tagging. This means the POS tag at position t depends only on a very limited local context.

P(t_t | t_t−1)   or   P(t_t | t_t−1, t_t−2)

In other words, Hidden Markov Models (HMMs) assume:

• The current POS tag depends only on a limited number of previous tags (usually one in bigram HMMs, two in trigram HMMs).
• They cannot look far ahead or far behind in a sentence.

Example:

“The book you gave me yesterday was interesting.”

To correctly tag the word “was”, the model benefits from understanding long-distance syntactic relationships. However, HMMs cannot effectively capture such long-range dependencies.

Why neural models perform better

Modern neural POS taggers such as BiLSTM, Transformer, and BERT:

• Capture long-range dependencies across the sentence
• Use bidirectional context (both left and right)
• Learn character-level and subword features
• Handle unknown and rare words more effectively

Hidden Markov Model (HMM) – MCQs, Notes & Practice Questions | ExploreDatabase

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Key Concepts Illustrated in the Figure

1. Visible states (Observations)
   Visible states are the observed outputs of an HMM, such as words in a sentence. In the above figure, 'cat', 'purrs', etc are observations.
2. Hidden states
   Hidden states are the unobserved underlying states (e.g., POS tags - 'DT', 'N', etc in the figure) that generate the visible observations.
3. Transition probabilities
   Transition probabilities define the likelihood of moving from one hidden state to another. In the figure, this is represented by the arrows from one POS tag to the other. Example: P(N -> V) or P(V | N).
4. Emission probabilities
   Emission probabilities define the likelihood of a visible observation being generated by a hidden state. In the figure, this is represented by the arrows from POS tags to words. Example: P(cat | N).
5. POS tagging using HMM
   POS tagging using HMM models tags as hidden states and words as observations to find the most probable tag sequence.
6. Evaluation problem
   The evaluation problem computes the probability of an observation sequence given an HMM.
7. Forward algorithm
   The forward algorithm efficiently solves the evaluation problem using dynamic programming.
8. Decoding problem
   The decoding problem finds the most probable hidden state sequence for a given observation sequence.

1. In POS tagging using HMM, the hidden states represent:

A. Words in the sentence
B. POS tags assigned to each word
C. Dependency relations
D. Lemmatized tokens

Correct Answer: B

In HMM-based POS tagging, tags are hidden states and words are observed symbols.

2. The most suitable algorithm for decoding the best POS sequence in HMM tagging is:

A. Forward algorithm
B. Backward algorithm
C. Baum–Welch algorithm
D. Viterbi algorithm

Correct Answer: D

Viterbi decoding finds the most probable hidden tag sequence.

3. Transition probabilities in HMM POS tagging define:

A. P(word | tag)
B. P(tag_t | tag_t−1)
C. P(sentence | corpus)
D. P(tag | word frequency)

Correct Answer: B

Transition probability models tag-to-tag dependency. That is, the probability of a tag t given another tag t-1 which is previous tag. It is calculated using Maximum Likelihood Estimation (MLE) as follows;

Maximum Likelihood Estimation (MLE)

When the state sequence is known (for example, in POS tagging with labeled training data), the transition probability is estimated using Maximum Likelihood Estimation.

a_ij = Count(t_i → t_j) / Count(t_i)

Where:

• Count(t_i → t_j) is the number of times a POS tag t_i is immediately followed by a POS tag t_j in the training data.
• Count(t_i) is the total number of appearence of tag t_i in the entire training data.

This estimation ensures that the transition probabilities for each state sum to 1.

For example, the transition probability P(Noun | Det) will be 6/10 or 0.6 if in the training corpus the tag sequence "Det Noun" (Eg. like in "The/Det cat/Noun" - this is called tagged training data) occurs 6 times and the tag "Det" alone appears 10 times overall.

4. Emission probability in POS tagging refers to:

A. Probability of tag transition
B. Probability of sentence structure
C. Probability of word given a tag
D. Probability of next two tags jointly

Correct Answer: C

Emission probability is P(word | tag).

It answer the question "Given a particular POS tag, how likely is it that this tag generates (emits) a specific word?"

