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!

  • fugacity@kbin.social
    4·
    11 个月前

    P(passing) = 1- P(failure)
    P(failure) = P(failure first try)*P(failure second try)
    P(failure first try)=(3/4)^2
    P(failure second try)=(gonna post in reply)

    • fugacity@kbin.social
      4·
      11 个月前

      P(failure second try)=(2/3)^2 since you can eliminate one choice but 2 others are still wrong.

      To total:
      P(failure)=(3/4)2*(2/3)2=1/4
      1-1/4=0.75

      So the probability of passing is 0.75

      Edit:
      Remark: this problem is elegant if you attempt to calculate the passing as the complement of failure rather than enumerate all successes. Shouldn’t take more than 3 minutes with a clear head if you know the correct approach. If this was an college level intro probability exam question, it should be done the fast way since it’s meant to eat up your time otherwise.