Probabilities: 20 question, 5 options test

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babibooboo

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Oct 11, 2015
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Hello. I'm not a student anymore and this is more of a hobby but I need help figuring out the optimal chances if I randomly fill out a 20 question exam. The scenario of the exam is this:

Each question in this exam has 5 different answers (a, b, c, d and e) and only one of them is right for each question. The thing is, in these 20 questions, there are always 4 of each option that are correct. So in a perfect score (you get every question correct) there will always be 4 questions where the correct option is A, 4 questions where where the correct one is B etc etc. So, if I choose option A for every single question out of the 20 questions, I'd get 4 points out of 20 in this exam FOR SURE. I'll always get 4 points if I do that.

What I wanna know is:
1) how could I maximize my chances of getting more than 4 points out of 20?
2) What's the chance of getting every answer correct in this exam in comparison to other regular exams where every correct answer could be an A?
3) What's the probability of getting 8 points if I choose 10 As and 10 Bs?
4) What's the probability of getting 16 points if I choose 5As, 5Bs, 5Cs and 5Ds? And of getting 4 points if I do that?
5) Is the best and safest option always choosing A?

I'm almost completely out of my field since it's been ages since I've solved this kind of probability problem. I started out by trying to figure out how many scenarios there are of getting a perfect score. I did 20C4 + 16C4 + 12C4 + 8C4 = 7230. So the % of getting everything right would be 1/7230=0,01%. That seems a bit too much to me, is it correct?

The thinking process was that I'd first fit the 4 As and see how many combinations there are of me gtting the 4 As correct, then so on and so forth for the remaining questions. But now I'm completely lost. I don't even know if this 7230 number is correct or if I'm going about it the wrong way.

Honestly, I just want the answers but I get that this forum's rules are more about explaining how to get there. If someone could do both that'd be great! Just posting your calculations would be enough, even though I'd like to understand the thinking process behind it.

Thank you very much!
 
babibooboo said:
Hello. I'm not a student anymore and this is more of a hobby but I need help figuring out the optimal chances if I randomly fill out a 20 question exam. The scenario of the exam is this:

Each question in this exam has 5 different answers (a, b, c, d and e) and only one of them is right for each question. The thing is, in these 20 questions, there are always 4 of each option that are correct. So in a perfect score (you get every question correct) there will always be 4 questions where the correct option is A, 4 questions where where the correct one is B etc etc. So, if I choose option A for every single question out of the 20 questions, I'd get 4 points out of 20 in this exam FOR SURE. I'll always get 4 points if I do that.

What I wanna know is:
1) how could I maximize my chances of getting more than 4 points out of 20? ..... Study - do not depend on random selection
2) What's the chance of getting every answer correct in this exam in comparison to other regular exams where every correct answer could be an A?
3) What's the probability of getting 8 points if I choose 10 As and 10 Bs?
4) What's the probability of getting 16 points if I choose 5As, 5Bs, 5Cs and 5Ds? And of getting 4 points if I do that?
5) Is the best and safest option always choosing A?

I'm almost completely out of my field since it's been ages since I've solved this kind of probability problem. I started out by trying to figure out how many scenarios there are of getting a perfect score. I did 20C4 + 16C4 + 12C4 + 8C4 = 7230. So the % of getting everything right would be 1/7230=0,01%. That seems a bit too much to me, is it correct?

The thinking process was that I'd first fit the 4 As and see how many combinations there are of me gtting the 4 As correct, then so on and so forth for the remaining questions. But now I'm completely lost. I don't even know if this 7230 number is correct or if I'm going about it the wrong way.

Honestly, I just want the answers but I get that this forum's rules are more about explaining how to get there. If someone could do both that'd be great! Just posting your calculations would be enough, even though I'd like to understand the thinking process behind it.

Thank you very much!
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Obviously I'm not going to random my entire test. It's a huge important test for me which I am studying for. The fact that it is that important and difficult is why I'm maximizing my chances of getting a better grade. It's a 100 question exam, it's huge and has a huge bibliography. If it has this system of 4 right questions per sections, you're ******* certain I'm trying to maximize my shots when I'm not 100% sure of the correct answer.

I'm still asking you to help me figuring out a probability problem I'm really interested in knowing how to solve. My bad if this is mostly directed towards younger students and to help them understand math. Are you not helping me out of some fear of me not actually studying? Why is that important? I'm studying either way, I'm not some high school student trying to flake my way out of school. I respect if you feel morally obligated not to help me but please understand that's not in anyway what's happening.

Either way I was doing this jokingly on the side while many of us exam takers theorize about this. I was always genuinely interested in Math and I thought I could do this on my own, but it's waaaay past my current understanding. I appreciate your teaching sense and I understand you wouldn't feel right facilitating a student in his small test and enabling him for not studying, but please know you're certainly not on point in this case.

Thanks anyway, I hope you reconsider. I'm not a native english speaker so I'm sorry for if I'm not too clear.
 
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