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!

  • @[email protected]
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    fedilink
    411 months 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.