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!
But in order to get 100% in your second try, you can have either 0% or 50% in the first.
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.
An unknown factor is if you even get to make a second try at getting 100% if you already passed with 50% on the first test. If it is possible to redo a passed test, I still find it unlikely that anyone would do so given that they know that they don’t know the answers.
Including the edit that you’re not told which one was right in the first attempt with a 50% score, it makes a lot more sense to accept the first 50% pass. Choosing different answers for the second try would only give the maximum score of 50% again, while choosing completely random answers again would only give the same chance as the first attempt, in which 0% is still more likely than 100%
Similarly, if you do get 100% on the first attempt, why’d you want to try again… a lot of the answers here calculate the overall statistics when using both attempts regardless.