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Subject:
probability question
Category: Science Asked by: chris2002micrometer-ga List Price: $15.00 |
Posted:
15 May 2003 17:13 PDT
Expires: 14 Jun 2003 17:13 PDT Question ID: 204347 |
I teach and recently graded a test with 30 mult choice questions (a thru d) where two students not only had the same score, but also the same answers. Both missed seven with the same incorrect answers. My gut feel was a one in 16384 (4 to the 7th) odds that this was coincidental. How can I figure this out? What if the test had 100 questions? What if 5 incorrect answers matched but 2 were different? Can I prove anything? I let it slide this time, but would like to know if I should have said something. |
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Subject:
Re: probability question
Answered By: chis-ga on 15 May 2003 17:32 PDT Rated: |
Hello chris2002micrometer, It is impossible to find an exact probability of this occurring becuase we don't know the likelihood of the students getting an answer correct. Here is how to find the probability of this occurring naturally (in this approximation, I assume that the probability of a correct choice is 80%, which you can modify): [(30c7)(.8)^23(.2)^7]^2 = .02366 = 2.37% Note that 30c7 = 30 choose 7 = 30!/((23!)*7!) This is definitely very suspect as the probability is quite low. Is it possible that these seven questions were extremely hard? Were these kids sitting very close to each other? Are they friends who would likely cheat? If the test had 100 questions, the probability would be: [(100c7)(.8)^93(.2)^7]^2 = 1.99023*10^(-4) = .0199% To figure these probabilities, I use the following: [(Questions c questions the same)(prob. of getting it right)^(questions right)(probability of getting it wrong)^(questions wrong)]^2 (squared because it occurs twice) For five incorrect answers, but two being different: (30c7)(.8)^23(.2)^7 * (23c2)(.8)^21(.2)^2 * (7c5)(.8)^2(.2)^5 = 6.17496*10^(-5) = .0162% As a result, no, you cannot truly prove anything because it IS possible, but it is also extremely unlikely. For the first time, it was right to let it go, but if it occurs again, the probability would be so low, that it would be nearly impossible for it to be a random event. Let me know if I can help any more. Thanks, chis-ga | |
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chris2002micrometer-ga
rated this answer:
I got a lot of good insight into this from chis, et al. Worth every penny. |
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Subject:
Re: probability question
From: racecar-ga on 16 May 2003 12:20 PDT |
I would tend to go about the calculation in a different way. First, make a number of assumptions: IF THERE WERE NO CHEATING: 1) The probability of getting each question correct is the same for each question and each student and is 23/30. This may be unfair to the students, since, (a) as pointed out in the answer, some questions may be more difficult than others, (b) the students may have studied together, and so there may be a legitimate reason for correlation between the answers they know. 2) The probability of choosing each of the three incorrect answers to each question is the same, and is 7/90. Thus if the correct answer is B, the probabilities would be A: 7/90, B: 23/30, C: 7/90, and D: 7/90. Again, this may be unfair to the students, since some incorrect answers may be more attractive than others. 3) You, the teacher, looked through all the exams and chose the two most similar. Having made these assumptions, the correct question to ask is: GIVEN that one student chose the answers he did, what is the probability IF NO CHEATING OCCURED that the other student would choose the same answers. The answer is: The probability that the same 23 correct answers are chosen by the second student is (23/30)^23. The probability that the same 7 incorrect answers are chosen by the second student is (7/90)^7. So overall probablity is the product of these, and is 3.8 E -11 (call this P), or 1 chance in 26 billion. However, if there were N students in the class, there were [N choose 2] (call this M) possible pairs, each of which has a 1 in 26 billion chance of turning in identical tests. So the final answer is approximately PROBABILITY = M*P. The exact formula would be PROBABILITY = 1 - (1-P)^M, since (1-P)^M is the probability that all M pairs turn in different tests. But for any realistic class size, this is very nearly the same as M*P. So, let's say your class size is 30. M is then (30 choose 2), or 435, and PROBABILITY = M*P = 1.66 E -8 = 1 chance in 60 million. As pointed out earlier, in reality the probability may be somewhat higher, since the students may have had legitimate reasons for missing some of the same questions (were the same incorrect answers chosen by many of the other students?). However, regardless of these considerations, the probability is vanishingly small, and as they sat next to each other, the opportunity to cheat was there. The verdict is that cheating occured. You may be as sure of it as you are of anything else in your life, because from where I'm standing, the probability that you're insane and dreamed the whole episode is more than 1 in 60 million. (Just a joke--but you get my point :) ). |
