GATE DA Supervised Learning – Data Science & AI | Set 1

Original solution basis:
H(S1) = 0 (pure). S2 has 1 pos, 5 neg (n=6): p+=1/6, p-=5/6 => H(S2) = -(1/6)log2(1/6) - (5/6)log2(5/6) = 0.65 bit. Weighted child entropy = (4/10)*0 + (6/10)*0.65 = 0.39. IG = 1.0 - 0.39 = 0.61.

Deep Dive
The parent entropy is 1 bit. The left child is pure (4 positive, 0 negative), so H(S₁)=0. The right child has proportions 1/6 and 5/6, giving H(S₂)=-(1/6)log₂(1/6)-(5/6)log₂(5/6)$\approx 0.650$. Weighted child entropy=(4/10)(0)+(6/10)(0.650)$\approx 0.390$. Information Gain=1.000-0.390$\approx 0.610 bit, so$ the answer is 0.61.

Key Points to Remember:
• Weight child entropy by child size: 4/10 and 6/10.
• The pure left child has entropy zero.
• IG = parent entropy - weighted child entropy $\approx 0.61 bit$.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
Margin width = 2 / ||w|| = 0.5 => ||w|| = 2 / 0.5 = 4.

Deep Dive
The distance between the two margin hyperplanes is 2/||w||. Given 0.5=2/||w||, multiply both sides by ||w|| and divide by 0.5: ||w||=2/0.5=4. This uses the canonical SVM margin planes wᵀx+b=$\pm 1$.

Key Points to Remember:
• SVM margin width = 2/||w||.
• Rearrange to ||w||=2/(margin width).
• For a margin width of 0.5, the norm is 4.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
P('crypto'|Spam) = (19 + 1) / (4000 + 1000) = 20 / 5000 = 0.004.

Deep Dive
Add-one smoothing gives P(word|Spam)=(count(word,Spam)+1)/(total word count in Spam+V). Substitute count=19, total=4000, and V=1000: (19+1)/(4000+1000)=20/5000=0.004. The vocabulary-size adjustment in the denominator accounts for adding one pseudo-count to every vocabulary term.

Key Points to Remember:
• Add-one multinomial NB estimate = (count+1)/(class token total+V).
• Use the total token count for the class, not the number of documents.
• The smoothed probability is 0.004.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
n = 22, p = 1. Adjusted R^2 = 1 - [(1 - 0.80)(22 - 1) / (22 - 1 - 1)] = 1 - [0.20 * 21 / 20] = 1 - [0.21] = 0.79.

Deep Dive
For n=22 and p=1, adjusted R²=1-[(1-0.80)(22-1)/(22-1-1)]. This is 1-[0.20$\times 21/20]=1-0.21=0.79$. The adjustment is slightly lower than the ordinary R² because it accounts for the predictor and sample-size degrees of freedom.

Key Points to Remember:
• Adjusted R² = 1-[(1-R²)(n-1)/(n-p-1)].
• Here p=1 excludes the intercept.
• Adjusted R² is 0.79, slightly below R²=0.80.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
Stratified K-Fold ensures that each fold contains approximately the same percentage of samples of each target class as the complete dataset.

Deep Dive
Stratification constructs each fold so that its class proportions approximately match those of the complete dataset. In an imbalanced dataset, this helps ensure that rare classes are represented in both training and validation partitions. This makes fold-wise evaluation more stable and comparable than an unstratified split may be.

Key Points to Remember:
• Stratified K-fold preserves approximate class proportions in each fold.
• It is particularly useful when minority classes have few examples.
• Stratification improves representation; it does not itself fix class imbalance in model training.

Exam tip:
Identify the governing definition or theorem first, write the formula, substitute the given values, and check dimensions/signs before selecting the option.