Firstly, thank you all for taking the time to help. I'm trying to re-read the replies to understand them, so to provide that level straight off the bat - thank you very much!
Can we consider the 6 exams independent?
To be honest, whilst this seems like a very simple question to ask - I'm embarrassed to ask you to explain what you mean. Independent in what way?
I have given a great deal of thought to this question. And this may not be useful or even correct.
Like you I struggle with "How do I get the statistical likelihood that those 4 were passed by the same individual?"
It seems to me that we must assume that each test has an equal likelihood of pass or fail.
Thus each person can generate one of \(2^6=64\) possible outcomes. Only one of which is all six failures.
Therefore, it seems to me that the probability of failing all six tests is \(\frac{1}{64}\).
There are \(15\) ways to pass four tests and fail two: \(\dfrac{6!}{2!\cdot 4!}=15\).
Thus I surmise that the probability of failing two & passing four is \(\frac{15}{64}\)
Again that is on the assumption that all outcomes are equality likely.
If we can use that then we get \(\left(\frac{15}{64}\right)\cdot\left(\frac{1}{64}\right)^{24}\)
SEE HERE
That is minuscule!
I welcome corrections!
I am trying to follow the example given but am struggling - why is it 2^6 - because there is only 2 option - pass and fail? If so, please accept my apologies for what may have been my having wasted your time. I am very grateful for your help so I did not appreciate that I may have accidentally left some required information to calculate out of my original post. I'm actually not sure if I did, but I suspect I did.
Please don't die working on this problem!
You haven't explained the context of the question. Is this from real life, or is it an assignment? In the former case, things are much more complicated than in the latter! But in either case, we can't make unsupported assumptions, such as that taking an exam is the same as flipping a coin, or that all people are identical. For a good answer, I think we'd need to know something about the distribution of skill in the population, and things like that. So I'm not convinced this has an answer.
But let's try something. If 4 of 150 exams are marked Pass, and I randomly pick up 6 of them off the floor, what is the probability that I get all four Passes? That would be [MATH]\frac{{{4}\choose{4}}\cdot{{146}\choose{2}}}{{{150}\choose{6}}} = 0.00000074[/MATH]. So if the exams are no reflection of skill at all, then it is unlikely that one person gets all the passes.
Here's another try: Assuming the results of the exam are random and independent, it appears (empirically) that the probability of my passing any one exam is 4/150. The probability that I will pass 4 of 6 such exams is, by the binomial distribution, [MATH]{{6}\choose{4}}\left(\frac{4}{150}\right)^4\left(\frac{146}{150}\right)^2 = 0.0000072[/MATH]. Still very unlikely.
But I think it's also unlikely that 25 people are all identical in ability, or that an exam has random results!
I'll try not to! Owing to the replies above I'm glad I came to ask for help because it's way more complex than I thought!
It's real life. I changed the context and the numbers, however - and planned to sub in the correct numbering and re-calculate for my own use. I was not trying to be coy, but trying to remove the reasoning - I hope this doesn't further muddy the water:
It is actually part of a pay dispute. I would like to show the likelihood (I know it is highly implausible) of a work-related assessment.
Notwithstanding changing the context, I think I can see that I didn't provide enough information for a correct calculation. My apologies, I did not wish to waste your time trying to help me in the last calculation.
There are 100 members of staff who are all assessed annually. There are 5 assessment areas per person. Each assessment area has grading: A, B, C, D. Grades A or B are required to "pass", C and D are "fail".
Of 100 staff who were assessed by the same management, 99 staff achieved an A in all 5 areas. 1 staff member received 3 assessments of C. As these areas are equally weighted, that staff member then had 60% C rating and was therefore overall rated C.
Given the information above, I'm not sure which way to try to calculate it out.
I am trying to get the likelihood that of the 100 staff, all assessed on 5 areas with 4 grades in each:
what is the likelihood that of the 500 assessments (100 staff x 5 areas), there were only 3 which were not A and crucially, they were all the same person - taking that persons overall rating to C.
I hope this makes more sense now and thank you all again for taking the time to help.
e1
