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.

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?

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 ;)

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.

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)

~IIRC from my one stats class almost a decade ago, the math is pretty simple. If it’s truly a random guess then you have a 25% chance to get each question right, all you have to do is multiply them. (0.25)(0.25)=0.0625, you have a 6.25% of getting a 100. The other option to get a 50 is (0.25)(0.75)=0.1875, 18.75%. So there is a 75% chance of failing both.~

~If you get a second chance and can remember your two wrong answers, each problem now has only three options. To get each right, (0.33)(0.33)=0.1089, 10.89% chance. To get one right, (0.33)(0.67)=0.2211, 22.11% chance. 67% chance of failing both a second time.~

Something is amiss with my logic/math here but I’m too tired and am going to sleep now. With this logic, failing both the first time would be (0.75)(0.75)=0.5625 56.25%, the %s don’t add up to 100% so someone please correct me lol

Edit 2: thanks for the corrections everyone, I forgot that the order does matter in this equation

For a passing grade - the easy way to calculate it is by calculating the chance of failing and then subtracting it from 1. To fail, you need to get the first two questions wrong (3/4 times 3/4) and then get them both wrong again on the second try (2/3 times 2/3 since you are choosing from the remaining 3 choices for each). So 3/4 x 3/4 x 2/3 x 2/3 = 1/4 or 25%, so you have a 75% chance of passing.

I can’t do the math rn, but OIRC the probability is the number of favourable cases divided by the mber of possible cases

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%.