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### English Exam Probability

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Date: 12/17/2001 at 19:47:08
From: Hunter Brooks
Subject: Probability in an English Exam

My English teacher gave us a list of 25 "threads"- potential exam
questions that take some time to prepare for, and told us that 10 of
these will be on the actual exam. Of these, we can take any 2 and
write essays on them. How many threads should we prepare to be
reasonably (i.e. at least 95%) confident that we won't be forced to
write on a thread we haven't thought about before? I can do the math
if we only had to write on one thread but can't figure out where to
begin calculating it for two.

Thanks!
```

```
Date: 12/19/2001 at 06:21:52
From: Doctor Mitteldorf
Subject: Re: Probability in an English Exam

Dear Hunter,

What an interesting and practical problem! I'd think about it this
way:

Thinking approximately, there's a 10/25 = 2/5 chance that each
question you prepare will be on the test, and 3/5 that it won't. Say
you prepare 6 of them. The probability that none of the 6 will be on
the test is (3/5)^6, or 0.0467. The probability that exactly 1 will be
on the test is (3/5)^5 times (2/5)^1 times the number of choices for
which one will be the one on the test, which is 6. (This corresponds
to a binomial formula: (3/5)^5 * (2/5)^1 * C(6,1). In case you're not
familiar with the notation, C(6,1) is the number of ways to choose 1
thing out of a total of 6. It's just 6.)

So the approximate calculation says that the chance of only one of
your 6 questions being on the test is 0.1867, and that the chance of
none being on the test is 0.0467, so 0.233 is the probability that
either of these two will be the case, and 1-0.233 = 0.767 is the
probability that everything will be fine. This calculation for 6
indicates that 7 or 8 questions is probably the right number to
prepare.

To find the answer precisely, you should count combinations, and we'll
see a lot more of those C(n,k) formulas. Start with the idea that
there are 10 threads on the test and 15 not on the test, and of those
you are going to pick 7 threads to prepare. The total number of ways
to pick 7 threads is C(25,7). The number of ways to be completely
unlucky (no overlap) is C(15,7). So the probability of this is
C(15,7)/C(25,7)=0.0134. The number of ways to be "partially unlucky"
(overlap of one thread) is: 10*C(15,6). This comes from the fact that
there are 10 ways to choose the lucky thread, and C(15,6) ways to
choose the remaining 6 which you study out of the 15 that aren't going
to be on the test. 10*C(15,6)/C(25,7)=0.1041, so the total probability
for overlap of 0 or 1 is 0.1175.

It looks as though you'll have to study at least 8 threads to get to
your 95% assurance. Can you finish the calculation?

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations
High School Probability

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