How many exams do I need to complete if I have a pool of 10 practice exams to choose from, which 2 of these will appear on the exam, and I want to see at least one of the ones I completed, on exam day.
I know the answer is 9 applying the pigeonhole principle, but cannot show it is.
The number of possible combinations is $\binom{10}2$ for how many different possibilities could be presented on exam day.
I am thinking that the worst case is that if you study 8, there is 1 option where you could be given those two on the exam, hence the answer is 9.
I am hoping to see if someone can help approach this mathematically to show the answer is indeed 9 (either through Binomial or some sort of cdf function?)
Thank you
Of 10 practice exam papers, if 2 of them will appear in the real exam, then if you practice 9 of them. 1 paper in the real exam can be the 1 you didn't practice but the other paper has to be from the 9 you practiced.
If you practice $n$ papers, the probability that the $2$ that appear in the real exam is not among these $n$ = $\frac{^{10 - n}C_2}{^{10}C_2}$
The probability that the 2 questions will be from the $n$ you did practice = $P(E) = 1 - \frac{^{10 - n}C_2}{^{10}C_2}$
With 8 questions $P(E) = 1 - \frac{^2C_2}{^{10}C_2} = \frac{44}{45}$
With 9 questions, $P(E) = 1 - \frac{^1C_2}{45}$
$^1C_2 = 0$???