An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly

(a) all 5 problems?

So the sample space, $\displaystyle S$ is $\displaystyle \binom{10}{5}=252$

$\displaystyle S_5$ will be the event that the student doesn't know all 5 problems. $\displaystyle S_4$ that he doesn't know only 4 questions, and so on.

$\displaystyle S_0=S-(S_1+S_2+S_3)$

Since the student knows 7 problems the event that the test contains more then 3 unknown problems has to be a null set as there are at most 3 unknown problems.

OK so there are two pools of problems, the seven that he knows and the 3 that he doesn't. So $\displaystyle \binom{7}{n_1}$ and $\displaystyle \binom{3}{n_2}$ where $\displaystyle n_1+n_2=5$ and only four valid combinations that sum to five, $\displaystyle (n_1,n_2)=\{(2,3)(3,2)(4,1)(5,0)\}$. So,

$\displaystyle \binom{7}{2}\binom{3}{3}+\binom{7}{3}\binom{3}{2}+ \binom{7}{4}\binom{3}{1}-231$

So $\displaystyle S_0=252-231$ and $\displaystyle \frac{21}{252}=0.0833\overline{3}$

This is the listed answer but does this process look good/correct?

Also $\displaystyle \frac{\binom{7}{2}}{\binom{10}{5}}=0.0833\overline {3}$

(b) at least 4 of the problems?

So this should just be

$\displaystyle S_0+S_1=S-(S_2+S_3)$

$\displaystyle 252-\binom{7}{2}\binom{3}{3}+\binom{7}{3}\binom{3}{2}= 126$

Lastly the probablility is the quotient of the $\displaystyle S_0+S_1$ and $\displaystyle S$

$\displaystyle 126/252=.5$

Which is the given answer. Could someone be kind enough to double check this for me?