Question 575425: I am an instructor at a high school and just gave a fifty item multiple choice exam (a,b,c,d). Five students who sit together at a table got the same 8 of the 50 questions wrong. What is the probability of that happening? I want to be able to support the fact that I believe they cheated. Thanks!
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Found 2 solutions by stanbon, richard1234:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I am an instructor at a high school and just gave a fifty item multiple choice exam (a,b,c,d). Five students who sit together at a table got the same 8 of the 50 questions wrong. What is the probability of that happening? I want to be able to support the fact that I believe they cheated. Thanks!
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The probability I give the same answer as you on one question is 1/2
The probability that 6 others give the same answer is (1/2)^6
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The probability 8 of us give the same answer is (1/2)^7
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The probability that happens on 8 questions is (1/2)^7^8 = (1/2)^56
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BUT, if we assume the 8 questions could hardly be answered incorrectly
by anyone taking your class, the probability does not change, but the
reality of the situation would force us to reasses our suspicion.
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In other words the theoretical probability does not always tell the
whole story about events. It only tells us something about RANDOM
events.
Question: What is the probability your car will follow some particular
car at a particular red light tomorrow morning? Small, right?
But what is the probability you will follow some car? Hugh, right?
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Recommend: Look beyond the probabilities.
Cheers,
Stan H.
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! I'll assume that those five students (A,B,...E) got the other 42 correct.
Suppose we fix student A's set of responses. We want to find the probability that student B got the exact same set of responses, by chance. If the probability of obtaining a question is 1/5, it is pretty simple:
 = (\frac{1}{5})^{42} (\frac{4}{5})^8) , we don't multiply by 50C8 because it's not exactly a binomial distribution; order of the problems is important. This probability is roughly 7.37*10^(-31), pretty small. Since we also have students C,D,E, raise this number to the fourth power; you'll get a pretty minuscule number.
However, the probability of a correct answer is not 20%. Assume the probability of a correct answer is p, then
 = p^{42}(1-p)^8)
If p = .99, then P(B=A) = 6.56*10^(-17). You can use calculus or inequalities to optimize P(B = A) but it will still be small. Note that you have to raise this probability to the fourth power to account for four students who are supposedly independent of each other. They're probably cheating.
There are definitely exceptions though, e.g. the eight questions were extremely difficult while the other 42 were extremely easy. Then I wouldn't be all that suspicious.
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