Question

Consider that 20 stories of Author X and 10 stories of Author Y were kept together without mentioning the names of the authors. A classifier was then asked to predict the author (X or Y) of each of these stories. Let, out of X's stories, 6 were classified as that of Y. On the other hand, out of Y's stories, 2 were classified as that of X.

Considering X and Y as two classes, which of the following statements is/are true?

• A. Accuracy of the classifier is 11/15.
• B. Precision of Class X is higher than the Precision of Class Y.
• C. Recall of Class X is higher than the Recall of Class Y.
• D. Accuracy of the classifier is 14/15.

Hint:

Build the 2x2 confusion matrix from the story counts first (14, 6, 2, 8), then compute accuracy, precision per class, and recall per class from it.

Answer 1

The Correct Option is A, B

Step 1: Build the confusion matrix from the given numbers.
There are 20 stories by X and 10 by Y, 30 in total. Out of X's 20 stories, 6 are wrongly called Y, so \(20-6=14\) are correctly called X. Out of Y's 10 stories, 2 are wrongly called X, so \(10-2=8\) are correctly called Y.

	Predicted X	Predicted Y
Actual X (20 stories)	14	6
Actual Y (10 stories)	2	8

Step 2: Compute the overall accuracy.
Accuracy is the fraction of all stories correctly classified:
\[ \text{Accuracy} = \frac{14 + 8}{30} = \frac{22}{30} = \frac{11}{15} \]
This matches statement (A), so (A) is true, and it also rules out statement (D), which claims 14/15.

Step 3: Compute Precision for class X and class Y.
Precision of a class is (correctly predicted as that class) divided by (all stories predicted as that class).
For X: predicted X totals \(14 + 2 = 16\) (the 14 correct plus the 2 Y stories wrongly called X), so
\[ \text{Precision}_X = \frac{14}{16} = 0.875 \]
For Y: predicted Y totals \(8 + 6 = 14\) (the 8 correct plus the 6 X stories wrongly called Y), so
\[ \text{Precision}_Y = \frac{8}{14} \approx 0.571 \]
Since \(0.875 > 0.571\), Precision of X is indeed higher than Precision of Y, so statement (B) is true.

Step 4: Compute Recall for class X and class Y.
Recall of a class is (correctly predicted as that class) divided by (all stories that actually belong to that class).
\[ \text{Recall}_X = \frac{14}{20} = 0.7, \qquad \text{Recall}_Y = \frac{8}{10} = 0.8 \]
Here Recall of X (0.7) is lower than Recall of Y (0.8), not higher, so statement (C) is false.

Final Answer:
Statements (A) and (B) are true; (C) and (D) are false.\[ \boxed{\text{(A) and (B)}} \]