# More WSD numericals: given likelihoods, unequal priors and repeated clues

Subject: Natural Language Processing

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/nlp/#tee/formulas/f-wsd/u5-bayes-practice

Canonical: https://deep-learning-study-room.vercel.app/read/nlp/tee/formulas/f-wsd--u5-bayes-practice/

Formulas explained through full worked examples

**Question. For bass, let P(Fish)=0.6 and P(Music)=0.4. Given Fish likelihoods P(caught|Fish)=0.5, P(lake|Fish)=0.4 and Music likelihoods P(caught|Music)=0.05, P(lake|Music)=0.1, classify context {caught,lake} and normalize.**

#### Given

Priors and likelihoods are provided directly; no smoothing or training-count estimation is requested.

#### To find

The winning sense and its posterior.

#### Formula and why it applies



$$
\operatorname{score}(s)=P(s)P(\text{caught}\mid s)P(\text{lake}\mid s)
$$

, because the two features are treated as conditionally independent.

#### Step 1. Fish multiplication



$$
0.6\times0.5=0.3
$$

, then 

$$
0.3\times0.4=0.12
$$

.

#### Step 2. Music multiplication



$$
0.4\times0.05=0.02
$$

, then 

$$
0.02\times0.1=0.002
$$

.

#### Step 3. Total evidence



$$
0.12+0.002=0.122
$$

.

#### Step 4. Normalize



$$
P(Fish\mid C)=0.12/0.122\approx0.983607
$$





$$
P(Music\mid C)=0.002/0.122\approx0.016393.
$$



**Answer.** Fish, with model posterior approximately 0.983607.

**Question. Disambiguate bank from the repeated context tokens {loan, loan, river}. Priors are Finance=2/3 and River=1/3. Vocabulary V=4. Finance has N=8 total tokens, loan count 3 and river count 0; River has N=4, loan count 0 and river count 2. Apply add-one smoothing.**

#### Given

Multinomial counts are supplied. The two appearances of loan must both contribute.

#### To find

Scores and normalized Finance posterior.

#### Formula and why it applies

Naive Bayes selects the sense with the largest prior × likelihood product. The context evidence denominator is identical across candidate senses, so it can be omitted for comparison.



$$
\operatorname{score}(s)=P(s)\prod_j P(w_j\mid s).
$$



For add-one multinomial smoothing:



$$
P(w\mid s)=\frac{C(w,s)+1}{N_s+V}.
$$





$$
C(w,s)
$$

 counts occurrences of word 

$$
w
$$

 in training contexts of sense 

$$
s
$$

; 

$$
N_s
$$

 is the total number of word occurrences in those contexts; 

$$
V
$$

 is the shared vocabulary size. It is not the number of training sentences. Repeated query words contribute repeated factors.

#### Step 1. Finance denominator



$$
8+4=12
$$

. 

$$
P(loan\mid F)=(3+1)/12=1/3
$$

; 

$$
P(river\mid F)=(0+1)/12=1/12
$$

.

#### Step 2. River denominator



$$
4+4=8
$$

. 

$$
P(loan\mid R)=(0+1)/8=1/8
$$

; 

$$
P(river\mid R)=(2+1)/8=3/8
$$

.

#### Step 3. Finance product



$$
\operatorname{score}(F)=\frac23\times\frac13\times\frac13\times\frac1{12}=\frac1{162}\approx0.00617284.
$$



#### Step 4. River product



$$
\operatorname{score}(R)=\frac13\times\frac18\times\frac18\times\frac38=\frac1{512}\approx0.001953125.
$$



#### Step 5. Compare and normalize



$$
P(F\mid C)=\frac{1/162}{1/162+1/512}=\frac{512}{674}=\frac{256}{337}\approx0.759644.
$$



**Answer.** Finance wins. Repeated loan contributes a squared likelihood; do not reduce the query to a unique-word set.

### Answer-writing order in an exam

Write the class priors, vocabulary size, class token totals, a likelihood table, both score products and the final class. Keep fractions until the comparison where possible. If the question only asks classification, normalization is optional; if it asks posterior probabilities, normalization is required.

Repeated word aaye toh factor bhi repeat. Sirf word-set bana diya toh multinomial answer badal jayega.