Emission probability calculation: Out of the total number of times a tag appears in the training data (eg. NOUN), how many times it appears as the tag of a given word (eg. "cat/NOUN").

5. Which problem does Baum–Welch training solve in HMM POS tagging?

A. POS disambiguation
B. Optimal decoding
C. Learning parameters from unlabeled text
D. Removing rare tags

Correct Answer: C

Baum–Welch (EM) learns transition and emission probabilities without labeled data.

Baum–Welch Method

The Baum–Welch method is an algorithm used to train a Hidden Markov Model (HMM) when the true state (tag) sequence is unknown.

What does the Baum–Welch method do?

It estimates (learns) the transition and emission probabilities of an HMM from unlabeled data.

In Simple Terms

• You are given only observation sequences (e.g., words)
• You do not know the hidden state sequence (e.g., POS tags)
• Baum–Welch automatically learns the model parameters that best explain the data

The Baum–Welch method is used to train an HMM by estimating transition and emission probabilities from unlabeled observation sequences using EM. Baum-Welch is a special case of Expectation-Maximization (EM) algorithm.

6. If an HMM POS tagger has 50 tags and a 20,000-word vocabulary, the emission matrix size is:

A. 20,000 × 20,000
B. 50 × 20,000
C. 20,000 × 50
D. 50 × 50

Correct Answer: B

Rows correspond to tags and columns to words.

In an HMM POS tagger, the emission matrix represents:

P(word | tag)

So its dimensions are:

• Rows = number of tags
• Columns = vocabulary size

Given:

• Number of tags = 50
• Vocabulary size = 20,000

Emission matrix size:

50 × 20,000

7. A trigram HMM POS tagger models:

A. P(tag_t | tag_t−1)
B. P(tag_t | tag_t−1, tag_t−2)
C. P(word | tag)
D. P(word_t | word_t−1)

Correct Answer: B

Trigram models capture dependency on two previous tags.

Trigram Model

A trigram model assumes that the probability of a tag (or word) depends on the previous two tags.

P(t_i | t_i−1, t_i−2)

In POS tagging using an HMM:

• Transition probabilities are computed using trigrams of tags
• The model captures more context than unigram or bigram models

Example:

If the previous two tags are DT and NN, the probability of the next tag VB is:

P(VB | DT, NN)

Note: In practice, smoothing and backoff are used because many trigrams are unseen.

8. Data sparsity in emission probabilities mostly occurs due to:

A. Too many hidden states
B. Unseen words in test data
C. Transition collapse
D. Deterministic decoding

Correct Answer: B

Unseen words lead to zero emission probabilities without smoothing.

Data sparsity in emission probabilities means that many valid word–tag combinations were never seen during training, so their probabilities are zero or unreliable.

Data sparsity may occur due to one or more of the following;

• Natural language has a very large vocabulary.
• Training data is finite.
• New or rare words often appear during test time.

As a result, many words in the test data were never observed with any tag during training.

9. A common solution for unknown words in HMM POS tagging is:

A. Random assignment
B. Laplace or Good–Turing smoothing
C. Removing words
D. Ignoring emissions

Correct Answer: B

Smoothing assigns non-zero probabilities to unseen events.

Refer here for more information about Laplace smoothing.

10. POS tagging is considered a:

A. Generation task
B. Knowledge-graph task
C. Sequence labeling task
D. Alignment task

Correct Answer: C

Each token is labeled sequentially → classic sequence labeling.

POS Tagging as a Sequence Labeling Task

POS tagging is a sequence labeling task because the goal is to assign a label (POS tag) to each element in a sequence (words in a sentence) while considering their order and context.

What is Sequence Labeling?

In sequence labeling, we:

• Take an input sequence: w1, w2, …, wn
• Produce an output label sequence: t1, t2, …, tn

Each input item receives one corresponding label, and the labels are not independent of each other.

