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IIT M DEGREE AN EXAM QDB2 03 Aug 2025 - 03 Aug 2025

MLT Quiz 215 questions40 marks7 MCQ · 6 SA · 2 MSQafternoon / QDB2
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Q1MCQ4 marks
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Q2MCQ3 marks
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Q3MCQ3 marks
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  1. Setting 3, because it has the lowest training error, indicating the best fit to thetraining data.
  2. Setting 2, because it has the lowest validation error and a relatively lowtraining error, indicating a good generalization.
  3. Setting 1, because it has the highest number of iterations and therefore likelyto be more stable.
  4. Setting 4, because it uses the highest learning rate and converges the fastest.
Q4MCQ3 marks
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Q5MCQ2 marks
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  1. TRUE
  2. FALSE
Q6MCQ1 mark
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  1. TRUE
  2. FALSE
Q7MSQ2 marksselect all that apply
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Q8MSQ2 marksselect all that apply
Which among the following are true?
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Q9SA3 marks
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Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
Q10MCQ3 marks
Which among the following represents the correct decision boundary for the decision tree? Note: Here + represents the label 1 and − represents the label 0.
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Q11SA2 marks
What will be the prediction for the new datapoint (2, 2)?
Comprehension passage - the questions below it refer to this
Based on the above data, answer the given subquestions. question figure
Q12SA2 marks
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Q13SA3 marks
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Comprehension passage - the questions below it refer to this
Suppose you want to build a decision tree to help diagnose a certain disease based on patient data. You have a dataset of 1000 patients labeled as either “positive” or “negative”. Here, positive represents patients with the disease and negative represents no disease. One of the feature used to diagnose is headache. Below is the breakdown of the patients reporting headache: Headache present: 300 positive, 50 negative Headache not present: 100 positive, 550 negative question figure Suppose you want to build a decision tree to help diagnose a certain disease based on patient data. You have a dataset of 1000 patients labeled as either “positive” or “negative”. Here, positive represents patients with the disease and negative represents no disease. One of the feature used to diagnose is headache. Below is the breakdown of the patients reporting headache: Headache present: 300 positive, 50 negative Headache not present: 100 positive, 550 negative Based on the above data, answer the given subquestions.
Q14SA3 marks
What is the entropy of the root node? Enter the answer correct to two decimal places.
Q15SA4 marks
What is the information gain for choosing the feature “headache” for splitting at the root node? Enter the answer correct to two decimal places.

Answer key

  1. Q1 (MCQ, 4 marks): A
  2. Q2 (MCQ, 3 marks): C
  3. Q3 (MCQ, 3 marks): B
  4. Q4 (MCQ, 3 marks): C
  5. Q5 (MCQ, 2 marks): A
  6. Q6 (MCQ, 1 marks): A
  7. Q7 (MSQ, 2 marks): A, C
  8. Q8 (MSQ, 2 marks): A, C
  9. Q9 (SA, 3 marks): 4.2
  10. Q10 (MCQ, 3 marks): B
  11. Q11 (SA, 2 marks): 0
  12. Q12 (SA, 2 marks): 0.22 to 0.26
  13. Q13 (SA, 3 marks): 1
  14. Q14 (SA, 3 marks): 0.95 to 0.99
  15. Q15 (SA, 4 marks): 0.34 to 0.38

Concepts this paper tests

Auto-generated summary from the source site. Handy as a checklist; the lessons above explain each idea from scratch.

Summary of Core Topics, Concepts, Principles, and Equations

1. Maximum Likelihood Estimation (MLE)

w^MLE=arg⁡max⁡w∏i=1nP(yi∣xi,w)\hat{w}_{MLE} = \arg \max_{w} \prod_{i=1}^{n} P(y_i | x_i, w)

2. Maximum A Posteriori (MAP) Estimation

w^MAP=arg⁡max⁡wP(w∣y,x)=arg⁡max⁡wP(y∣x,w)P(w)\hat{w}_{MAP} = \arg \max_{w} P(w | y, x) = \arg \max_{w} P(y | x, w) P(w)

3. Linear Regression

y=wTx+ϵy = w^T x + \epsilon

where ϵ∼N(0,σ2)\epsilon \sim \mathcal{N}(0, \sigma^2).

4. Regularization

w^=arg⁡min⁡w{∥y−XTw∥2+λ∥w∥2}\hat{w} = \arg \min_{w} \left\{ \| y - X^T w \|^2 + \lambda \| w \|^2 \right\}

where λ\lambda is the regularization parameter.

5. Decision Trees

6. Naive Bayes Classifier

P(y∣x)∝P(y)∏i=1nP(xi∣y)P(y | x) \propto P(y) \prod_{i=1}^{n} P(x_i | y)

where P(xi∣y)P(x_i | y) is modeled as a Gaussian distribution.

7. Model Evaluation