# Simple probability for test studying strategy

I have come accross a relatively simple problem for which I cannot find the proper solution. Here goes: A student has 87 facts to remember the night before the test and is only able to remember 47 of them. The test contains 25 questions. What grade can most likely expect?
Thank you for your time.

Clarification. 1)This is the information as I have found it. Feel free to interperate the question any way you see fit, as I have no additional information, but would like to come to some answer, and thus better understand the problem.

2) I am assuming that if a question is studied, then the taker will get the question right on the test, and each question is worth an equal amount. This may or may not be the case, as I am not taking the test myself and have gathered the basic information from a facebook post.

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Should we assume that remembering a particular fact corresponds (one-to-one) to getting a particular question correct? –  Jubbles Oct 25 '11 at 2:32
What's the grading schematic? What type of questions does the test have? –  Doug Spoonwood Oct 25 '11 at 2:51
It the test is multiple choice, one has a chance of guessing the ones one forgot... –  Ross Millikan Oct 25 '11 at 2:55

## 2 Answers

I'm assuming here that a fact will appear in a maximum of one question and if the student knows the fact he is guaranteed to get the question right, while if he does not know the fact he is guaranteed to get the question wrong (if you're looking for a different interpretation let me know).

We want to find the expected value for the number of questions the student gets right. This is just the weighted average of all of the outcomes, so we need to find the probability of the student getting each grade (0-25). For each of these possibilities, the probability of him getting the grade is simply the number of ways that we can choose that many facts that he knows out of the number of ways the test can be made. In math terms, if $P(n)$ is the probability that the student gets $n$ questions right then $$P(n) = \frac{{47\choose{n}} {{87-47}\choose{25-n}}}{87\choose25}$$ This equation comes from the fact that $47\choose{n}$ is the number of ways we can pick the $n$ questions the student knows, ${87-47}\choose{25-n}$ is the number of ways we can pick the $25-n$ questions the student doesn't know, and $87\choose25$ is the total number of ways we can pick a 25 question test out of 87 questions (facts). Now to find the expected value $E$, $$E = \sum P(n)*n = \frac{1175}{87}$$ So we expect the student will get about 13.5 out of 25 on the test.

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I interpreted "most likely" to refer to the grade/number of correct questions with the highest probability, not to the expected value. –  joriki Oct 25 '11 at 3:06
Why not simply $\frac{47}{87}\cdot 25$? –  Brian M. Scott Oct 25 '11 at 3:47
Think of it as drawing a sample of $25$ from a box of $87$ tickets, $47$ of which bear the number $1$ while the rest bear the number $0$; your score is the sum of the numbers drawn. Each ticket appears in $25/87$-th of the samples, so each $1$ ticket contributes $\frac{25}{87}\binom{87}{25}$ to the sum of all $\binom{87}{25}$ scores, and each $0$ ticket contributes nothing. The sum of all $\binom{87}{25}$ scores is therefore $47\cdot\frac{25}{87}\binom{87}{25}$, and the mean score is $47\cdot\frac{25}{87}$. –  Brian M. Scott Oct 25 '11 at 4:13
@discipulus: I am impressed that you got the right answer by doing the computation the hard way. (I am assuming that you actually computed and added the $nP(n)$.) There is a manipulation of the sum that makes it collapse. But there is a fancier way of doing things, either as explained by Brian M. Scott, or by more explicitly using indicator random variables. In general, if you draw $k$ objects with or without replacement from a collection of $N$ objects, $G$ of which are good, then the expected number of good objects drawn is $k\frac{G}{N}$. –  André Nicolas Oct 25 '11 at 7:10
@discipulus: You don't have to worry about duplicates and such because of linearity of expectation. –  joriki Oct 25 '11 at 7:37

There is no information in the question about grades, so the best we can do is to determine the most likely number of correctly answered test questions. I'll assume that the $25$ questions are randomly drawn with uniform distribution without replacement from $87$ possible questions about the $87$ facts, and the student answers a question correctly if and only if she remembered the corresponding fact. (It would have been preferable if you had stated such assumptions in the question rather than leaving them to our fancy.)

The student learns $47$ out of $87$ facts. The number of ways in which $25$ questions can be chosen from $87$ is $\binom{87}{25}$, and the number of ways they can be chosen such that the student knows the answers to $n$ of them is $\binom{47}{n}\binom{40}{25-n}$. Since $\binom{87}{25}$ doesn't depend on $n$, the most likely number of correct questions is the one that maximizes $\binom{47}{n}\binom{40}{25-n}$, which is

$$\frac{47!}{n!(47-n)!}\frac{40!}{(25-n)!(15+n)!}\;.$$

The ratio between the values for $n$ and $n+1$ is

$$\frac{(n+1)(15+n+1)}{(47-n)(25-n)}\;.$$

The most likely value occurs where this crosses $1$:

$$\frac{(n+1)(15+n+1)}{(47-n)(25-n)}=1\;,$$

$$(n+1)(15+n+1)=(47-n)(25-n)\;,$$

$$17n+16=-72n+1175\;,$$

$$89n=1159\;,$$

$$n=13.02\dots\;.$$

Since the ratio is $1$ for $n=13.02\dots$, it is less than $1$ for $n=13$ and greater than $1$ for $n=14$, so the maximum occurs at $n=14$. This is confirmed by a table of the ratios. Thus, the most likely number of questions for the student to get right is $14$.

[Edit:]

Several comments have argued that the question might be intended to ask for the expected value. (I think asking what grade we can "most likely expect" would be a terrible way to ask for the expected value.) For this interpretation, Brian has given the direct solution in a comment on discipulus' answer. The reason the answer is so straightforward in this case is linearity of expectation. The chance of getting a question right is the same for all questions, so the expected number of correctly answered questions is simply the number of questions times the chance of getting one of them right. The latter is $47/87$, since the student knows the answer to $47$ out of the $87$ questions, so the result is $25\cdot47/87=1175/87\approx13.5$.

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It wouldn’t greatly surprise me if we got all of the information that was contained in the question as the OP found it. –  Brian M. Scott Oct 25 '11 at 3:03
@Brian: Still, it would be preferable if the OP remarked on this, e.g. suggesting which interpretation to choose, asking for opinions which interpretation might be intended, or asking for answers under all possible interpretations. The fact that there is no reflection on this in the question suggests to me that the OP was not aware that there was information missing. –  joriki Oct 25 '11 at 3:07
It seems to me that the question is asking for the expected value (or at least I've seen similarly worded questions that were), which is not necessarily the same as the most likely outcome. I agree that it's not clear though, and this interpretation is useful as well. –  smackcrane Oct 25 '11 at 3:08
@Joriki, you are correct. This is the information as I have found. –  wespiserA Oct 25 '11 at 4:02