Karnataka Teacher Eligibility Test (TET)
Foundation & Exam Strategy
Topic: Marking Scheme
The Karnataka TET is a multiple‑choice exam administered by the Karnataka State Council of Educational Research and Training (SCERT).
This section focuses on the marking scheme that is used to compute the final score, the underlying probability theory, and how to devise optimal test‑taking strategies.
1. First Principles of the Marking Scheme
| Item | Marks Awarded | Marks Deductions | Notes |
|---|---|---|---|
| Correct answer | +4 | – | Full credit. |
| Uncertain / Wrong answer | 0 | –1 | Negative marking to discourage random guessing. |
| Unanswered | 0 | – | No penalty. |
Total marks: 4 × (number of correct answers).
Maximum possible score: 4 × N (where N is the number of questions).
Minimum possible score: 0 (if all answers are blank).
2. Deeper Theoretical Analysis
2.1 Expected Value of Random Guessing
Assume each question has four options (A, B, C, D) and the candidate guesses uniformly at random.
- Probability of a correct guess: ( p = \frac{1}{4} )
- Probability of a wrong guess: ( q = \frac{3}{4} )
Expected score per guessed question:
[ E[\text{score per question}] = (+4) \cdot p + (-1) \cdot q = 4 \cdot \frac{1}{4} - 1 \cdot \frac{3}{4} = 1 - 0.75 = +0.25 ]
Interpretation: Random guessing yields a positive expected score of +0.25 marks per question.
Thus, random guessing is statistically beneficial relative to leaving the question blank (expected 0).
2.2 Variance and Risk
Variance of score per question:
[ \operatorname{Var} = (4)^2 \cdot p + (-1)^2 \cdot q - (E)^2 = 16 \cdot \frac{1}{4} + 1 \cdot \frac{3}{4} - (0.25)^2 = 4 + 0.75 - 0.0625 = 4.6875 ]
Standard deviation ≈ 2.167 marks per question.
Risk considerations:
- If you are highly confident (p ≈ 0.9), you should answer, because the expected reward is (4 \times 0.9 - 1 \times 0.1 = 3.5) marks.
- If you are uncertain (p ≈ 0.25), the expected reward is (4 \times 0.25 - 1 \times 0.75 = 0).
- If p < 0.25, you should skip the question to avoid negative expected value.
2.3 Optimal Guessing Strategy
For each question, compute the confidence level ( p ) (your subjective probability of being correct).
- If ( p > 0.25 ) → Answer.
- If ( p = 0.25 ) → Either answer or skip (expected score 0).
- If ( p < 0.25 ) → Skip.
In practice, you cannot compute exact p, so you adopt a threshold rule:
- High certainty (e.g., you can eliminate at least two options): answer.
- Moderate certainty (eliminate one option): answer.
- Low certainty (no elimination): skip.
3. Edge Cases & Practical Considerations
| Situation | Strategy | Why it works |
|---|---|---|
| Question with 3 options (rare) | Treat as 4‑option with one blank possibility | Same calculation: (p) reduces to (\frac{1}{3}); still above 0.25 → answer. |
| All questions answered | Might exceed maximum possible score (if you get all correct) | No penalty for extra marks; but you cannot get more than 4×N. |
| Tie in total marks | Use time taken or number of correct answers as tie‑breakers in internal rankings (if used by SCERT) | Not relevant for final score but important for internal ranking. |
| Negative total score | Not possible due to minimum 0; but if you answer many wrong you might get close to 0 | Avoid answering if highly uncertain. |
| Differential penalty | Some exam boards use –1.5 or –0.5; Karnataka TET uses –1 | Be aware of the exact penalty while calculating expected values. |
4. Advanced Analytical Tools
4.1 Bayesian Update for Confidence
If you have prior knowledge of a question’s difficulty, you can update your confidence (p):
[ p_{\text{posterior}} = \frac{p_{\text{prior}} \times \text{likelihood}{\text{correct}}} {p{\text{prior}}\times\text{likelihood}{\text{correct}} + (1-p{\text{prior}})\times\text{likelihood}_{\text{wrong}}} ]
Use this to refine your threshold strategy during the exam.
4.2 Marking Scheme as a Linear Function
Let (C) be the number of correct answers, (W) wrong answers, (U) unanswered.
[ \text{Score} = 4C - W ] subject to (C + W + U = N).
Because (U = N - C - W), you can express the score purely in terms of (C) and (W):
[ \text{Score} = 4C - W = 4C - (N - C - U) = 5C + U - N \