# Q4B · A second HMM sequence

Subject: Natural Language Processing

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/nlp/#tee/mocks/timed-reexam-2026/reexam-2026-q4b

Canonical: https://deep-learning-study-room.vercel.app/read/nlp/tee/mocks/timed-reexam-2026--reexam-2026-q4b/

TEE · actual previous paper · 100 marks · 180 minutes

Question. Consider the following POS tagged corpus. Apply Hidden Markov Model (HMM) approach to:  
1. Dog/Noun eats/Verb bread/Noun  
2. The/Det cat/Noun sleeps/Verb  
3. A/Det dog/Noun eats/Verb fish/Noun  
i. Construct the emission probability matrix.  
ii. Construct the transmission probability matrix.  
iii. Calculate the probability of the given sequence of events: The/Det dog/Noun eats/Verb bread/Noun.  
Note: Include start and end characters for all the sentences. [10 marks]

### Answer

The repeated verb eats has a different emission count from the final paper; recalculate it rather than copying the previous numerical result.

### Given

| Sentence | Tagged sequence |
| --- | --- |
| 1 | dog/Noun eats/Verb bread/Noun |
| 2 | the/Det cat/Noun sleeps/Verb |
| 3 | a/Det dog/Noun eats/Verb fish/Noun |

Lowercase the corpus. Test sequence: the/Det dog/Noun eats/Verb bread/Noun. Add Start and End markers.

### To find

Emission and transition matrices, then the probability of the supplied tagged sequence.

### Formula: which one and why



$$
P(w\mid t)=\frac{C(t,w)}{C(t)}
$$





$$
P(t_j\mid t_i)=\frac{C(t_i,t_j)}{\sum_kC(t_i,t_k)}
$$



Emission divides the word’s tag-specific count by the total tokens with that tag. Transition divides the tag-pair count by all transitions leaving the previous tag, including End.



$$
P(\mathbf w,\mathbf t)=P(t_1\mid\mathrm{Start})\prod_iP(w_i\mid t_i)\prod_{i=2}^{n}P(t_i\mid t_{i-1})P(\mathrm{End}\mid t_n)
$$



### Solution

### Step 1. Count tag totals and word frequencies

Det: the=1, a=1, total2. Noun: dog=2, bread=1, cat=1, fish=1, total5. Verb: eats=2, sleeps=1, total3.

### Step 2. Construct emission matrix

| Tag / word | a | bread | cat | dog | eats | fish | sleeps | the |
| --- | --- | --- | --- | --- | --- | --- | --- | --- |
| Det | 1/2 | 0 | 0 | 0 | 0 | 0 | 0 | 1/2 |
| Noun | 0 | 1/5 | 1/5 | 2/5 | 0 | 1/5 | 0 | 0 |
| Verb | 0 | 0 | 0 | 0 | 2/3 | 0 | 1/3 | 0 |

### Step 3. Count transitions from the three chains

Start → Noun → Verb → Noun → End  
Start → Det → Noun → Verb → End  
Start → Det → Noun → Verb → Noun → End

| From | Det count | Noun count | Verb count | End count | Total |
| --- | --- | --- | --- | --- | --- |
| Start | 2 | 1 | 0 | 0 | 3 |
| Det | 0 | 2 | 0 | 0 | 2 |
| Noun | 0 | 0 | 3 | 2 | 5 |
| Verb | 0 | 2 | 0 | 1 | 3 |

### Step 4. Divide outgoing counts

| From / to | Det | Noun | Verb | End |
| --- | --- | --- | --- | --- |
| Start | 2/3 | 1/3 | 0 | 0 |
| Det | 0 | 1 | 0 | 0 |
| Noun | 0 | 0 | 3/5 | 2/5 |
| Verb | 0 | 2/3 | 0 | 1/3 |

All nonterminal rows sum to one. No smoothing is requested.

### Step 5. Write every sequence factor

| Event | Probability |
| --- | --- |
| Start → Det | 2/3 |
| Det emits the | 1/2 |
| Det → Noun | 1 |
| Noun emits dog | 2/5 |
| Noun → Verb | 3/5 |
| Verb emits eats | 2/3 |
| Verb → Noun | 2/3 |
| Noun emits bread | 1/5 |
| Noun → End | 2/5 |

### Step 6. Multiply all factors



$$
P=\frac23\times\frac12\times1\times\frac25\times\frac35\times\frac23\times\frac23\times\frac15\times\frac25
$$



Start with subtotal 1. At each row, multiply the previous subtotal by exactly one next factor. Simplify the fraction after multiplication, but do not round it.

| Factor being included | Previous subtotal × next factor = new subtotal |
| --- | --- |
| 1. Start → Det | 

$$
1\times\frac{2}{3}=\frac{2}{3}
$$

 |
| 2. Det emits the | 

$$
\frac{2}{3}\times\frac{1}{2}=\frac{1}{3}
$$

 |
| 3. Det → Noun | 

$$
\frac{1}{3}\times1=\frac{1}{3}
$$

 |
| 4. Noun emits dog | 

$$
\frac{1}{3}\times\frac{2}{5}=\frac{2}{15}
$$

 |
| 5. Noun → Verb | 

$$
\frac{2}{15}\times\frac{3}{5}=\frac{2}{25}
$$

 |
| 6. Verb emits eats | 

$$
\frac{2}{25}\times\frac{2}{3}=\frac{4}{75}
$$

 |
| 7. Verb → Noun | 

$$
\frac{4}{75}\times\frac{2}{3}=\frac{8}{225}
$$

 |
| 8. Noun emits bread | 

$$
\frac{8}{225}\times\frac{1}{5}=\frac{8}{1125}
$$

 |
| 9. Noun → End | 

$$
\frac{8}{1125}\times\frac{2}{5}=\frac{16}{5625}
$$

 |

The last subtotal is the complete product. Only now convert the fraction to a decimal.



$$
P=\frac{16}{5625}\approx0.0028444444
$$



### Final answer

**P = 16/5625 ≈ 0.0028444444** for these words and specified tags jointly. The two occurrences of eats give P(eats | Verb) = 2/3.

**Yaad rakho:** Same tag structure ho sakta hai, lekin word emission counts dobara count karna.

**Boundary reminder:** Include P(Det | Start) = 2/3 and P(End | Noun) = 2/5. The handwritten key’s final line instead inserts the overall Det fraction 2/10 and omits End; the complete boundary-aware probability requested here is 0.0028444444.