Subject:
Re: probability question
From: fstokens-ga on 16 May 2003 12:20 PDT |
Of the other students who missed those 7 questions, did the other students answer them in the same incorrect way? For probability analysis, you generally assume that each wrong answer is equally likely, but on multiple choice tests there are often some wrong answers that are closer to the right answer than others. If most of the people in the class who got these answers wrong answered them the same way, then there is no evidence that these two students cheated (though there may be evidence for cheating on a wider scale!). On the other hand, if these 2 students answered those question wrong in a different way than most other students who got them wrong, then you have some strong evidence for collusion. |
Subject:
Re: probability question
From: racecar-ga on 16 May 2003 13:58 PDT |
Just wanted to add what I think about the other cases you mentioned. If the test had 100 questions, and the same 7 incorrect answers were chosen by the students, P = (93/100)^93 * (7/100)^7 = 9.7 E -12 = 1 in 104 billion. M is the same as before, and for 30 students, PROBABILITY = 4.2 E -9 = 1 in 240 million. These numbers are not dramatically different from the 30 question case. However, if you had indeed given a similar test, but with 100 questions, it is likely that the students in question would have gotten about 77 right and 23 wrong. If this had occured, and the same 23 incorrect responses had been chosen, P = (77/100)^77 * (23/300)^23 = 4.03 E -35 = 1 in 2.5 E 34. So (for 30 students) PROBABILITY = 1.8 E -32, or 1 in 5.7 E 31. This is about 1000 times less likely than buying 4 super lotto tickets, and winning 4 jackpots in a row. Now for the case with 30 questions, the same 7 of which are answered incorrectly, with 5 of those 7 answered the same. P = (23/30)^23 * (7/90)^5 * (21/90)^2 * (7 choose 5) The first factor is for the correct answers, the second for the incorrect ones which are answered the same, and the third for the incorrect ones which are answered differently. This third factor would be (14/90)^2 if you wanted the probability that EXACTLY 5 of the the incorrect answers matched, but I think it's more appropriate to find the probability that AT LEAST 5 matched. The fourth factor is to account for the number of possible ways the second student could match 5 out of 7 of the other student's incorrect responses. Thus P = 7.2 E -9 = 1 in 140 million, and PROBABILITY (for 30 students) = 3.1 E -6 = 1 in 320,000. In this last case, it might almost be reasonable to chalk the similarity up to freak coincidence and the fact that the students studied together. I'd still fail 'em though. :) |
Subject:
Re: probability question
From: chris2002micrometer-ga on 16 May 2003 17:10 PDT |
Racecar - Where did the 90 come from? "2) The probability of choosing each of the three incorrect answers to each question is the same, and is 7/90." |
Subject:
Re: probability question
From: racecar-ga on 16 May 2003 21:56 PDT |
The probability 7/90 comes from the assumption that each of the three incorrect answers has an equal probability of being chosen. Given that 23 of the 30 questions were answered correctly, I guessed that the probability of a correct answer is 23/30. The probability of an incorrect answer is then 7/30. There are three possible incorrect answers to each question, so I assigned a probability of 7/90 to each of them. |
Subject:
Re: probability question
From: racecar-ga on 16 May 2003 22:28 PDT |
In regard to your clarification request: It seems you want to know the answer to the following question: Given that the students missed the same seven questions, and assuming the choice on those seven was random, what is the probability that a given number of answers will match? This question can be answered fairly easily, but I don't think it relates very well to the situation you described. It is unlikely that the students would know the answers to the same 23 questions. I have to go now, but I'll answer the above question later. |
Subject:
Re: probability question
From: racecar-ga on 16 May 2003 22:50 PDT |
