Probability of questions being on an exam

My girlfriend has an exam in her international development class tomorrow. She's been given $60$ terms to study (each takes a long time to learn thoroughly). Of those $60$ terms, $10$ will be on the exam, and she must discuss $3$ of them. Now, she's been spending a lot of time trying to figure out the probability of knowing $x$ of the $3$ questions, time that could be spent studying.

Funnily enough, I actually have a probability and statistics for computer science final coming up next week and I have no idea how to figure out this probability. Given that she studied $n$ of $60$ terms, what is the probability that she will know (a) $0$, (b) $1$, (c) $2$, (d) $3$ of the $10$ terms that appear on the exam?

Solution 1:

This is no more than a generalization of Ross Millikan's answer. What you/she needs in called hypergeometric probability. There are $n$ total unique items (questions in your case), of then $m<n$ are sampled without replacement (i.e. they are not returned to the set). There are $x<m<n$ special items in this sample (in your case $0 \leq x \leq 3$. We use some special mesh size $r$ to select $r$ items out of $m$. We want to know the probability to have $x$ special items in this mesh.

Clearly there are $\binom{n}{m}$ ways to select $m$ items out of $n$. Since we want $x$ special items in the mesh (And the rest does not matter), there are $\binom{r}{x} \binom{n-r}{m-x}$ ways of doing so. Hence, the probability to catch $x$ special items is $$ P(X=x)=\frac{\binom{r}{x} \binom{n-r}{m-x}}{\binom{n}{m}} $$

Solution 2:

The chance that she will know none of them is the chance that none of the $10$ on the test matches one she studied. The test makers have to pick $10$ times without hitting. So it is $\dfrac {(60-n)(59-n)\ldots(51-n)}{60\cdot 59 \ldots 51}=\dfrac {(60-n)!50!}{60!(50-n)!}$

The chance she will know all three is picking $3$ of the $n$ she studied and $7$ of the others, so $\dfrac {{n \choose 3}{60-n \choose 7}}{{60 \choose 10}}$ You can probably see how to do $1$ and $2$.