POS Tagging as Sequence Labeling

Input sequence → words in a sentence

The / cat / sleeps

Output sequence → POS tags

DT / NN / VBZ

Each word must receive exactly one POS tag, and the choice of tag depends on:

• The current word (emission probability)
• The neighboring tags (context / transition probability)

POS Tagging using HMM Solved exercises 02 – Multiple Choice Questions (MCQs) with Answers

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Important Definitions

Bigram Model

A bigram model is a probabilistic model that assumes each element (such as a POS tag) depends only on the immediately preceding element. In POS tagging, it is based on the first-order Markov assumption.

Mathematically:

P(t_i | t₁, …, t_i−1) ≈ P(t_i | t_i−1)

Transition Probability

A transition probability is the probability of one POS tag following another POS tag in a sequence.

Mathematically:

P(t_j | t_i)

It indicates how likely a tag t_j occurs after tag t_i.

Emission Probability

An emission probability is the probability that a given POS tag generates (emits) a specific word.

Mathematically:

P(w | t)

It represents how likely a word w is produced by a POS tag t.

11. Using the POS-tag HMM below, what is the most likely tag for the word "apple"?

Tags: Noun (N), Verb (V)
Emission Probabilities:

Word	P(word|N)	P(word|V)
eat	0.05	0.60
apple	0.70	0.10

Assume equal prior tag probability.

Noun (N) Verb (V) Both equal probability Cannot be inferred

Answer: A
Explanation: P(apple|N)=0.70 ≫ P(apple|V)=0.10. With equal priors, **Noun** wins.

12. In the following transition matrix, what is the probability of the POS sequence N → V → N?

From \ To	N	V
N	0.30	0.70
V	0.50	0.50

Initial probability P(N)=0.6

0.6 × 0.70 × 0.30 = 0.126 0.6 × 0.30 × 0.70 = 0.126 0.6 × 0.70 × 0.50 = 0.21 0.6 × 0.50 × 0.30 = 0.09

Answer: C
Explanation: P(N→V)=0.70, P(V→N)=0.50 ⇒ 0.6 × 0.70 × 0.50 = 0.21.

Given the transition probability matrix and as per the bigram model, the probability of the POS sequence N → V → N can be calculated by simply multiplying the bigram probabilities P(V | N) and P(N | V). (bigram probability is P(tag_j | tag_i)). The following are two valid calculations;

• Without START state: P(N → V → N) = P(V | N) X P(N | V) = 0.70 X 0.50 = 0.35
• With START state: P(N → V → N) = P(N | START) X P(V | N) X P(N | V) = 0.6 X 0.70 X 0.50 = 0.21

In this question, you are given bigram transition probabilities and initial (START) probability.

13. Using HMM emissions below, what is the most probable tag for "runs"?

Word	P(word|Verb)	P(word|Noun)
runs	0.65	0.20
dog	0.10	0.75

Prior: P(N)=0.4, P(V)=0.6

Verb Noun Both equal Need transition probability

Answer: A
Explanation: 0.65×0.6 = 0.39 ≫ 0.20×0.4 = 0.08 ⇒ Verb.

The question asks for the tag (Verb or Noun) that is more likely to have generated the word "runs".

How do we decide the tag?

In Hidden Markov Models (HMMs), for a single word, we compare:

P(tag) × P(word | tag)

Step-by-step calculation

1. Probability that runs is a Verb

P(V) × P(runs | V) = 0.6 × 0.65 = 0.39

2. Probability that runs is a Noun

P(N) × P(runs | N) = 0.4 × 0.20 = 0.08

Compare the values

Since 0.39 > 0.08, Verb is the more probable tag.

14. You observe the sequence "cats chase mice". Which tag is most probable for chase?

Word	Noun	Verb
cats	0.65	0.05
chase	0.10	0.80
mice	0.70	0.05

The table can be read as P(cats | Noun) = 0.65 and P(cats | Verb) = 0.05 and so on

Noun Verb Equally probable None

Answer: B
Explanation: P(chase|Verb)=0.80 ≫ P(chase|Noun)=0.10.

You have an observation sequence "cats chase mice" and need to determine which part-of-speech tag (Noun or Verb) is most probable for the word "chase".