Ok, I'm back. The probability that exactly N of the answers match is (7 choose N) * (1/4)^N * (3/4)^(7-N) Here's a table of approximate values: N Prob. 0 .1335 1 .3115 2 .3115 3 .1730 4 .0577 5 .0115 6 .0013 7 .00006 Once again, I do not think this is a reasonable answer to the question 'given no cheating, what's the probability of 2 students choosing the same 7 wrong answers out of 30 questions. As you say, some of the answers will be known, and some guessed, but why should both students know the answers to the same questions? If the missing knowledge were due to 'tardiness or whatever' why should both students have been 'present and attentive' at all the same times? It would be suspicious that both students answered the same 23 questions correctly even if they left the other 7 blank. By the way, how many students are in the class? |
Subject:
Re: probability question
From: chris2002micrometer-ga on 17 May 2003 04:53 PDT |
Racecar - * (3/4)^(7-N) - Yep, that's what I left out. There were only 4 students and the two in question missed the same lecture. The other two students answered the 7 questions differently and more were correct. I guess with all the assumptions made here and inside knowledge, it is a dice-roll problem. Thanks for your input. |
Subject:
Re: probability question
From: khane96-ga on 23 May 2003 14:51 PDT |
I do not think you have a probability problem here, even though probas can help you solve it, logic is the way to go. 1) Let us admit they both cheated : Well the leat you could say is that they are not clever, if they did work together by exchanging informations a smart way to do the thing would be to pick a different answer for every question they were not sure of. Furthermore it is hard to believe that they did agree with one another on every question they got right. They got 4/5 of answers right, meaning that combinig their informations they have an way above average knowledge for the test. It is rare that you fail to give the good answer while being absolutly sure you are right, especially 7 times in a row. So if they were working together, we can assume that they would have hesitated on at leat one of the questions, then the smart thing to do is to pick an answer randomly for one of the two and to let the other choose a different answer. So working together and being smart about cheating all the wrong answers would have been different. so if they worked together : -They were smart enough to get 4/5 of the test right. -they were smart enough not to get caught while cheating -they were to dumb about cheating to choose different answers on question they weren't sure about. That seems really unlikely. 2)let us admit they did not cheat Questions answered right doesn't bring any information, no matter how suspicious it looks. The reason is knowing the answer gives you a 100% probability of answering right. No matter if you know the answer by yourself, if you bindly copy the answer from someone who knows, take it from the book or else. So let us foccus on answers that were wrong. Since they did not cheat, anyone of them has a fair knowledge on the subjects th etest is about. This will most probably allow them to exclude 1 answer from the list(unless the test requires no logic at all and is pure "by heart"). So basically let us say they have 1/3 to go wrong on questions they do not know about for sure(quite optimistic choice). So knowing that, statistically how many unsure questions do you need to answer to get 7 wrongs ? 7 questions 7 wrongs : p=(2/3)^7 (=0.06) 8 questions 7 wrongs : p=((2/3)^7x1/3)x8 (=0.16) 9 questions 7 wrongs : p=((2/3)^7x(1/3)^2)x9x8/2 (=0.23) 10 questions 7 wrongs : p=((2/3)^7x(1/3)^3)x10x9x8/6 (=0.26) 11 q 7 w : p=0.23 12 q 7 w : p=0.19 13 q 7 w : p=0.13 So let us assume that someone that got 7 answers wrong was actually hesitating on 9, 10 or 11 answers (very optimistic choice) So what are the probabilities that two students hestating on 9, 10 or 11 answers gets the same answers (still keeping in mind that one answer is obviously wrong each time.)?. for 9 answers there are 3^9 possibilities (19683) for 10 answers : 59049 for 11 answers : 177147 so even while being really optimist there is still only about one chance out of 60,000 that they did not cheat. Very unlikely 3) One of them cheated, and the other either didn't realize and/or did not interfere with his neighboor in any way. Well that is really, really probable. And it would explain everything, the cheater not having the smallest clue about what was wrong or what was right simply did copy whatever he saw without question, hence the similarities going as far as mistakes. |
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