English Exam ProbabilityDate: 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/ |
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