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English Exam Probability

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.


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 

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 

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   
Associated Topics:
High School Permutations and Combinations
High School Probability

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