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Hi, I am new here, and my name is Jonas. I'm a CS major at a university in the Northeast US. I'm a senior and wrapping up degree requirements which include a science track. I chose Chemistry because Physics was full.

The chemistry exams are multiple choice (because you couldn't grade 300 exams in a timely fashion any other way), choices A-E, and have 25 questions.

It turns out that the answers to these exams have, it seems to me, an unlikely distribution of A's, B's, C's, D's, and E's.

I was wondering how I would find out the likelihood that an exam has no more than 6 of the same choices and no less than 4.

I was thinking it'd be easier to find out 1 - (probability of 7 or greater of the same selections + probability of 3 or less of the same answers). So, to do that I just do 1 - ((1/5 * summation from k=7 to k= 25 of (25 choose k)) + (1/5 * summation from k = 0 to k = 3 of (25 choose k)).

However, it seems that this would be the likelihood that there are no more than 6, no less than 4 of one

Here is what made me so curious. The reason there are four forms per exam is to prevent cheating. Both the person to the right of you and the left of you will be working from different exams, as will the person in front and behind of you will. Could someone point me in the right direction here?

Thanks!

http://img600.imageshack.us/img600/1001/47634578.png [Broken]'

Here, I have an excel table with the probabilities I mentioned. The probability that 1 choice (A-E) will have 4, 5, or 6 occurrences is .546042. The probability that all would, should at best be .048543 (that's .54602 ^ 5). But this discounts the fact that there are only 25 questions, so it is impossible for all choices to have 6 occurrences. I am not sure how to rectify this, but maybe you guys will be.

http://img577.imageshack.us/img577/3518/binomialdistributions.png [Broken]

The chemistry exams are multiple choice (because you couldn't grade 300 exams in a timely fashion any other way), choices A-E, and have 25 questions.

It turns out that the answers to these exams have, it seems to me, an unlikely distribution of A's, B's, C's, D's, and E's.

I was wondering how I would find out the likelihood that an exam has no more than 6 of the same choices and no less than 4.

I was thinking it'd be easier to find out 1 - (probability of 7 or greater of the same selections + probability of 3 or less of the same answers). So, to do that I just do 1 - ((1/5 * summation from k=7 to k= 25 of (25 choose k)) + (1/5 * summation from k = 0 to k = 3 of (25 choose k)).

However, it seems that this would be the likelihood that there are no more than 6, no less than 4 of one

*specific*choice (A-E), rather than all choices. Is this correct, or am I totally off track?Here is what made me so curious. The reason there are four forms per exam is to prevent cheating. Both the person to the right of you and the left of you will be working from different exams, as will the person in front and behind of you will. Could someone point me in the right direction here?

Thanks!

http://img600.imageshack.us/img600/1001/47634578.png [Broken]'

Here, I have an excel table with the probabilities I mentioned. The probability that 1 choice (A-E) will have 4, 5, or 6 occurrences is .546042. The probability that all would, should at best be .048543 (that's .54602 ^ 5). But this discounts the fact that there are only 25 questions, so it is impossible for all choices to have 6 occurrences. I am not sure how to rectify this, but maybe you guys will be.

http://img577.imageshack.us/img577/3518/binomialdistributions.png [Broken]

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