Nobody around me knows how to calculate it except my teacher

Probability theory and statistics

Nobody around me knows how to calculate it except my teacher

Postby Guest » Wed Mar 18, 2020 4:47 pm

Hello there,
at first I would like to tell that it´s quite hard for me to write this in English but I have no other choice. I got this example from my math teacher because she accused me and my friends of cheating during the test. Everyone got own math problem which we have to solve - then she will delete the bad mark. I would ask in some math forum in my country but I´m afraid she could see it. She made up these math problems specially for us - so there are no similar math problems on the Internet. She would know that it is from me. I spend a lot of time trying to calculate it but I´m completely lost. :oops:
The example: There is 20 math problems and in every (of those 20) math problems you can make 3 mistakes and count it in wrong (different) way. There are 3 girls and the girls "accidentally" (they cheated) made the same 6 mistakes in those 20 math problems. What is the probability?

So that's it. I hope someone here could help me. I don't even know if I should count it like - 60 possible mistakes, from this I choose 6 mistakes.. So 60nCr6? I don't think so.. And what with those 3 girls? Where to multiply it?
If you can count it for me, please tell me even the process how you get there. I would really like to understand it.
Thank you!
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Re: Nobody around me knows how to calculate it except my tea

Postby Guest » Fri Mar 20, 2020 4:46 pm

Don't feel bad. The question isn't well formed in my opinion. Probabilities assume random distributions but errors on a math test are not random errors, they are errors in process/logic. Putting that aside, we'll assume each of the 20 question in fact had 3 wrong options and 1 correct option and options would be chosen at random by any person taking the test. Then, the probability of making a specific choice (correct or not) on any given question is 1/4. For two persons to answer all questions the same way (with completely random selections) would require they match on question 1 AND they match on question 2 AND they match on question 3 AND...(all the way to question 20). In considering compound probabilities the AND is represented by multiplying the two individual probabilities. So, the chance of two persons matching on question 1 is (1/4)*(1/4)....to match on question 2 is also (1/4)*(1/4)...and so on. But, to match on question 1 AND question 2 has a chance of (1/16)*(1/16).

So, to match all 20 with complete random choices would be [tex](1/16)^{20}[/tex]...and that's just two persons. Integrating the 3rd person is left for you.
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Re: Nobody around me knows how to calculate it except my tea

Postby Guest » Sun Mar 22, 2020 1:30 pm

Guest wrote:Don't feel bad. The question isn't well formed in my opinion. Probabilities assume random distributions but errors on a math test are not random errors, they are errors in process/logic. Putting that aside, we'll assume each of the 20 question in fact had 3 wrong options and 1 correct option and options would be chosen at random by any person taking the test. Then, the probability of making a specific choice (correct or not) on any given question is 1/4. For two persons to answer all questions the same way (with completely random selections) would require they match on question 1 AND they match on question 2 AND they match on question 3 AND...(all the way to question 20). In considering compound probabilities the AND is represented by multiplying the two individual probabilities. So, the chance of two persons matching on question 1 is (1/4)*(1/4)....to match on question 2 is also (1/4)*(1/4)...and so on. But, to match on question 1 AND question 2 has a chance of (1/16)*(1/16).

So, to match all 20 with complete random choices would be [tex](1/16)^{20}[/tex]...and that's just two persons. Integrating the 3rd person is left for you.



Hi, thank you! I think I understand it - when three girls match on all the twenty answers it's \frac{1}{64}^{20}. But what I don't understand is - how to integrate into it those 6 same mistakes. Because there are 20 math problems (and in every math problem there is 1 good answer and 3 possible WRONG answers - and every of the girls made in 20 math problems 6 mistakes and the mistakes are the same (because they cheated). In those 6 wrong math problems there were 3 possibilities how to do it wrong - but they chose the same option. I can´t really explain it in case you don´t understand - so in attachment I painted it. :mrgreen:
So once more - if you can look at it and help me?
I considered something like - {20\14}*{60\6} and then?
Attachments
problem.jpg
problem.jpg (72.12 KiB) Viewed 112 times
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