Question

Maria is taking a multiple-choice test with 5 options for each question, and 3 marks for each correct answer. The test is negatively marked, which means she is deducted points for each answer she gets wrong. She will not be deducted any marks if she does not attempt the question. Maria is trying to figure out if it is beneficial to guess the answer to a question once she has narrowed down the number of potentially correct answers. Assume that once Maria has narrowed down the questions, all remaining answers are equally likely to be correct from Maria’s point of view. A guess is worthwhile if it is a better option than not attempting the question.

On the first question, Maria will lose 1 mark if she is incorrect. What is the minimum number of answers Maria will need to eliminate for a guess to be worthwhile

Answer 1

Maria must eliminate at least 2 of the incorrect options for a guess to have a positive expected value, making the guess statistically advantageous over not attempting the question.

Step-by-step explanation:

The question involves determining whether it is statistically beneficial for Maria to guess an answer on a multiple-choice test after she has narrowed down the possible options. Since Maria gains 3 marks for a correct answer but loses 1 mark for an incorrect one, we need to calculate the expected value of making a guess to find out the minimum number of answers she needs to eliminate for a guess to be worthwhile.

Let's assume Maria has narrowed down to n possible correct answers. The probability of guessing correctly is 1/n and the probability of guessing incorrectly is (n-1)/n. The expected value (EV) of a guess can be calculated by multiplying the probabilities by their respective outcomes: EV = (1/n)(+3 points) + ((n-1)/n)(-1 point).

For the guess to be worthwhile, the EV needs to be greater than 0. Simplifying the equation: 3/n - (n-1)/n > 0, this yields 3 - (n-1) > 0, which simplifies to n < 4. Therefore, Maria needs to eliminate at least 2 of the incorrect answers for a guess to have a positive expected value and hence be worthwhile considering.

Answer 2

Maria should eliminate at least 2 options to make guessing worthwhile on a negatively marked multiple-choice test. By removing 2 out of 5 options, the probability of guessing correctly (1/n) outweighs the penalty for a wrong guess, leading to a positive expected value.

Step-by-step explanation:

The question deals with a concept in probability where Maria needs to decide whether guessing on a multiple-choice question is beneficial after eliminating some incorrect options. We can calculate the expected value (EV) of guessing versus not attempting to determine the minimum number of options Maria needs to eliminate for guessing to be worthwhile. For guessing to be beneficial, the EV must be greater than zero.

Let's say 'n' is the number of options Maria thinks could be correct. If Maria guesses and answers correctly, she earns 3 marks. If she guesses incorrectly, she loses 1 mark. The probability of guessing correctly is 1/n and the probability of being wrong is (n-1)/n.

The expected value of guessing (EV) is:
EV = (3 × (1/n)) - (1 × ((n-1)/n)).
For guessing to be worthwhile, EV needs to be > 0.
We solve for 'n' to find that she must eliminate at least 2 options for the guess to have a positive expected value.