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Recently one of my friends asked me this deceivingly simple question:

I have a midterm with a predetermined list of 15 possible questions. Out of the 15 questions only 7 will actually appear on the test. Out of the 7 that appear on the test I will only have to answer 5. How many answers should I memorize to make sure I get 100% on the test?

I think I have convinced myself that it is pretty easy to show that he needs to memorize exactly 13 questions - 15-(7-5) - in order to guarantee he will know at least 5 of the questions on the midterm and thus get 100%.

However I am having trouble generalizing the math to answer some more interesting questions. Specifically:

How many answers should I memorize such that 75% of the time I will get 100% on the test?

Or:

How many answers should I memorize such that on average I will get at least an 80% on the test?

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Let $N = 15$ be the total population of questions, $n = 7$ the number that will appear on the test, and $m = 5$ the number of questions you need to answer. There are $\binom{N}{n}$ possible tests; I'll assume the test is selected uniformly at random without replacement.

Say you memorize $K$ of the questions. Then, the number of questions on the test that you know the answer to, call it $k$, is hypergeometric, with population size $N$, "success" population size $K$, and number of draws $n$ (like the names in the Wikipedia article).

How many answers should I memorize such that 75% of the time I will get 100% on the test?

You get 100% on the test if $k \ge m$, so the probability of doing so is the complement of the CDF at $m-1$. Unfortunately, the expression for this is not pretty:

$$ {{{n \choose {m}}{{N-n} \choose {K-m}}}\over {N \choose K}} \,_3F_2\!\!\left[\begin{array}{c}1,\ m-K,\ m-n \\ m+1,\ N+m+1-K-n\end{array};1\right] \ge .75 $$

where $_3F_2$ is the generalized hypergeometric function.

This probably needs to be solved numerically.

How many answers should I memorize such that on average I will get at least an 80% on the exam?

Your score on the exam can be written as $S := \max\left(1, \frac{k}{m}\right)$. To get rid of the max, you can do: $$ \begin{align*} \mathbb{E}[S] &= \Pr[k \le m] \, \mathbb{E}[S \mid k \le m] + \Pr[k > m] \, \mathbb{E}[S \mid k > m] \\ &= \Pr[k \le m] \, \mathbb{E}[k \mid k \le m] + \Pr[k > m] \end{align*} $$ where the two probability terms are again gross hypergeometric cdfs, and I don't know if there's a nice form for $\mathbb{E}[k \mid k \le m]$. Maybe if you derive it the way you derive $\mathbb{E}[S]$ it'll mostly cancel out with $\Pr[T \le k]$ or something; not sure.

For smallish numbers like the ones given, you could compute $\mathbb{E} S$ exactly based on the hypergeometric pmf, but you'd probably have to do a binary search or something to find the exact cutoff.

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    $\begingroup$ Thinking a little more about this I think your answer is double counting. Lets say we know 6 answers (1-6 WLOG). Your solution is showing the tests {1,2,3,4,5}U{6,7} and {1,2,3,4,6}U{5,7} as distinct test. I think the solution requires inclusion-exclusion. $\endgroup$
    – Matt
    Jul 23 '14 at 18:14
  • $\begingroup$ Ugh, you're right about the double-counting. I rewrote the answer to make everything in terms of the hypergeometric, which means no closed forms for these two questions. If you can come up with a better form based on inclusion-exclusion, that'd be interesting, but I suspect it'll reduce to the same sum as the $_3F_2$ call in the hypergeometric cdf. $\endgroup$
    – Danica
    Jul 24 '14 at 1:31
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The objective of a test is to encourage you to learn and understand the material. If you do that you shouldn't have to memorize anything

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  • $\begingroup$ Completely agreed. And that is exactly what I told my friend when he asked. Although I have to admit it is an extremely simple question to ask - and one I think many students can relate to - that leads to some complex ideas in combinatorics and probability. $\endgroup$
    – Matt
    Jul 23 '14 at 16:38

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