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?

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    $\begingroup$ I don't see why this has been voted down. It does present a useful probabilistic question and has background for why he asked it. I am voting it up. My only comment is that $x \in \[0,3\]$ is not the correct notation but rather $x \in \{0,1,2,3\}$, assuming that is what you are asking. I am voting up. $\endgroup$ – Jebruho Dec 7 '12 at 7:13
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    $\begingroup$ Interesting applied problem! For solving the problem, it is natural to assume that all $\binom{60}{10}$ choices by the examiners are equally likely. This is unlikely, since probably there will be an attempt to space the terms more or less evenly across the curriculum. It is a hard problem to decide by how much this might affect our conclusions. $\endgroup$ – André Nicolas Dec 7 '12 at 7:50
  • $\begingroup$ Doesn't this fit better on Cross Validated? $\endgroup$ – gerrit Dec 7 '12 at 14:53

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 $2$.

  • $\begingroup$ I think something might be wrong. When I compute $\dfrac{\binom{60}{3}\binom{60-60}{7}}{\binom{60}{10}}$ I get 0.. Also, subbing in $n=30$ is giving me ~10%. $\endgroup$ – connorbode Dec 7 '12 at 9:08
  • $\begingroup$ Is it perhaps 1 minus that answer? (as $n$ decreases, the probability decreases.. Shouldn't it be the opposite?) $\endgroup$ – connorbode Dec 7 '12 at 9:09
  • $\begingroup$ @somekindarukus: for the first, you have the top values in the numerator wrong. For the third-the probability of knowing $0$ should decrease as $n$ increases, and it does, as there are more terms to multiply. It is also $\dfrac {{60-n \choose 10}}{{60 \choose 10}}$ $\endgroup$ – Ross Millikan Dec 7 '12 at 17:45
  • $\begingroup$ @RossMillikan Perhaps it should be noted that the probability $\dfrac{\binom{n}{3}\binom{60-n}{7}}{\binom{60}{10}}$ that you gave is the probability that she will know exactly three terms; thus it makes sense that for large values of $n$, that probability decreases, because she will likely know more than three terms if she has studied a large number of the terms. $\endgroup$ – Zilliput Apr 3 '13 at 4:46
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    $\begingroup$ The probability that she will know at least three terms is $\dfrac{\binom{n}{3}\binom{60 - n}{7}+\binom{n}{4}\binom{60 - n}{6}+\binom{n}{5}\binom{60 - n}{5}+\binom{n}{6}\binom{60 - n}{4}+\binom{n}{7}\binom{60 - n}{3}+\binom{n}{8}\binom{60 - n}{2}+\binom{n}{9}\binom{60 - n}{1}+\binom{n}{10}}{\binom{60}{10}}$ $\endgroup$ – Zilliput Apr 3 '13 at 5:23

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 $$ P(X=x)=\frac{\binom{r}{x} \binom{n-r}{m-x}}{\binom{n}{m}} $$


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