My sister, while administering a college exam, thought two students seated next to one another were communicating during the test.There are of course confounding variables, including different question difficulties and different plausibility of incorrect answers. Nonetheless, she'd like a reasonable approximation of the math.
The test had twenty-six questions, each with answers A through D. Of the twenty-six, Student A answered sixteen correctly and ten incorrectly. Student B answered sixteen correctly, eight incorrectly, and left two blank.
Of the questions that Student B answered, every single correct and incorrect answer matched the answers given by Student A. Assuming they did not cheat, what is the likelihood of them giving these twenty-four identical answers?
Answers in the comments section please, and feel free to communicate with the person sitting next to you.
6 comments:
The questions were likely of varying difficulty, and/or grouped around differing bodies of knowledge.
Nevertheless, if this college girl really wants a mathematical equation using the absurd assumptions that all questions are equally difficult and completely unrelated to each other, try
(16/26)^16 * (10/26)^10 = 0.00002%
Why wouldn't it be .25^24? One test would would be the standard and the likelyhood of the second student getting the same answer on each particular question in 1 in 4. They only have 24 of the same answer so it would then have to be repeated 1 in 4 for 24 successive times.
---Jimmy Beckham
Jimmy, the problem with that is that correct answers are likelier than incorrect answers. They're not drawing equal lots out of a hat. WC is close to the mark.
And W.C., this very problem was at the root of a bunch of Chicago teachers getting fired for cheating on standardized tests. And it was a mathematical model that sealed their fate.
Don't hassle her. She's a professor, not a "college girl."
Charlie,
I ran your question by a friend who's a math professor. His response:
I'll have you know -- I went to a math department lunch and discussed this with one of them
Annnnd...well, I don't have a helluvalot of certainty.
First off, let's just pitch the two unanswered questions, tho the fact that they responded "differently" to those two might actually help exculpate them We're just railroading these poor students, eh?
Btw, to answer your question -- I think you probably meant 0.25^24 -- the chance the second guy matches the first guy is a random 25%, and he needs to do it 24 times. But that assumes they're just randomly picking answers, when some thought probably went into their selections.)
Let's be generous and assume they're C students -- 70% of the time they are right, 30% of the time they are wrong, with their wrong answers distributed evenly among the three wrong choices. Based on these details, we get the following probabilities:
P(both get one right): 0.7 * 0.7 = 49%
P(one is right, one is wrong): 0.7 * 0.3 * 2 = 42%
P(both are wrong): 0.3 * 0.3 = 9%
P(both give same wrong answer): this should be one-third of the time they're both wrong -> 3%
P(both give the same exact answer for any given question): P(both right) + P(both wrong) = 52%
(Now this depends a lot on how "smart" they are. You and I could take a pre-algebra quiz and put down the exact same set of 50 answers w/o collusion, but that's because our P(right) numbers would be a great deal higher than 70%. Still, 70% is probably close enough to their actual percentage of correctness for it not to be laughably wrong.)
Anyhow, according to these numbers, 0.58^24 (2 in a million) is certainly a reasonable estimate of this probability. If they were 80% students, then it would be 0.653^24 (3.6 in every 100000 times); if 90% students, 0.813^24 ~ 0.7%. Hell, if they were 95% students, 0.903^24 would still only happen 8.6% of the time randomly.
This doesn't actually take the fact that they got 16 questions right into account, so let me try to mock that one up. We've got sixteen 49% occurrances and eight 3% occurrances, and we can arrange them C(24,16) ways. So it should look something like...
C(24,16)*.49^16*.03^10 = 735471*0.000011*6.5E-13 = 5.33E-12, or 0.000000000533% chance they weren't cheating. Seems unlikely, no?
We're possibly not being very sporting here, and maybe they're actually 66.7% students (reflected in the 16/24). I think this is probably the best-case, least-likely cheating scenario for them. It comes to 735471*(4/9)^16*(1/27)^8 = 3.96E-08. Still pretty slim, methinks.
50/50, either they were or they weren't.
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