The table provides emission probabilities — the probability of observing a particular word given a specific POS tag

For the word "chase", we look at its emission probabilities:

• P(chase | Noun) = 0.10 (probability of observing "chase" if it's a Noun)
• P(chase | Verb) = 0.80 (probability of observing "chase" if it's a Verb)

Since 0.80 > 0.10, the word "chase" is much more likely to be a Verb than a Noun.

15. Given transitions and equal priors, what is P(N→N→V)?

From \ To	N	V
N	0.55	0.45
V	0.30	0.70

P(N)=0.5

0.5 × 0.55 × 0.45 = 0.12375 0.5 × 0.45 × 0.55 = 0.12375 0.5 × 0.55 × 0.55 = 0.15125 0.5 × 0.45 × 0.45 = 0.10125

Answer: A
Explanation: 0.5 × 0.55 × 0.45 = 0.12375.

16. Which tag is more probable for "light" given context? Previous tag = ADJ

ADJ NOUN Insufficient data Both equal

Answer: C
Explanation: Insufficient data.

What is required to answer this type of question?

To find the most probable tag for a given word, even when the previous tag is known, you need:

P(t_i | t_i−1) and P(w_i | t_i)

• Transition probability → models tag sequence likelihood. (Example: P(NOUN | ADJ), P(ADJ | ADJ))
• Emission probability → models word–tag compatibility. (Example: P(light | NOUN), P(light | ADJ))

Both are required for a valid probabilistic decision in an HMM. But none of these probabilities are provided in the question.

17. In an HMM-based POS tagger, how is the tag of the first word in a sentence primarily determined?

Emission probability only Transition probability from the previous word Transition probability from the START state Random selection

Answer: C
Explanation: In an HMM, the best tag for a sentence start is mainly determined by the START → tag transition probability, and Determiners most frequently follow the START state in training data.
At the start of a sentence, there is no previous real tag. Instead, HMM uses a special START state. So for the first word, the tag is chosen mainly using: P(tag∣START) This probability is learned from training data by counting which tags most often begin sentences.

18. Compute likelihood of sequence DT → N → V given the initial P(DT) = 0.55.

From \ To	DT	N	V
DT	0.10	0.75	0.15
N	0.05	0.60	0.35
V	0.10	0.20	0.70

0.55×0.75×0.35 = 0.144 0.55×0.10×0.35 = 0.01925 0.55×0.75×0.15 = 0.0618 0.55 × 0.15 × 0.70 = 0.05775

Answer: A
Explanation: 0.55×0.75×0.35 = 0.144.

What does “likelihood of the sequence DT → N → V” mean?

It means: What is the probability that the HMM generates this tag sequence?

In an HMM, the likelihood of a tag sequence is computed by multiplying the relevant transition probabilities.

Formula to be used

P(DT → N → V) = P(DT | START) × P(N | DT) × P(V | N)

Substituting values

= 0.55 × 0.75 × 0.35 = = 0.144

19. Which tag is most probable for the word "book" given emissions P(book | Noun) = 0.45 and P(book | Verb) = 0.40? Transition probabilities for fixed previous tag VERB are as follows: P(N | V) = 0.55 and P(V | V) = 0.45.

Noun Verb Both equal Not enough data

Answer: A
Explanation: 0.55×0.45 = 0.2475 greater thatn 0.45×0.40 = 0.18 ⇒ Noun.

What does HMM compute?

For each possible tag t, an HMM compares:

P(t | previous tag) × P(word | t)

Step 1: Probability of Noun

P(N) = P(V → N) × P(book | N) = 0.55 × 0.45 = 0.2475

Step 2: Probability of Verb

P(V) = P(V → V) × P(book | V) = 0.45 × 0.40 = 0.18

Since probability of NOUN is greater than that of VERB, NOUN is probable.

20. If emission probability for unknown words is smoothed using Laplace smoothing, what happens?

Unknown word gets zero probability Unknown word receives a small non-zero probability Transition matrix removed HMM becomes rule-based

Answer: B
Explanation: Laplace smoothing prevents zero probabilities.

What is Laplace smoothing?

Laplace smoothing (add-one smoothing) is used to handle unseen events in probabilistic models.

In POS tagging with HMMs, an unknown word is a word that did not appear in the training data.

