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

Original solution basis:
p+ = 0.4, p- = 0.6. Gini = 1 - (0.$4^2 + 0$.$6^2)$ = 1 - (0.16 + 0.36) = 1 - 0.52 = 0.48.

Deep Dive
The class probabilities are 0.4 and 0.6. Gini=1-(0.4²+0.6²)=1-(0.16+0.36)=0.48. The value is zero for a pure node and reaches 0.5 for a balanced binary node.

Key Points to Remember:
• Gini=1-$\Sigma pᵢ²$.
• Use proportions, not raw counts, in the formula.
• For a binary node, Gini is at most 0.5.

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

Original solution basis:
Precision = 40 / (40 + 10) = 0.8. Recall = 40 / (40 + 10) = 0.8. F1 = 2 * 0.8 * 0.8 / (0.8 + 0.8) = 0.8.

Deep Dive
Precision=TP/(TP+FP)=40/50=0.8. Recall=TP/(TP+FN)=40/50=0.8. Therefore F1=2(0.8)(0.8)/(0.8+0.8)=0.8. The equal precision and recall make the harmonic mean equal to the same value.

Key Points to Remember:
• Precision=0.8 and recall=0.8.
• F1=2PR/(P+R).
• The given confusion matrix contains 50 actual positives and 50 predicted positives.

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

Original solution basis:
F1 = 2 * (0.75 * 0.50) / (0.75 + 0.50) = 2 * 0.375 / 1.25 = 0.75 / 1.25 = 0.6.

Deep Dive
Substitute P=0.75 and R=0.50 into F1=2PR/(P+R). The numerator is 2$\times 0.75\times 0.50=0.75$ and the denominator is 1.25. Thus F1=0.75/1.25=0.60.

Key Points to Remember:
• Use the harmonic-mean formula, not the arithmetic mean.
• F1=2PR/(P+R).
• The result must lie between precision and recall; here it is 0.60.

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

Original solution basis:
P(C3, X) = P(C3) * P(X|C3) = 0.3 * 0.4 = 0.12.

Deep Dive
The joint probability is prior multiplied by likelihood: P(C₃,X)=P(C₃)P(X|C₃)=0.3$\times 0.4=0.12$. This is an unnormalized class score; it is not yet the posterior probability because the evidence P(X) has not been used as a denominator.

Key Points to Remember:
• Joint = prior $\times likelihood$.
• P(C₃,X)=0.3$\times 0.4=0.12$.
• Normalize across all classes to obtain a posterior probability.

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

Original solution basis:
Joints: P(C1,X)=0.16, P(C2,X)=0.10, P(C3,X)=0.12. Sum = 0.38. Posterior P(C3|X) = 0.12 / 0.38 = 0.3157 ~ 0.32.

Deep Dive
Compute each class joint: C₁ gives 0.2$\times 0.8=0.16; C₂ gives 0.5\times 0.2=0.10; C₃ gives 0.3\times 0.4=0.12$. The evidence is their sum, 0.16+0.10+0.12=0.38. Hence P(C₃|X)=0.12/0.38$\approx 0.3158$, which rounds to 0.32.

Key Points to Remember:
• Posterior = joint for the class / sum of all class joints.
• The normalizing evidence is 0.38.
• Round only after completing the division: 0.3158… → 0.32.

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