I need help with the following problem:
A student that studies for an exam from the previously given exam gives an answer on a question in two ways: if the question appeared at previous exams he’ll always give a correct answer; if the question is a new one he’ll pick uniformly from the given m options.
a. A teacher picks questions for an exam by blindly picking them from a pool of overall n questions, from which k are older. What is the probability of the student to get 95 points on the exam (each question is worth 5 points)
My answer: $$X - total \,\, of \,\,right \,\, answers\\ Y - total \,\,of \,\,new\,\, questions \,\,out\,\, of\,\, 20\,\, questions \,\,in \,\,the\,\, exam \\ Y \sim Bin(20, \frac{n-k}{n})\\ X \sim Bin(Y, \frac{1}{m})\\ $$ $$ P(1 \,\,mistake) = P(Y=20)\times P(X=19|Y=20)+P(Y=19)\times P(X=18|Y=19)\\+P(Y=18)\times P(X=17|Y=18)+P(Y=17)\times P(X=16|Y=17)\\+...+P(Y=1)\times P(X=0|Y=1)=\\ \binom{20}{20}(\frac{n-k}{n})^{20}(1-\frac{n-k}{n})^{0}\binom{20}{19}(\frac{1}{m})^{19}(1-\frac{1}{m})^{1}+\\ \binom{20}{19}(\frac{n-k}{n})^{19}(1-\frac{n-k}{n})^{1}\binom{19}{18}(\frac{1}{m})^{18}(1-\frac{1}{m})^{1}+...\\ +\binom{20}{1}(\frac{n-k}{n})^{1}(1-\frac{n-k}{n})^{19}(\frac{1}{m})^{0}(1-\frac{1}{m})^{1}$$
I'm not so sure of my answer, is it correct?
There are 20 questions, of which $Y = \binom{20}{n-k}$ will be new. Because the probability of getting old questions right is $1$, we can focus only on this new questions.
We can think of the exam as a process that occurs in $Y$ iterations, were we need a number of successes $Y - 1$ to get $95$ points. At the same time, the probability of success on each iteration is $\frac{1}{m}$. Hopefully you can see the exam fits a binomial distribution.
Now, you did good in noticing that $Y$ can be any number between $0$ and $20$, and that we must account for all those possibilities (which are exclusive). At the same time, each exam has a different probability of occuring (the probability that the professor picked the number of new questions occurring).
Therefore, the probability of obtaining $95$ points (only one wrong answer) in an examn with $Y$ new questions, where there is a probability of $\frac{1}{m}$ of getting each of those questions right, is
$$ P(\text{1 mistake}) = P[N=20, x = 2, p=\frac{n-k}{n}] P[Y=2, x = 1, p=\frac{1}{m}] + P[N=20, x = 3, p=\frac{n-k}{n}] P[Y=3, x = 2, p=\frac{1}{m}]+...+P[N=20, x = 20, p=\frac{n-k}{n}]P[Y=20, x = 19, p=\frac{1}{m}] \\ =\binom{20}{2}(\frac{n-k}{n})^2(1-\frac{n-k}{n})^{18} \times \binom{2}{1} (\frac{1}{m})^1(1-\frac{1}{m})^{2 - 1} + \binom{20}{3}(\frac{n-k}{n})^3(1-\frac{n-k}{n})^{17} \times \binom{3}{2} (\frac{1}{m})^2(1-\frac{1}{m})^{3 - 2} +...+ \binom{20}{20}(\frac{n-k}{n})^{20}(1-\frac{n-k}{n})^{0} \times \binom{20}{19} (\frac{1}{m})^{19}(1-\frac{1}{m})^{20 - 19} \\ = (1-\frac{1}{m})\sum_{i=2}^{20}\binom{20}{i} (\frac{n-k}{n}^i)(1-\frac{n-k}{n})^{20-i} \binom{i}{i-1}(\frac{1}{m})^{i-1}$$
which looks pretty much like what you have. So your answer seems correct to me.