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!
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:
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%
First time you picked both right, so didn’t need the second time: 6.25%
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%
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.
But you are assuming that he cannot remember his answers from the first try, and whether they were right or wrong.
his edit does not assume that; it’s the cleanest way of doing the problem
I’am considering that, which is why I subtracted one from the number of possibilities in the second try.
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.