Hi,

I just did a test which had two multiple choice questions. Each question was worth one point. Getting them both right would result in getting a 100% score. Suffice it to say, getting just one question right would give you 50% and with that a passing grade.

So you have two multiple choice questions. Both of which are unrelated to the other. Each question has four possible answers. When you finish the test. You get to have one more try. The questions and possible answers remain the same.

Let’s say you use both tries and you remember your previous two respected answers. What would your odds be, if you were to brute force guess your way through this test, to get a passing grade or a 100%?

Edit: Both questions only have one correct answer.

IMPORTANT EDIT: YOU DO NOT KNOW WHICH ANSWER YOU HAD RIGHT OR WRONG THE SECOND TIME AROUND. You only know how many questions you got right. But you don’t know which. Sorry for the confusion!

  • Nibodhika@lemmy.world
    11·
    2 years ago

    The simple version of the answer is: each question has a 1/4 chance of getting right, and since they’re independent and you can mark two answers you have 2/4 or 1/2 of getting each correct, which gives you a combined chance of 25% for the entire test. The correct analysis is the combination of chances of:

    1. First time you picked a wrong answer on both (3/4 * 3/4) and second time you eliminated one answer from each and picked the correct one (1/3 * 1/3): 6.25%

    2. First time you picked both right, so didn’t need the second time: 6.25%

    3. First time you picked the first one right, but the second one wrong (1/4 * 3/4) and second time you picked the correct one on the second one (1/3): 6.25%

    4. Same as above but for the second question: 6.25%

    Which is also 25% btw, the other analysis is also correct, it’s just an alternate problem with the same chances as this one.

    Edit: sorry, didn’t read the part about getting one question right would be a passing grade, so that’s easier, to get a non passing grade you need to mark wrong both questions the first time (3/4 * 3/4) and mark both wrong the second time around (2/3 * 2/3) any other combination provides at least one correct answer, this has a 25% chance, so you have a 75% chance of getting at least one question right.

      • Nibodhika@lemmy.world
        1·
        2 years ago

        I’am considering that, which is why I subtracted one from the number of possibilities in the second try.

          • Nibodhika@lemmy.world
            1·
            2 years ago

            Yes, I took that into consideration, those are my scenarios 1 (0% on the first try), and 3 and 4 (both with 50% on the first try). Scenario 2 has 100% in the first try, thus accounting for all the possible ways to get to 100% in up to two tries.

  • NeoNachtwaechter@lemmy.world
    11·
    2 years ago

    Let’s say you use both tries and you remember your previous two respected answers.

    Important question here:

    If your first try gives you a grade of 50%, do they tell you which one of your answers was the right one and which one was the wrong one?

  • ilmagico@lemmy.world
    3·
    2 years ago

    The probability of getting both right the first time is easy: 0.25*0.25 = 0.625 or 6.25%

    The probability of getting exactly one right is: either you get the 1st one right and miss the second, or vice versa. Thats 0.25*0.75 + 0.75*0.25 = 0.5*0.75 = 0.375 or 37.5%, so the probability of getting at least 50% is 0.375 + 0.0625 = 0.4375 or 43.75%, even without retries, so pretty good odds. The probability of missing both is 1 - 0.4375 = 0.5625 (or 0.75*0.75).

    When you retry, there’s two possibilities:

    1. You missed both: now your probability of getting at least one of them right is: (1/3)(1/3) + 2*(1/3)(2/3) =~ 55.55%
    2. You got only one wrong: you just need to guess the other, so it’s 100% for you to get at least one, and 1/3 (33.33%) to get both

    So, including a retry, you either:

    1. Guess them both the first try: 0.0625 or 6.25%
    2. Guess one of them, then guess both: 0.375*(1/3) = 0.125
    3. Guess one of them, then still guess only one: 0.375*(2/3) = 0.25 or 25%
    4. Guess none first, then guess one: 0.5625*2*(1/3)(2/3) = 0.25 or 25%
    5. Guess none first, then guess both: 0.5625*(1/3)*(1/3) = 0.0625 or 6.25%
    6. Guess none, then still guess none: 0.5625*(2/3)*(2/3) = 0.25 or 25%

    So, probability of a passing grade is 75%. Not a very good test if it’s so easy to pass by random guessing ;)

  • unclad8226@lemmy.ml
    2·
    2 years ago

    Let’s assume you get every answer wrong every time. For the first try you have a 75% chance of getting each question wrong. So this is 0.75x0.75 for both questions being wrong. This is a 56.25% chance of being incorrect on the first test.

    The second test you now have 3 possible answers for each question since you can now eliminate the incorrect answers from the previous test. You now have a 66.6% chance of getting each question wrong. This is now 0.66x0.66 to get both wrong, so a 43.56% chance of failing a second time.

    Now let’s find the chance that you fail both the first and second attempt. This is 0.5625x0.4356 which gives 24.5% chance of failing both. We can do 1-0.245 to find the chance of passing, which gives a 75.5% chance of passing on one of the two attempts.

    Been a long time since I’ve done something like this, so please correct if wrong. You should be able to do the opposite and calculate all the different ways of passing a and total to 100%, but that is longer than this and I cannot be bothered to check.

  • TheMurphy@lemmy.world
    11·
    2 years ago

    GPT-4 answered this. I will link it’s calculations down below in the next comment.

    The probability of passing the test by brute force guessing (i.e., getting at least one question right) is approximately 68.36%. This includes the scenarios where you get one question right and one wrong, or both questions right.

    Additionally, the probability of getting a perfect score (i.e., both questions right) by guessing is approximately 19.14%.

    • TheMurphy@lemmy.world
      2·
      2 years ago

      To calculate the odds of getting a passing grade (at least one question correct) or a perfect score (both questions correct) through brute-force guessing, let’s break down the scenario:

      1. Each question has 4 possible answers, and only one is correct.
      2. You have two attempts to answer each question.
      3. You remember your previous answers and do not repeat them.

      First, we’ll calculate the probability of guessing at least one question correctly. There are two scenarios where you pass:

      • Scenario 1: You get one question right and one wrong.
      • Scenario 2: You get both questions right.

      For each question:

      • Probability of getting it right in one of the two tries = ( 1 - ) (Probability of getting it wrong twice)
      • Probability of getting it wrong in one try = ( \frac{3}{4} ) (since there are 3 wrong answers out of 4)
      • Probability of getting it wrong twice = ( \left( \frac{3}{4} \right)^2 )

      So, the probability of getting at least one question right in two attempts is ( 1 - \left( \frac{3}{4} \right)^2 ).

      For two questions:

      • Scenario 1 (One Right, One Wrong):
        • Probability of getting one question right (as calculated above) multiplied by the probability of getting the other question wrong twice.
      • Scenario 2 (Both Right):
        • Probability of getting each question right (as calculated above) and multiplying these probabilities together.

      The overall probability of passing (getting at least one question right) is the sum of the probabilities of these two scenarios.

      Now, let’s calculate these probabilities.