Thread: Odds of passing a test.

1. Odds of passing a test.

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, $S$ is $\binom{10}{5}=252$

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

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

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 $\binom{7}{n_1}$ and $\binom{3}{n_2}$ where $n_1+n_2=5$ and only four valid combinations that sum to five, $(n_1,n_2)=\{(2,3)(3,2)(4,1)(5,0)\}$. So,

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

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

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

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

(b) at least 4 of the problems?

So this should just be
$S_0+S_1=S-(S_2+S_3)$

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

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

$126/252=.5$

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

2. Re: Odds of passing a test.

Originally Posted by bkbowser
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?
Why would the answer is as simple as $\frac{\dbinom{7}{5}}{\dbinom{10}{5}}~?$

Does that give the same answer? Why or why not?

3. Re: Odds of passing a test.

Originally Posted by Plato
Why would the answer is as simple as $\frac{\dbinom{7}{5}}{\dbinom{10}{5}}~$

Does that give the same answer? Why or why not?

It literally gives the same answer, but I can't construct a good argument around that binomial quotient.

I had just noticed, during the process of making my solution, that that particular fraction spit out the same number and I was curious to see if I missed something or not.