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:
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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%
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First time you picked both right, so didn’t need the second time: 6.25%
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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%
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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.
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.
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.
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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?
No they don’t! I edited it in the post. Thanks for pointing that out.
During standardized testing, because I hated math, I always just bubbled in my answers at random on the Scantron and still got average marks for it like half the time.
Later in high school I also learned that in a multiple choice questionnaire, C is the most commonly correct answer so if you just answer C on everything your odds of getting a passing grade are pretty high. Legit were told to just answer C when we didn’t know in a class preparing us for an academic decathlon.
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:
- 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%
- 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:
- Guess them both the first try: 0.0625 or 6.25%
- Guess one of them, then guess both: 0.375*(1/3) = 0.125
- Guess one of them, then still guess only one: 0.375*(2/3) = 0.25 or 25%
- Guess none first, then guess one: 0.5625*2*(1/3)(2/3) = 0.25 or 25%
- Guess none first, then guess both: 0.5625*(1/3)*(1/3) = 0.0625 or 6.25%
- 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 ;)
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)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=0.75So 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.
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.
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~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
There are a couple of errors, that I could spot, but I just woke up so my math might also be wrong hahaha.
(0.25)(0.25)=0.0625, you have a 6.25% of getting a 100.
Correct
The other option to get a 50 is (0.25)(0.75)=0.1875, 18.75%.
Almost, that’s only true if the first one is the one you get right, but you can also get the second one right (3/4 * 1/4), meaning that it’s double your answer, or 37.5% chance of getting at least one right.
So there is a 75% chance of failing both.~
56.25% as per above
~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.
That assumes you got both wrong the first time around.
To get one right, (0.33)(0.67)=0.2211, 22.11% chance. 67% chance of failing both a second time.~
Same as above, also same as getting the second one right.
You only accounted for the situations with one correct answer in the case where it is the first question.
Your calculations to get 100% are right but you are off for the 50% and. You are only considering one specific outcome. But it doesn’t matter if the first question is wrong or the second so the chance is 0.250.75+0.750.25 which is 37.5 or double your answer. We can double check it by looking at it from the other direction.
The chance of failure is 0.75*0.75= 56.25%.
So there is a 43.75% of passing the first go around. Split between a 6.25% to get 100 and 37.5% to get 50.
Same mistake for the second calculations. 44.22% is the chance to get 50%
I see some correct solutions for the 50% case here already, so this reply is going for a perfect score within two tries.
There are 16 ways to answer the quiz, one of which is correct. Assuming you don’t repeat your previous answers, two attempts give you a 2/16 or 1/8 chance that one of them is perfect.
Now if you get feedback between your attempts, you should be able to do better. Let’s see by how much and break it into cases:
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Your first guess is already perfect. This happens 1/16 of the time. No further guessing is needed.
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Your first guess is 50% correct. This happens 3/8 of the time. Picking one of the unguessed answers improves your score to 100% 1/6 of the time.
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Your first guess is completely wrong. This happens 9/16 of the time. Picking different answers for both questions wins 1/9 of the time.
So the overall chance of a perfect score is the weighted sum of these cases or 1/16 + (3/8 * 1/6) + (9/16 * 1/9) = 3/16.
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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%.
GPT4 is pulling these numbers out of its … I mean, out of thin air ;)
GPT-4 is a language model, and while it was an interesting take, it appears to be the wrong tool for the job.
The answer is wrong and without any documentation or proof showing the line of thought to determine the result it’s just a useless number.
Math is not really about the result. It is about understanding the process. Having an AI do that is completely against the purpose of asking this kind of questions in the first place. OP doesn’t need to know if the chance is 68% or 75%, but rather how to figure it out.
To calculate the odds of getting a passing grade (at least one question correct) or a perfect score (both questions correct) through brute-force guessing, let’s break down the scenario:
- Each question has 4 possible answers, and only one is correct.
- You have two attempts to answer each question.
- You remember your previous answers and do not repeat them.
First, we’ll calculate the probability of guessing at least one question correctly. There are two scenarios where you pass:
- Scenario 1: You get one question right and one wrong.
- Scenario 2: You get both questions right.
For each question:
- Probability of getting it right in one of the two tries = ( 1 - ) (Probability of getting it wrong twice)
- Probability of getting it wrong in one try = ( \frac{3}{4} ) (since there are 3 wrong answers out of 4)
- Probability of getting it wrong twice = ( \left( \frac{3}{4} \right)^2 )
So, the probability of getting at least one question right in two attempts is ( 1 - \left( \frac{3}{4} \right)^2 ).
For two questions:
- Scenario 1 (One Right, One Wrong):
- Probability of getting one question right (as calculated above) multiplied by the probability of getting the other question wrong twice.
- Scenario 2 (Both Right):
- Probability of getting each question right (as calculated above) and multiplying these probabilities together.
The overall probability of passing (getting at least one question right) is the sum of the probabilities of these two scenarios.
Now, let’s calculate these probabilities.
phew… the human race is safe (for now) 😅
That’s just what thaiy wants you to think!
