Negative Marking in the Last Ten Minutes
Ten minutes remain in a 200-question MCA CET mock test. The marking scheme is +1 for a correct answer, −0.25 for a wrong answer, and 0 for an unattempted question. The candidate has attempted 150 questions and is confident that 118 of them are correct. Twelve questions are still untouched. Of these, she is certain about 4 and has no clue about the remaining 8, each of which has four options with exactly one correct answer. Her last mock score of 128 corresponded to the percentile she needs. She can comfortably read and answer about one question every 45 seconds, so all 12 are within reach. A friend advises her to leave the 8 unknown questions untouched because 'negative marking will cancel out the gains', while another advises her to attempt everything. She wants to decide on expected value rather than instinct, and she also wants to know what would change if the penalty were −1 instead of −0.25.
1. Compute the candidate's current score, then decide on expected-value grounds whether she should attempt the 8 unknown questions. State the highest final score she can possibly reach.
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Step 1 - current score. Attempted = 150, correct = 118, so wrong = 150 − 118 = 32. Score = 118 − 0.25 × 32 = 118 − 8 = 110.
Step 2 - the four certain questions. 4 × 1 = +4, giving 114 before any guessing.
Step 3 - expected value of one random guess (4 options). E = 1 × (1/4) + (−0.25) × (3/4) = 0.25 − 0.1875 = +0.0625.
Step 4 - all eight guesses. 8 × 0.0625 = +0.5, so expected final score = 114.5, against 114 for leaving them blank. Attempting is better in expectation, though the gain is small.
Step 5 - ceiling. Best case: 110 + 4 + 8 = 122. Since 122 < 128, the target is arithmetically unreachable no matter what she does. The rational goal therefore shifts from 'hit 128' to 'maximise the score', which still means attempting all 8.
Judgement note: the friend's claim that negative marking 'cancels out' the gains is false here because 0.25 < 1/3; the penalty is milder than fair odds. The claim would only be right if the penalty were 1/3 or harsher.
2. At what accuracy does a four-option guess break even under the −0.25 penalty? How does the recommendation change if the penalty is −1?
Read the worked answer
General break-even. Let p be the probability of answering correctly. Expected gain per question = p × 1 − (1 − p) × c, where c is the penalty.
Case c = 0.25: p − 0.25(1 − p) = 0 → 1.25p = 0.25 → p = 0.20. Blind guessing gives p = 0.25 > 0.20, so each guess is worth +0.0625. Attempt all 8: expected 114.5 instead of 114.
Case c = 1: p − (1 − p) = 0 → 2p = 1 → p = 0.50. Blind guessing at p = 0.25 gives E = 0.25 − 0.75 = −0.5 per question, i.e. −4 across 8 questions, expected score 110 instead of 114. Here she should leave all 8 blank unless she can eliminate at least two options and raise p above 0.5.
Judgement point. Expected value is the right default, but it is not the only criterion. If a hard cutoff is already unreachable (maximum 122 < 128), extra variance has no downside for qualification, so guessing costs nothing strategically. If the cutoff were reachable only through guessing, the variance of a 0.5-point expected gain across 8 questions would still make guessing a lottery, not a plan - the honest answer is that the 114-point base, not the guesses, is what needs fixing.