quickly who is good with probability

[Agro]

I had a mini-test in a class today and the professor accused us (the whole class) of cheating because of some weird discrepancy in marks between 2 groups of people that did the test. Basically they ran one session in the morning and another session in the afternoon, and there were way more people in the afternoon who got the correct answer for a question than people in the morning.

Morning session: 10/35 people got correct answer

Afternoon session: 26/31 people got correct answer

The question was a multiple choice question.

What I want to know is, can anyone get an estimate of the probability of something like this happening, assuming there was no cheating? Would I be correct in thinking that the approximate chance of a student getting the correct answer was 36/66 (therefore approx 0.545)? I don't actually plan on refuting the accusation with statistics but it'd be fun to know for the lulz

help me out you math nerds i did probability last year but forgot everything

[Grace]

how many possible answers to the multiple choice questions?

edit: okay let x = number of answers to the multiple choice question

(n C R) is the binomial expansion coefficient ((n!)/(R!(n-R)!))

Probability of the morning group = (35 C 10)((1/x)^10)((1-(1/x))^25)

Probability of the afternoon group = (31 C 26)((1/x)^26)((1-(1/x))^5)

Overall probability = (66 C 36)((1/x)^36)((1-(1/x))^30)

using the formula for binomial probabilities: (n C R)(p^R)((1-p)^(n-R)) where p = 1/x (chance of randomly getting the answer right with x possible answers)

this shows the overall chance of the groups getting the exact sample set given.

edit round 2: unsure though if that's quite what you wanted, but shrug

I don't remember probability that well because I did it two years ago and I don't remember everything

Edited November 28, 2012 by Manix

[Agro]

oh there were 4 possible answers

but we can't assume that everyone guessed (because they didn't)

[Grace]

then you can't really use probability on it (unless there's something I've forgotten) :/

[Aere]

Also depends on if the multiple choice answers all had a 25% chance of being chosen. This is usually not the case, most come down to a one vs one.

Yeah this isn't anything you can reliably calculate.

[Raven]

Looking at those numbers there's a fair chance that the people leaving the first test informed the second group what to revise before they completed the test. All it takes is one person to spill the beans and before you know it everyone's revising the right stuff.

[dondon151]

geez agro why didn't you just ask me

this is a statistics question, not a probability question

first let's define p = probability of any student getting the answer correct and (1-p) = the converse

from your total sample size of n = 66, 36 got the correct answer, so we'll estimate p = 36/66 = 0.545

next we need to estimate the variance. because the choices here are binomial and p is close to 0.5, we'll assume the worst case per-sample variance of 0.25 (because var = p*(1-p), which has a maximum at p = 0.5).

using the per-sample variance, the distribution variance is equal to var/sqrt(n) = 0.25/sqrt(66) = 0.03. the standard deviation is sqrt(var) = 0.173.

in your case, the morning class observed p = 0.286 and the afternoon class observed 0.839. using the statistical model above, both observations fall within 2 standard deviations of the mean! so we can actually say with close to ~90% confidence (can't be assed to look up a Z table and calculate the actual confidence) that a random sampling of students with sample size n = 66 falls within the margin of estimates defined by your observed p between the morning and afternoon classes. note that this does not mean that we are 90% sure that no one cheated (in fact, it suggests the converse, that maybe there is only about a ~10% certainty that there was no cheating); it only suggests that the observed discrepancy is probable.

furthermore, your morning and afternoon classes have n < 66, meaning that the variance is even higher. and keep in mind that this is assuming purely random guessing with p = 0.545.

Edited November 28, 2012 by dondon151

[Agro]

omg thank you don <3

[Integrity]

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[Gold Vanguard]

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