Without smoothing:

• P(unknown word | tag) = 0
• This would make the entire sequence probability zero, even if all other probabilities are high.

What does Laplace smoothing do?

Laplace smoothing adds 1 to all word counts:

As a result:

• Even unseen words receive a small, non-zero probability
• No emission probability becomes zero

Therefore, the correct answer is: Option B

Top 10 Advanced HMM for POS Tagging — Important MCQs (2nd Order, Trigram, Smoothing, Viterbi)

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Advanced POS Tagging with Hidden Markov Models — MCQ Introduction

Master POS tagging with this focused MCQ set on Hidden Markov Models (HMM): second-order & trigram models, Viterbi decoding, Baum-Welch training, smoothing techniques, and practical tips for supervised & unsupervised learning.

Part-of-speech (POS) tagging is a cornerstone task in Natural Language Processing (NLP) that assigns grammatical categories (noun, verb, adjective, etc.) to each token in a sentence. This question set concentrates on classical statistical taggers built with Hidden Markov Models (HMMs), which model tag sequences as hidden states and observed words as emissions.

HMM-based POS taggers remain valuable for their interpretability and efficiency. They are particularly useful when you need:

• Lightweight, fast taggers for resource-constrained systems
• Explainable probabilistic models for linguistic analysis and teaching
• Strong baselines before moving to neural models like BiLSTM or Transformer taggers

This MCQ collection targets advanced HMM concepts — including second-order (trigram) models, the Viterbi decoding algorithm, Forward–Backward / Baum–Welch for unsupervised learning, and various smoothing strategies to handle rare or unseen words. Each question includes a concise explanation to help you understand not only the correct choice but why it matters for real-world POS tagging.

What you’ll learn

1. How higher-order HMMs (trigram / second-order) capture broader tag context.
2. Why supervised training requires labeled word–tag corpora and how unsupervised EM works.
3. The purpose of smoothing (Laplace, Good-Turing) to avoid zero probabilities.
4. Trade-offs: model complexity, overfitting, and inference cost when increasing hidden states.

Use these MCQs to prepare for exams, interviews, or to evaluate your grounding before progressing to neural POS taggers. Scroll down to start the questions and test your understanding of HMM-based POS tagging fundamentals and advanced techniques.

11. A second-order POS HMM considers:

One previous tag Two previous tags No previous tag All future tags

Answer: B
Explanation:

Higher-order HMM models allow P(tᵢ | tᵢ₋₁, tᵢ₋₂) improving context.

A second-order POS HMM (Hidden Markov Model) is also called a trigram HMM.

It means: P(tᵢ | tᵢ₋₁, tᵢ₋₂).

This tells us: The probability of the current tag depends on two previous tags. Therefore, the HMM "looks back" two steps in the tag sequence.

Example: Let us take a simple sentence "dog chase cats". For the last word “cats”, whose tag is t₃, and given the previous two tags: t₂ = VERB (tag for “chase”) t₁ = NOUN (tag for “dogs”), you should write:

𝑃(𝑡_cats ∣ 𝑡_chase = VERB, 𝑡_dogs = NOUN)

12. Input for supervised POS HMM training must contain:

Raw sentences only Word-tag annotated sentences Part-of-speech dictionary Dependency trees

Answer: B
Explanation:

Supervised models need labeled corpora to learn emission + transition probabilities.

Supervised POS HMM — Why labeled data is required?

A supervised POS HMM needs already labeled data so it can learn probabilities:

Transition probabilities

P(t_k | t_k-1) or P(t_k | t_k-1, t_k-2)

These require tag sequences.

Emission probabilities

P(w_k | t_k)

These require each word paired with its correct tag.

Therefore, supervised training must have sentences where every word already has a POS tag.

Example training line:

Dogs/NOUN chase/VERB cats/NOUN

13. Decoding in POS HMM refers to:

Parameter estimation Tokenizing words Selecting most likely tag sequence Expanding vocabulary

Answer: C
Explanation:

Decoding maps observed words to best hidden tag sequence using Viterbi.

