Probability of test passes

[jimbob77]

I have X students that take tests of equal difficulty
There are Y tests in total and you can either pass or fail
I have data on all of the X students but so far not every student has taken every test - some have taken all Y, some have taken y-1, some y-2, etc. down to 1.

I want to know how I’d calculate

1.) the probability that a future student passes all Y tests
2.) the expected total number of fails for a future student

To work it out I thought about just treating each exam as an independent event and using a flat probability of passing each exam (total passes / total attempts) for any student. However, that doesn’t capture the fact that some students are better than others / 1 fail increases likelihood of another (which is what the data shows)

 

[AngleWyrm]

replace the word 'student' with 'unfair coin'

 

[hlsmith]

Are you just treating tests as binary pass/fail?

 

[jimbob77]

Are you just treating tests as binary pass/fail?

yes

 

[hlsmith]

I did not stare too hard at your question, but if there are 4 tests and say 80% of people have completed all 4 tests and the outstanding 20% is representative of those that completed the tests, you can just frame out the probabilities based on the 80%.

E.g.,
75% of people that passed 3/3 initial tests passed the fourth test.
50% of people that passed 2/3 initial tests passed the fourth test.
20% of people that passed 1/3 initial tests passed the fourth test.

 

[AngleWyrm]

I suspect without knowing that the percentage of pass/fail is different for different students, but the questions are generalized to a generic future student. So this seems like a measure of the test rather than the student.

If we lump all test results into one set, and call them the student body's performance against a given test, then it could be looked at as a Bernoulli Trial. If the data from the student body does indeed look like a bell curve.

Which can answer these questions:

• The number of pass results in the first n test results, which has a binomial distribution B(n, p)
• The number of fail results needed to get r pass results, which has a negative binomial distribution NB(r, p)
• The number of fail results needed to get one pass result, which has a geometric distribution NB(1, p), a special case of the negative binomial distribution

Another way to examine the results if there are many is to get the average and standard deviation for each student. This can then be looked at from either perspective: Student performance vs a test, or test performance vs students. I mention this example, not because it's directly applicable to the binary pass/fail problem, but because it gives a clear sense of the attribution of blame either to student or to test, which may or may not be correct.