SOLUTION: An instructor gives his class a set of 17 problems with the information that the next quiz will consist of a random selection of 6 of them. If a student has figured out how to do 1
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Question 368252: An instructor gives his class a set of 17 problems with the information that the next quiz will consist of a random selection of 6 of them. If a student has figured out how to do 13 of the problems, what is the probability the he or she will answer correctly
(a) all 6 problems?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You really cannot calculate this probability in any real sense. Just because you know that the student knows how to do 13 of the 17 problems correctly tells you nothing about how the student is likely to perform on the other four problems. If you add the condition that if one of the remaining 4 problems appears on the exam that the student is guaranteed to answer incorrectly, then proceed as follows:
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
\ =\ \left(6\cr 6\right\)\left(\frac{13}{17}\right)^6\left(\frac{4}{17}\right)^0)
Note that \ =\ 1\ \forall\ n\ \in\ \mathbb{N})
and 
Hence:
\ =\ \left(\frac{13}{17}\right)^6)
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
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