Amathematics competition uses the following scoring procedure to discourage students from guessing (choosing an answer randomly) on the multiple-choice questions. for each correct response, the score is 7. for each question left unanswered, the score is 2. for each incorrect response, the score is 0. if there are 5 choices for each question, what is the minimum number of choices that the student must eliminate before it is advantageous to guess among the rest?

For each correct answer, you have 7 points, for an unanswered you have 2 and for an incorrect you have 0.

to be advantageous to guess, the expected value must be greater than 2, so if we have 5 choices, we have 4 incorrect ones, and one correct, and the expected value is:

(4*0 + 1*7)/5 = 1.4

if we eliminate one choice, we have:

(3*0 + 1*7)/4 = 1.75

if we eliminate two choices, we have:

(2*0 + 1*7)/3 = 2.33

So if you eliminate two choices, the expected value is almost the same that if you do not answer the question, so if you do not know the correct answer, is almost the same to answer than not. Now assuming that the points must be whole numbers, we will need to remove another choice and get an expected value.

(1*0 + 1*7)/2 = 3.5

At this point, is better try to answer than not answer

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