In a POS HMM (Hidden Markov Model), decoding means: Finding the most probable sequence of POS tags for a given sequence of words. This is usually done using the Viterbi algorithm.

So, decoding = tagging = choosing the best tag path.

14. Forward-Backward algorithm is mainly used for:

Viterbi decoding Unsupervised HMM learning POS dictionary building Tokenization

Answer: B
Explanation:

It computes expected probabilities used in Baum-Welch EM training.

More information:

The Forward–Backward algorithm is the core of the Baum–Welch algorithm, which is used for: Unsupervised training of Hidden Markov Models (HMMs).

In unsupervised learning, the data has no POS tags, so the model must estimate: Transition probabilities, Emission probabilities

The Forward–Backward algorithm computes:

• Forward probabilities α
• Backward probabilities β
• Expected counts for transitions and emissions

These expected counts are then used to re-estimate HMM parameters. This is EM (Expectation–Maximization).

15. Smoothing in HMM prevents:

Overtraining Hidden state ambiguity Viterbi path errors Zero-probability transitions

Answer: D
Explanation:

Unseen tag-word or tag-tag pairs must not be zero → smoothing distributes probability.

In an HMM, probabilities are estimated from counts in the training data. If a transition or a word–tag pair never appears in training data, its probability becomes zero. This is dangerous because:

• A zero probability wipes out entire Viterbi paths
• The model cannot handle unseen words or unseen tag transitions

Smoothing (like Laplace, Good–Turing, Witten–Bell) adds a small nonzero probability to unseen events.

So smoothing prevents: Zero-probability transitions and zero-probability emissions.

16. A core limitation of POS HMM is that it:

Cannot tag new words Assumes Markov + word independence Requires deep networks Needs semantic embeddings

Answer: B
Explanation:

HMM relies only on previous tag and assumes words depend only on tag, limiting context.

Markov + word independence - A limiation in HMM

A standard POS HMM makes two strong assumptions.

• Markov Assumption (for tags): The current tag depends only on a small number of previous tags.
  • This ignores long-range syntactic dependencies (e.g., subject–verb agreement across clauses).
• Output Independence Assumption (for words): Words depend only on their own tag, not surrounding words.
  • This ignores context that modern taggers use (e.g., CRFs, BiLSTMs, Transformers).

These assumptions simplify the model, but they also severely limit accuracy compared to modern NLP models.

17. Increasing hidden states in POS HMM generally:

May cause overfitting Guarantees higher accuracy Does nothing to model quality Reduces computation

Answer: A
Explanation:

More states = more parameters → risk of overfitting & slower inference.

Increasing hidden states in POS HMM may cause overfitting

In a Hidden Markov Model (HMM) used for Part-of-Speech (POS) tagging, the "hidden states" correspond to the POS tags (like Noun, Verb, Adjective). Increasing the number of hidden states means using a more granular tagset (e.g., splitting "Noun" into "Singular Noun" and "Plural Noun") or simply increasing the model's capacity in an unsupervised setting.

Effect of increasing hidden states - Discussion

When you increase the number of states N:

• You must estimate many more parameters.
• But your dataset size stays the same.

So the model tries to estimate:

• Many more transition probabilities (N²),
• Many more emission probabilities (N × V).

With limited data, the HMM begins to:

• Fit the quirks/noise of the training data,
• Memorize rare patterns,
• Over-specialize to word sequences it has seen,
• Lose its ability to generalize to unseen text.

This phenomenon is overfitting.

18. Rare words are best handled using:

Laplace / Good-Turing smoothing Discarding them Forcing one tag Ignoring in training

Answer: A
Explanation:

Smoothing reallocates probability mass → better tagging for unseen/low-freq words.

19. A trigram HMM improves tagging by modeling:

No transition Two historical tags One future tag Word similarity

Answer: B
Explanation:

Trigram uses P(tᵢ | tᵢ₋₁, tᵢ₋₂) → better captures context patterns.

20. Unsupervised POS HMM accuracy increases with:

Random initialization Morphological features + smoothing Deleting rare words Using only transitions

Answer: B
Explanation:

Morphology aids tagging without labels → suffix, prefix, capitalization rules.