Most exams these days have multiple choice questions. When I was preparing for one such exam- EAMCET, one of our math lecturer had strongly recommended one technique- to verify if the equation/expression is satisfied by the options, instead of solving the equations. Though it is not in the best interest of the students, at least that involved a little bit of solving. But what if a student just randomly answers all questions?
Considering that these exams typically have more than a hundred questions, and that tens of thousands of students take these exams, it is safe to assume that when a typical student randomly answers a question having 'n' options, the probability that the answer will be correct is 1/n. In questions with 4 options, if a student randomly marks an option, the probability that it will be correct is 25%.
Now, consider the case where each correct answer is awarded 4 marks, and each wrong answer is awarded -1 marks. If a student randomly marks 1 out of the 4 options, the score he will gain is (4)x (1/4) + (-1) x (3/4). 1/4 is the probability that he is correct, 3/4 is the probability that he is wrong. So, this student will, on average, score 1/4 marks out of a maximum possible 4 marks. i.e, he scores 1/16 of the maximum.
Suppose the student puts in a little effort and somehow confidently eliminates 1 out of the 4 options, and then randomly marks 1 out of the remaining 3. Now, the probability of him being right is 1/3, and being wrong is 2/3. In this case, he scores 4x1/3 + (-1) x 2/3 = 2/3, out of a maximum of 4. So, he scores 1/6 of the total marks (in the previous case, he scored only 1/16).
If he confidently eliminates 2 out of the 4 options, he scores 4x1/2 + (-1) x1/2 = 1.5 out of 4, or 3/8 of the total marks.
If he confidently eliminates 3 out of the 4 options, he scores 4 out of 4 marks.
Comparing different other cases, like no penalty for wrong answers, 6 options, etc:
I suppose that most exams have either no penalty, or a +4,-1 system with 4 options. Having no penalty is a huge incentive for those who can't solve all the problems to randomly mark answers. They have nothing to lose. Considering that even the best of students often fail to score a 100% owing to various reasons such as time constraints or exam pressure, and that the cut-off scores (especially for reserved seats) are usually quite low, the current system is really not fair. If the boards decide to go for 6 options instead of 4 options, the equation drastically changes. If they move to 8 options and also impose penalty for wrong answers, no one will ever try to randomly mark answers. Also, students will be discouraged to verify the options by substituting into equations- simply because there are too many options to try and check.
It really surprises me that the IITs are really not considering something like this. Increasing the number of options is not a difficult thing. It takes a little bit of extra effort, but it would definitely limit the role of luck in the exams.
Considering that these exams typically have more than a hundred questions, and that tens of thousands of students take these exams, it is safe to assume that when a typical student randomly answers a question having 'n' options, the probability that the answer will be correct is 1/n. In questions with 4 options, if a student randomly marks an option, the probability that it will be correct is 25%.
Now, consider the case where each correct answer is awarded 4 marks, and each wrong answer is awarded -1 marks. If a student randomly marks 1 out of the 4 options, the score he will gain is (4)x (1/4) + (-1) x (3/4). 1/4 is the probability that he is correct, 3/4 is the probability that he is wrong. So, this student will, on average, score 1/4 marks out of a maximum possible 4 marks. i.e, he scores 1/16 of the maximum.
Suppose the student puts in a little effort and somehow confidently eliminates 1 out of the 4 options, and then randomly marks 1 out of the remaining 3. Now, the probability of him being right is 1/3, and being wrong is 2/3. In this case, he scores 4x1/3 + (-1) x 2/3 = 2/3, out of a maximum of 4. So, he scores 1/6 of the total marks (in the previous case, he scored only 1/16).
If he confidently eliminates 2 out of the 4 options, he scores 4x1/2 + (-1) x1/2 = 1.5 out of 4, or 3/8 of the total marks.
If he confidently eliminates 3 out of the 4 options, he scores 4 out of 4 marks.
Comparing different other cases, like no penalty for wrong answers, 6 options, etc:
|
|
4 options
+4, -0
|
4 options
+4,-1
|
4 options
+4,-2
|
4 options
+4,-3
|
6 options
+4,-1
|
8 options
+4,-1
|
|
Random
|
25%
|
7%
|
-12.5%
|
-31%
|
-4.2%
|
-9.4%
|
|
Eliminate 1
|
33%
|
17%
|
0%
|
-66%
|
0%
|
-7.1%
|
|
Eliminate 2
|
50%
|
38%
|
25%
|
12.5%
|
6.25%
|
-4.2%
|
|
Eliminate 3
|
100%
|
100%
|
100%
|
100%
|
17%
|
0%
|
I suppose that most exams have either no penalty, or a +4,-1 system with 4 options. Having no penalty is a huge incentive for those who can't solve all the problems to randomly mark answers. They have nothing to lose. Considering that even the best of students often fail to score a 100% owing to various reasons such as time constraints or exam pressure, and that the cut-off scores (especially for reserved seats) are usually quite low, the current system is really not fair. If the boards decide to go for 6 options instead of 4 options, the equation drastically changes. If they move to 8 options and also impose penalty for wrong answers, no one will ever try to randomly mark answers. Also, students will be discouraged to verify the options by substituting into equations- simply because there are too many options to try and check.
It really surprises me that the IITs are really not considering something like this. Increasing the number of options is not a difficult thing. It takes a little bit of extra effort, but it would definitely limit the role of luck in the exams.
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